DFA Loop Invariants, Pumping Quantifiers and Every Legal Adversarial Split

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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261015-advance-20261118-r07-pumping-lemma-regular-languages-foundations · 2026-10-15 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.

Field order uses a source-bound presentation supplement; canonical JSONL and mathematical answers are unchanged.

Field-order and source-binding data

1. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=1, word=[a, b], |x|=0, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(1, [a, b], 2, 0, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (1, [a, b], 2, 0, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

2. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=2, word=[a, a, b], |x|=1, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(2, [a, a, b], 3, 1, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (2, [a, a, b], 3, 1, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

3. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=3, word=[a, a, a, b], |x|=2, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(3, [a, a, a, b], 4, 2, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (3, [a, a, a, b], 4, 2, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

4. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=4, word=[a, a, a, a, b], |x|=3, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(4, [a, a, a, a, b], 5, 3, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (4, [a, a, a, a, b], 5, 3, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

5. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=5, word=[a, a, a, a, a, b], |x|=4, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(5, [a, a, a, a, a, b], 6, 4, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (5, [a, a, a, a, a, b], 6, 4, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

6. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=6, word=[a, a, a, a, a, a, b], |x|=5, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(6, [a, a, a, a, a, a, b], 7, 5, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (6, [a, a, a, a, a, a, b], 7, 5, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

7. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=7, word=[a, a, a, a, a, a, a, b], |x|=6, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(7, [a, a, a, a, a, a, a, b], 8, 6, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (7, [a, a, a, a, a, a, a, b], 8, 6, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

8. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=8, word=[a, a, a, a, a, a, a, a, b], |x|=7, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(8, [a, a, a, a, a, a, a, a, b], 9, 7, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (8, [a, a, a, a, a, a, a, a, b], 9, 7, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

9. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=9, word=[a, a, a, a, a, a, a, a, a, b], |x|=8, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(9, [a, a, a, a, a, a, a, a, a, b], 10, 8, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (9, [a, a, a, a, a, a, a, a, a, b], 10, 8, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

10. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=10, word=[a, a, a, a, a, a, a, a, a, a, b], |x|=9, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(10, [a, a, a, a, a, a, a, a, a, a, b], 11, 9, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (10, [a, a, a, a, a, a, a, a, a, a, b], 11, 9, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

11. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=11, word=[a, a, a, a, a, a, a, a, a, a, a, b], |x|=10, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(11, [a, a, a, a, a, a, a, a, a, a, a, b], 12, 10, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (11, [a, a, a, a, a, a, a, a, a, a, a, b], 12, 10, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

12. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=12, word=[a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=11, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(12, [a, a, a, a, a, a, a, a, a, a, a, a, b], 13, 11, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (12, [a, a, a, a, a, a, a, a, a, a, a, a, b], 13, 11, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

13. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=13, word=[a, a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=12, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(13, [a, a, a, a, a, a, a, a, a, a, a, a, a, b], 14, 12, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (13, [a, a, a, a, a, a, a, a, a, a, a, a, a, b], 14, 12, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

14. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=14, word=[a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=13, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(14, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 15, 13, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (14, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 15, 13, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

15. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=15, word=[a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=14, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(15, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 16, 14, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (15, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 16, 14, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

16. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=16, word=[a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=15, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(16, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 17, 15, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (16, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 17, 15, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

17. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=17, word=[a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=16, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(17, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 18, 16, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (17, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 18, 16, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

18. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=18, word=[a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=17, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(18, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 19, 17, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (18, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 19, 17, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

19. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=19, word=[a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=18, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(19, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 20, 18, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (19, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 20, 18, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

20. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=20, word=[a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=19, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(20, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 21, 19, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (20, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 21, 19, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

21. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=21, word=[a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=20, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(21, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 22, 20, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (21, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 22, 20, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

22. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=22, word=[a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=21, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(22, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 23, 21, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (22, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 23, 21, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

23. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=23, word=[a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=22, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(23, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 24, 22, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (23, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 24, 22, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

24. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=24, word=[a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=23, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(24, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 25, 23, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (24, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 25, 23, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

25. Pumping Quantifiers Place the Word before the Split and the Pump Count after It

Use positive p=25, word=[a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], |x|=24, |y|=1; w=xyz. Give (p, word, word_length, x_length, y_length, z_length, word_long_enough, xy_within_p, y_nonempty, quantifier_order).

View answer and explanation

(25, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 26, 24, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i)

Check |w|≥p, |xy|≤p, |y|>0 and the remaining z length. The required statement is exists p, for every sufficiently long member word, there exists a split, for every i≥0. A legal split alone does not establish its all-i language property or regularity. The exact result, retaining the requested order, is (25, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 26, 24, 1, 1, Yes, Yes, Yes, exists_p_forall_w_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-quantifier-audit · candidate

26. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=1, with accepted word length=1. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(1, 1, [0, 0], [0, 1, 0], 0, 1, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (1, 1, [0, 0], [0, 1, 0], 0, 1, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

27. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=2, with accepted word length=4. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(2, 4, [0, 1, 0], [0, 2, 0], 0, 2, 2, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (2, 4, [0, 1, 0], [0, 2, 0], 0, 2, 2, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

28. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=3, with accepted word length=9. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(3, 9, [0, 1, 2, 0], [0, 3, 0], 0, 3, 6, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (3, 9, [0, 1, 2, 0], [0, 3, 0], 0, 3, 6, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

29. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=4, with accepted word length=16. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(4, 16, [0, 1, 2, 3, 0], [0, 4, 0], 0, 4, 12, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (4, 16, [0, 1, 2, 3, 0], [0, 4, 0], 0, 4, 12, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

30. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=5, with accepted word length=5. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(5, 5, [0, 1, 2, 3, 4, 0], [0, 5, 0], 0, 5, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (5, 5, [0, 1, 2, 3, 4, 0], [0, 5, 0], 0, 5, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

31. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=6, with accepted word length=12. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(6, 12, [0, 1, 2, 3, 4, 5, 0], [0, 6, 0], 0, 6, 6, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (6, 12, [0, 1, 2, 3, 4, 5, 0], [0, 6, 0], 0, 6, 6, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

32. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=7, with accepted word length=21. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(7, 21, [0, 1, 2, 3, 4, 5, 6, 0], [0, 7, 0], 0, 7, 14, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (7, 21, [0, 1, 2, 3, 4, 5, 6, 0], [0, 7, 0], 0, 7, 14, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

33. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=8, with accepted word length=32. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(8, 32, [0, 1, 2, 3, 4, 5, 6, 7, 0], [0, 8, 0], 0, 8, 24, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (8, 32, [0, 1, 2, 3, 4, 5, 6, 7, 0], [0, 8, 0], 0, 8, 24, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

34. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=9, with accepted word length=9. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(9, 9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 0], [0, 9, 0], 0, 9, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (9, 9, [0, 1, 2, 3, 4, 5, 6, 7, 8, 0], [0, 9, 0], 0, 9, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

35. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=10, with accepted word length=20. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(10, 20, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [0, 10, 0], 0, 10, 10, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (10, 20, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0], [0, 10, 0], 0, 10, 10, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

36. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=11, with accepted word length=33. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(11, 33, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0], [0, 11, 0], 0, 11, 22, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (11, 33, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0], [0, 11, 0], 0, 11, 22, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

37. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=12, with accepted word length=48. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(12, 48, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [0, 12, 0], 0, 12, 36, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (12, 48, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0], [0, 12, 0], 0, 12, 36, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

38. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=13, with accepted word length=13. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(13, 13, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 0], [0, 13, 0], 0, 13, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (13, 13, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 0], [0, 13, 0], 0, 13, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

39. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=14, with accepted word length=28. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(14, 28, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [0, 14, 0], 0, 14, 14, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (14, 28, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0], [0, 14, 0], 0, 14, 14, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

40. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=15, with accepted word length=45. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(15, 45, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 0], [0, 15, 0], 0, 15, 30, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (15, 45, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 0], [0, 15, 0], 0, 15, 30, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

41. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=16, with accepted word length=64. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(16, 64, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0], [0, 16, 0], 0, 16, 48, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (16, 64, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0], [0, 16, 0], 0, 16, 48, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

42. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=17, with accepted word length=17. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(17, 17, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 0], [0, 17, 0], 0, 17, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (17, 17, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 0], [0, 17, 0], 0, 17, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

43. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=18, with accepted word length=36. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(18, 36, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 0], [0, 18, 0], 0, 18, 18, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (18, 36, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 0], [0, 18, 0], 0, 18, 18, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

44. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=19, with accepted word length=57. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(19, 57, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 0], [0, 19, 0], 0, 19, 38, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (19, 57, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 0], [0, 19, 0], 0, 19, 38, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

45. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=20, with accepted word length=80. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(20, 80, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 0], [0, 20, 0], 0, 20, 60, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (20, 80, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 0], [0, 20, 0], 0, 20, 60, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

46. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=21, with accepted word length=21. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(21, 21, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 0], [0, 21, 0], 0, 21, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (21, 21, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 0], [0, 21, 0], 0, 21, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

47. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=22, with accepted word length=44. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(22, 44, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 0], [0, 22, 0], 0, 22, 22, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (22, 44, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 0], [0, 22, 0], 0, 22, 22, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

48. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=23, with accepted word length=69. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(23, 69, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 0], [0, 23, 0], 0, 23, 46, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (23, 69, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 0], [0, 23, 0], 0, 23, 46, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

49. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=24, with accepted word length=96. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(24, 96, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0], [0, 24, 0], 0, 24, 72, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (24, 96, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 0], [0, 24, 0], 0, 24, 72, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

50. A Complete Unary DFA Loop Gives an Invariant for Every Pump Count

Use the unary DFA for lengths divisible by modulus=25, with accepted word length=25. Give (modulus, accepted_unary_word_length, first_modulus_plus_one_states, repeated_state_positions_and_state, x_length, y_length, z_length, all_pump_counts_invariant, loop_qualified).

View answer and explanation

(25, 25, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 0], [0, 25, 0], 0, 25, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes)

Follow the first modulus+1 states, repeat state0 and choose x empty, y of length modulus, and z the remainder. For every i≥0 the pumped length is i*modulus+|z|, still divisible by the modulus. This symbolic loop invariant, not finitely many pump tests, supports the whole family. The exact result, retaining the requested order, is (25, 25, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 0], [0, 25, 0], 0, 25, 0, for_all_i_length=(i*modulus)+z_is_0_mod_modulus, Yes).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-dfa-loop-witness · candidate

51. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=1 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(1, a^1b^1, [[0, 1, 0, a^0b^1, No]], 1, 1, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (1, a^1b^1, [[0, 1, 0, a^0b^1, No]], 1, 1, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

52. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=2 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(2, a^2b^2, [[0, 1, 1, a^1b^2, No], [0, 2, 0, a^0b^2, No], [1, 1, 0, a^1b^2, No]], 3, 3, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (2, a^2b^2, [[0, 1, 1, a^1b^2, No], [0, 2, 0, a^0b^2, No], [1, 1, 0, a^1b^2, No]], 3, 3, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

53. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=3 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(3, a^3b^3, [[0, 1, 2, a^2b^3, No], [0, 2, 1, a^1b^3, No], [0, 3, 0, a^0b^3, No], [1, 1, 1, a^2b^3, No], [1, 2, 0, a^1b^3, No], [2, 1, 0, a^2b^3, No]], 6, 6, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (3, a^3b^3, [[0, 1, 2, a^2b^3, No], [0, 2, 1, a^1b^3, No], [0, 3, 0, a^0b^3, No], [1, 1, 1, a^2b^3, No], [1, 2, 0, a^1b^3, No], [2, 1, 0, a^2b^3, No]], 6, 6, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

54. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=4 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(4, a^4b^4, [[0, 1, 3, a^3b^4, No], [0, 2, 2, a^2b^4, No], [0, 3, 1, a^1b^4, No], [0, 4, 0, a^0b^4, No], [1, 1, 2, a^3b^4, No], [1, 2, 1, a^2b^4, No], [1, 3, 0, a^1b^4, No], [2, 1, 1, a^3b^4, No], [2, 2, 0, a^2b^4, No], [3, 1, 0, a^3b^4, No]], 10, 10, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (4, a^4b^4, [[0, 1, 3, a^3b^4, No], [0, 2, 2, a^2b^4, No], [0, 3, 1, a^1b^4, No], [0, 4, 0, a^0b^4, No], [1, 1, 2, a^3b^4, No], [1, 2, 1, a^2b^4, No], [1, 3, 0, a^1b^4, No], [2, 1, 1, a^3b^4, No], [2, 2, 0, a^2b^4, No], [3, 1, 0, a^3b^4, No]], 10, 10, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

55. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=5 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(5, a^5b^5, [[0, 1, 4, a^4b^5, No], [0, 2, 3, a^3b^5, No], [0, 3, 2, a^2b^5, No], [0, 4, 1, a^1b^5, No], [0, 5, 0, a^0b^5, No], [1, 1, 3, a^4b^5, No], [1, 2, 2, a^3b^5, No], [1, 3, 1, a^2b^5, No], [1, 4, 0, a^1b^5, No], [2, 1, 2, a^4b^5, No], [2, 2, 1, a^3b^5, No], [2, 3, 0, a^2b^5, No], [3, 1, 1, a^4b^5, No], [3, 2, 0, a^3b^5, No], [4, 1, 0, a^4b^5, No]], 15, 15, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (5, a^5b^5, [[0, 1, 4, a^4b^5, No], [0, 2, 3, a^3b^5, No], [0, 3, 2, a^2b^5, No], [0, 4, 1, a^1b^5, No], [0, 5, 0, a^0b^5, No], [1, 1, 3, a^4b^5, No], [1, 2, 2, a^3b^5, No], [1, 3, 1, a^2b^5, No], [1, 4, 0, a^1b^5, No], [2, 1, 2, a^4b^5, No], [2, 2, 1, a^3b^5, No], [2, 3, 0, a^2b^5, No], [3, 1, 1, a^4b^5, No], [3, 2, 0, a^3b^5, No], [4, 1, 0, a^4b^5, No]], 15, 15, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

56. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=6 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(6, a^6b^6, [[0, 1, 5, a^5b^6, No], [0, 2, 4, a^4b^6, No], [0, 3, 3, a^3b^6, No], [0, 4, 2, a^2b^6, No], [0, 5, 1, a^1b^6, No], [0, 6, 0, a^0b^6, No], [1, 1, 4, a^5b^6, No], [1, 2, 3, a^4b^6, No], [1, 3, 2, a^3b^6, No], [1, 4, 1, a^2b^6, No], [1, 5, 0, a^1b^6, No], [2, 1, 3, a^5b^6, No], [2, 2, 2, a^4b^6, No], [2, 3, 1, a^3b^6, No], [2, 4, 0, a^2b^6, No], [3, 1, 2, a^5b^6, No], [3, 2, 1, a^4b^6, No], [3, 3, 0, a^3b^6, No], [4, 1, 1, a^5b^6, No], [4, 2, 0, a^4b^6, No], [5, 1, 0, a^5b^6, No]], 21, 21, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (6, a^6b^6, [[0, 1, 5, a^5b^6, No], [0, 2, 4, a^4b^6, No], [0, 3, 3, a^3b^6, No], [0, 4, 2, a^2b^6, No], [0, 5, 1, a^1b^6, No], [0, 6, 0, a^0b^6, No], [1, 1, 4, a^5b^6, No], [1, 2, 3, a^4b^6, No], [1, 3, 2, a^3b^6, No], [1, 4, 1, a^2b^6, No], [1, 5, 0, a^1b^6, No], [2, 1, 3, a^5b^6, No], [2, 2, 2, a^4b^6, No], [2, 3, 1, a^3b^6, No], [2, 4, 0, a^2b^6, No], [3, 1, 2, a^5b^6, No], [3, 2, 1, a^4b^6, No], [3, 3, 0, a^3b^6, No], [4, 1, 1, a^5b^6, No], [4, 2, 0, a^4b^6, No], [5, 1, 0, a^5b^6, No]], 21, 21, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

57. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=7 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(7, a^7b^7, [[0, 1, 6, a^6b^7, No], [0, 2, 5, a^5b^7, No], [0, 3, 4, a^4b^7, No], [0, 4, 3, a^3b^7, No], [0, 5, 2, a^2b^7, No], [0, 6, 1, a^1b^7, No], [0, 7, 0, a^0b^7, No], [1, 1, 5, a^6b^7, No], [1, 2, 4, a^5b^7, No], [1, 3, 3, a^4b^7, No], [1, 4, 2, a^3b^7, No], [1, 5, 1, a^2b^7, No], [1, 6, 0, a^1b^7, No], [2, 1, 4, a^6b^7, No], [2, 2, 3, a^5b^7, No], [2, 3, 2, a^4b^7, No], [2, 4, 1, a^3b^7, No], [2, 5, 0, a^2b^7, No], [3, 1, 3, a^6b^7, No], [3, 2, 2, a^5b^7, No], [3, 3, 1, a^4b^7, No], [3, 4, 0, a^3b^7, No], [4, 1, 2, a^6b^7, No], [4, 2, 1, a^5b^7, No], [4, 3, 0, a^4b^7, No], [5, 1, 1, a^6b^7, No], [5, 2, 0, a^5b^7, No], [6, 1, 0, a^6b^7, No]], 28, 28, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (7, a^7b^7, [[0, 1, 6, a^6b^7, No], [0, 2, 5, a^5b^7, No], [0, 3, 4, a^4b^7, No], [0, 4, 3, a^3b^7, No], [0, 5, 2, a^2b^7, No], [0, 6, 1, a^1b^7, No], [0, 7, 0, a^0b^7, No], [1, 1, 5, a^6b^7, No], [1, 2, 4, a^5b^7, No], [1, 3, 3, a^4b^7, No], [1, 4, 2, a^3b^7, No], [1, 5, 1, a^2b^7, No], [1, 6, 0, a^1b^7, No], [2, 1, 4, a^6b^7, No], [2, 2, 3, a^5b^7, No], [2, 3, 2, a^4b^7, No], [2, 4, 1, a^3b^7, No], [2, 5, 0, a^2b^7, No], [3, 1, 3, a^6b^7, No], [3, 2, 2, a^5b^7, No], [3, 3, 1, a^4b^7, No], [3, 4, 0, a^3b^7, No], [4, 1, 2, a^6b^7, No], [4, 2, 1, a^5b^7, No], [4, 3, 0, a^4b^7, No], [5, 1, 1, a^6b^7, No], [5, 2, 0, a^5b^7, No], [6, 1, 0, a^6b^7, No]], 28, 28, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

58. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=8 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(8, a^8b^8, [[0, 1, 7, a^7b^8, No], [0, 2, 6, a^6b^8, No], [0, 3, 5, a^5b^8, No], [0, 4, 4, a^4b^8, No], [0, 5, 3, a^3b^8, No], [0, 6, 2, a^2b^8, No], [0, 7, 1, a^1b^8, No], [0, 8, 0, a^0b^8, No], [1, 1, 6, a^7b^8, No], [1, 2, 5, a^6b^8, No], [1, 3, 4, a^5b^8, No], [1, 4, 3, a^4b^8, No], [1, 5, 2, a^3b^8, No], [1, 6, 1, a^2b^8, No], [1, 7, 0, a^1b^8, No], [2, 1, 5, a^7b^8, No], [2, 2, 4, a^6b^8, No], [2, 3, 3, a^5b^8, No], [2, 4, 2, a^4b^8, No], [2, 5, 1, a^3b^8, No], [2, 6, 0, a^2b^8, No], [3, 1, 4, a^7b^8, No], [3, 2, 3, a^6b^8, No], [3, 3, 2, a^5b^8, No], [3, 4, 1, a^4b^8, No], [3, 5, 0, a^3b^8, No], [4, 1, 3, a^7b^8, No], [4, 2, 2, a^6b^8, No], [4, 3, 1, a^5b^8, No], [4, 4, 0, a^4b^8, No], [5, 1, 2, a^7b^8, No], [5, 2, 1, a^6b^8, No], [5, 3, 0, a^5b^8, No], [6, 1, 1, a^7b^8, No], [6, 2, 0, a^6b^8, No], [7, 1, 0, a^7b^8, No]], 36, 36, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (8, a^8b^8, [[0, 1, 7, a^7b^8, No], [0, 2, 6, a^6b^8, No], [0, 3, 5, a^5b^8, No], [0, 4, 4, a^4b^8, No], [0, 5, 3, a^3b^8, No], [0, 6, 2, a^2b^8, No], [0, 7, 1, a^1b^8, No], [0, 8, 0, a^0b^8, No], [1, 1, 6, a^7b^8, No], [1, 2, 5, a^6b^8, No], [1, 3, 4, a^5b^8, No], [1, 4, 3, a^4b^8, No], [1, 5, 2, a^3b^8, No], [1, 6, 1, a^2b^8, No], [1, 7, 0, a^1b^8, No], [2, 1, 5, a^7b^8, No], [2, 2, 4, a^6b^8, No], [2, 3, 3, a^5b^8, No], [2, 4, 2, a^4b^8, No], [2, 5, 1, a^3b^8, No], [2, 6, 0, a^2b^8, No], [3, 1, 4, a^7b^8, No], [3, 2, 3, a^6b^8, No], [3, 3, 2, a^5b^8, No], [3, 4, 1, a^4b^8, No], [3, 5, 0, a^3b^8, No], [4, 1, 3, a^7b^8, No], [4, 2, 2, a^6b^8, No], [4, 3, 1, a^5b^8, No], [4, 4, 0, a^4b^8, No], [5, 1, 2, a^7b^8, No], [5, 2, 1, a^6b^8, No], [5, 3, 0, a^5b^8, No], [6, 1, 1, a^7b^8, No], [6, 2, 0, a^6b^8, No], [7, 1, 0, a^7b^8, No]], 36, 36, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

59. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=9 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(9, a^9b^9, [[0, 1, 8, a^8b^9, No], [0, 2, 7, a^7b^9, No], [0, 3, 6, a^6b^9, No], [0, 4, 5, a^5b^9, No], [0, 5, 4, a^4b^9, No], [0, 6, 3, a^3b^9, No], [0, 7, 2, a^2b^9, No], [0, 8, 1, a^1b^9, No], [0, 9, 0, a^0b^9, No], [1, 1, 7, a^8b^9, No], [1, 2, 6, a^7b^9, No], [1, 3, 5, a^6b^9, No], [1, 4, 4, a^5b^9, No], [1, 5, 3, a^4b^9, No], [1, 6, 2, a^3b^9, No], [1, 7, 1, a^2b^9, No], [1, 8, 0, a^1b^9, No], [2, 1, 6, a^8b^9, No], [2, 2, 5, a^7b^9, No], [2, 3, 4, a^6b^9, No], [2, 4, 3, a^5b^9, No], [2, 5, 2, a^4b^9, No], [2, 6, 1, a^3b^9, No], [2, 7, 0, a^2b^9, No], [3, 1, 5, a^8b^9, No], [3, 2, 4, a^7b^9, No], [3, 3, 3, a^6b^9, No], [3, 4, 2, a^5b^9, No], [3, 5, 1, a^4b^9, No], [3, 6, 0, a^3b^9, No], [4, 1, 4, a^8b^9, No], [4, 2, 3, a^7b^9, No], [4, 3, 2, a^6b^9, No], [4, 4, 1, a^5b^9, No], [4, 5, 0, a^4b^9, No], [5, 1, 3, a^8b^9, No], [5, 2, 2, a^7b^9, No], [5, 3, 1, a^6b^9, No], [5, 4, 0, a^5b^9, No], [6, 1, 2, a^8b^9, No], [6, 2, 1, a^7b^9, No], [6, 3, 0, a^6b^9, No], [7, 1, 1, a^8b^9, No], [7, 2, 0, a^7b^9, No], [8, 1, 0, a^8b^9, No]], 45, 45, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (9, a^9b^9, [[0, 1, 8, a^8b^9, No], [0, 2, 7, a^7b^9, No], [0, 3, 6, a^6b^9, No], [0, 4, 5, a^5b^9, No], [0, 5, 4, a^4b^9, No], [0, 6, 3, a^3b^9, No], [0, 7, 2, a^2b^9, No], [0, 8, 1, a^1b^9, No], [0, 9, 0, a^0b^9, No], [1, 1, 7, a^8b^9, No], [1, 2, 6, a^7b^9, No], [1, 3, 5, a^6b^9, No], [1, 4, 4, a^5b^9, No], [1, 5, 3, a^4b^9, No], [1, 6, 2, a^3b^9, No], [1, 7, 1, a^2b^9, No], [1, 8, 0, a^1b^9, No], [2, 1, 6, a^8b^9, No], [2, 2, 5, a^7b^9, No], [2, 3, 4, a^6b^9, No], [2, 4, 3, a^5b^9, No], [2, 5, 2, a^4b^9, No], [2, 6, 1, a^3b^9, No], [2, 7, 0, a^2b^9, No], [3, 1, 5, a^8b^9, No], [3, 2, 4, a^7b^9, No], [3, 3, 3, a^6b^9, No], [3, 4, 2, a^5b^9, No], [3, 5, 1, a^4b^9, No], [3, 6, 0, a^3b^9, No], [4, 1, 4, a^8b^9, No], [4, 2, 3, a^7b^9, No], [4, 3, 2, a^6b^9, No], [4, 4, 1, a^5b^9, No], [4, 5, 0, a^4b^9, No], [5, 1, 3, a^8b^9, No], [5, 2, 2, a^7b^9, No], [5, 3, 1, a^6b^9, No], [5, 4, 0, a^5b^9, No], [6, 1, 2, a^8b^9, No], [6, 2, 1, a^7b^9, No], [6, 3, 0, a^6b^9, No], [7, 1, 1, a^8b^9, No], [7, 2, 0, a^7b^9, No], [8, 1, 0, a^8b^9, No]], 45, 45, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

60. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=10 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(10, a^10b^10, [[0, 1, 9, a^9b^10, No], [0, 2, 8, a^8b^10, No], [0, 3, 7, a^7b^10, No], [0, 4, 6, a^6b^10, No], [0, 5, 5, a^5b^10, No], [0, 6, 4, a^4b^10, No], [0, 7, 3, a^3b^10, No], [0, 8, 2, a^2b^10, No], [0, 9, 1, a^1b^10, No], [0, 10, 0, a^0b^10, No], [1, 1, 8, a^9b^10, No], [1, 2, 7, a^8b^10, No], [1, 3, 6, a^7b^10, No], [1, 4, 5, a^6b^10, No], [1, 5, 4, a^5b^10, No], [1, 6, 3, a^4b^10, No], [1, 7, 2, a^3b^10, No], [1, 8, 1, a^2b^10, No], [1, 9, 0, a^1b^10, No], [2, 1, 7, a^9b^10, No], [2, 2, 6, a^8b^10, No], [2, 3, 5, a^7b^10, No], [2, 4, 4, a^6b^10, No], [2, 5, 3, a^5b^10, No], [2, 6, 2, a^4b^10, No], [2, 7, 1, a^3b^10, No], [2, 8, 0, a^2b^10, No], [3, 1, 6, a^9b^10, No], [3, 2, 5, a^8b^10, No], [3, 3, 4, a^7b^10, No], [3, 4, 3, a^6b^10, No], [3, 5, 2, a^5b^10, No], [3, 6, 1, a^4b^10, No], [3, 7, 0, a^3b^10, No], [4, 1, 5, a^9b^10, No], [4, 2, 4, a^8b^10, No], [4, 3, 3, a^7b^10, No], [4, 4, 2, a^6b^10, No], [4, 5, 1, a^5b^10, No], [4, 6, 0, a^4b^10, No], [5, 1, 4, a^9b^10, No], [5, 2, 3, a^8b^10, No], [5, 3, 2, a^7b^10, No], [5, 4, 1, a^6b^10, No], [5, 5, 0, a^5b^10, No], [6, 1, 3, a^9b^10, No], [6, 2, 2, a^8b^10, No], [6, 3, 1, a^7b^10, No], [6, 4, 0, a^6b^10, No], [7, 1, 2, a^9b^10, No], [7, 2, 1, a^8b^10, No], [7, 3, 0, a^7b^10, No], [8, 1, 1, a^9b^10, No], [8, 2, 0, a^8b^10, No], [9, 1, 0, a^9b^10, No]], 55, 55, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (10, a^10b^10, [[0, 1, 9, a^9b^10, No], [0, 2, 8, a^8b^10, No], [0, 3, 7, a^7b^10, No], [0, 4, 6, a^6b^10, No], [0, 5, 5, a^5b^10, No], [0, 6, 4, a^4b^10, No], [0, 7, 3, a^3b^10, No], [0, 8, 2, a^2b^10, No], [0, 9, 1, a^1b^10, No], [0, 10, 0, a^0b^10, No], [1, 1, 8, a^9b^10, No], [1, 2, 7, a^8b^10, No], [1, 3, 6, a^7b^10, No], [1, 4, 5, a^6b^10, No], [1, 5, 4, a^5b^10, No], [1, 6, 3, a^4b^10, No], [1, 7, 2, a^3b^10, No], [1, 8, 1, a^2b^10, No], [1, 9, 0, a^1b^10, No], [2, 1, 7, a^9b^10, No], [2, 2, 6, a^8b^10, No], [2, 3, 5, a^7b^10, No], [2, 4, 4, a^6b^10, No], [2, 5, 3, a^5b^10, No], [2, 6, 2, a^4b^10, No], [2, 7, 1, a^3b^10, No], [2, 8, 0, a^2b^10, No], [3, 1, 6, a^9b^10, No], [3, 2, 5, a^8b^10, No], [3, 3, 4, a^7b^10, No], [3, 4, 3, a^6b^10, No], [3, 5, 2, a^5b^10, No], [3, 6, 1, a^4b^10, No], [3, 7, 0, a^3b^10, No], [4, 1, 5, a^9b^10, No], [4, 2, 4, a^8b^10, No], [4, 3, 3, a^7b^10, No], [4, 4, 2, a^6b^10, No], [4, 5, 1, a^5b^10, No], [4, 6, 0, a^4b^10, No], [5, 1, 4, a^9b^10, No], [5, 2, 3, a^8b^10, No], [5, 3, 2, a^7b^10, No], [5, 4, 1, a^6b^10, No], [5, 5, 0, a^5b^10, No], [6, 1, 3, a^9b^10, No], [6, 2, 2, a^8b^10, No], [6, 3, 1, a^7b^10, No], [6, 4, 0, a^6b^10, No], [7, 1, 2, a^9b^10, No], [7, 2, 1, a^8b^10, No], [7, 3, 0, a^7b^10, No], [8, 1, 1, a^9b^10, No], [8, 2, 0, a^8b^10, No], [9, 1, 0, a^9b^10, No]], 55, 55, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

61. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=11 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(11, a^11b^11, [[0, 1, 10, a^10b^11, No], [0, 2, 9, a^9b^11, No], [0, 3, 8, a^8b^11, No], [0, 4, 7, a^7b^11, No], [0, 5, 6, a^6b^11, No], [0, 6, 5, a^5b^11, No], [0, 7, 4, a^4b^11, No], [0, 8, 3, a^3b^11, No], [0, 9, 2, a^2b^11, No], [0, 10, 1, a^1b^11, No], [0, 11, 0, a^0b^11, No], [1, 1, 9, a^10b^11, No], [1, 2, 8, a^9b^11, No], [1, 3, 7, a^8b^11, No], [1, 4, 6, a^7b^11, No], [1, 5, 5, a^6b^11, No], [1, 6, 4, a^5b^11, No], [1, 7, 3, a^4b^11, No], [1, 8, 2, a^3b^11, No], [1, 9, 1, a^2b^11, No], [1, 10, 0, a^1b^11, No], [2, 1, 8, a^10b^11, No], [2, 2, 7, a^9b^11, No], [2, 3, 6, a^8b^11, No], [2, 4, 5, a^7b^11, No], [2, 5, 4, a^6b^11, No], [2, 6, 3, a^5b^11, No], [2, 7, 2, a^4b^11, No], [2, 8, 1, a^3b^11, No], [2, 9, 0, a^2b^11, No], [3, 1, 7, a^10b^11, No], [3, 2, 6, a^9b^11, No], [3, 3, 5, a^8b^11, No], [3, 4, 4, a^7b^11, No], [3, 5, 3, a^6b^11, No], [3, 6, 2, a^5b^11, No], [3, 7, 1, a^4b^11, No], [3, 8, 0, a^3b^11, No], [4, 1, 6, a^10b^11, No], [4, 2, 5, a^9b^11, No], [4, 3, 4, a^8b^11, No], [4, 4, 3, a^7b^11, No], [4, 5, 2, a^6b^11, No], [4, 6, 1, a^5b^11, No], [4, 7, 0, a^4b^11, No], [5, 1, 5, a^10b^11, No], [5, 2, 4, a^9b^11, No], [5, 3, 3, a^8b^11, No], [5, 4, 2, a^7b^11, No], [5, 5, 1, a^6b^11, No], [5, 6, 0, a^5b^11, No], [6, 1, 4, a^10b^11, No], [6, 2, 3, a^9b^11, No], [6, 3, 2, a^8b^11, No], [6, 4, 1, a^7b^11, No], [6, 5, 0, a^6b^11, No], [7, 1, 3, a^10b^11, No], [7, 2, 2, a^9b^11, No], [7, 3, 1, a^8b^11, No], [7, 4, 0, a^7b^11, No], [8, 1, 2, a^10b^11, No], [8, 2, 1, a^9b^11, No], [8, 3, 0, a^8b^11, No], [9, 1, 1, a^10b^11, No], [9, 2, 0, a^9b^11, No], [10, 1, 0, a^10b^11, No]], 66, 66, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (11, a^11b^11, [[0, 1, 10, a^10b^11, No], [0, 2, 9, a^9b^11, No], [0, 3, 8, a^8b^11, No], [0, 4, 7, a^7b^11, No], [0, 5, 6, a^6b^11, No], [0, 6, 5, a^5b^11, No], [0, 7, 4, a^4b^11, No], [0, 8, 3, a^3b^11, No], [0, 9, 2, a^2b^11, No], [0, 10, 1, a^1b^11, No], [0, 11, 0, a^0b^11, No], [1, 1, 9, a^10b^11, No], [1, 2, 8, a^9b^11, No], [1, 3, 7, a^8b^11, No], [1, 4, 6, a^7b^11, No], [1, 5, 5, a^6b^11, No], [1, 6, 4, a^5b^11, No], [1, 7, 3, a^4b^11, No], [1, 8, 2, a^3b^11, No], [1, 9, 1, a^2b^11, No], [1, 10, 0, a^1b^11, No], [2, 1, 8, a^10b^11, No], [2, 2, 7, a^9b^11, No], [2, 3, 6, a^8b^11, No], [2, 4, 5, a^7b^11, No], [2, 5, 4, a^6b^11, No], [2, 6, 3, a^5b^11, No], [2, 7, 2, a^4b^11, No], [2, 8, 1, a^3b^11, No], [2, 9, 0, a^2b^11, No], [3, 1, 7, a^10b^11, No], [3, 2, 6, a^9b^11, No], [3, 3, 5, a^8b^11, No], [3, 4, 4, a^7b^11, No], [3, 5, 3, a^6b^11, No], [3, 6, 2, a^5b^11, No], [3, 7, 1, a^4b^11, No], [3, 8, 0, a^3b^11, No], [4, 1, 6, a^10b^11, No], [4, 2, 5, a^9b^11, No], [4, 3, 4, a^8b^11, No], [4, 4, 3, a^7b^11, No], [4, 5, 2, a^6b^11, No], [4, 6, 1, a^5b^11, No], [4, 7, 0, a^4b^11, No], [5, 1, 5, a^10b^11, No], [5, 2, 4, a^9b^11, No], [5, 3, 3, a^8b^11, No], [5, 4, 2, a^7b^11, No], [5, 5, 1, a^6b^11, No], [5, 6, 0, a^5b^11, No], [6, 1, 4, a^10b^11, No], [6, 2, 3, a^9b^11, No], [6, 3, 2, a^8b^11, No], [6, 4, 1, a^7b^11, No], [6, 5, 0, a^6b^11, No], [7, 1, 3, a^10b^11, No], [7, 2, 2, a^9b^11, No], [7, 3, 1, a^8b^11, No], [7, 4, 0, a^7b^11, No], [8, 1, 2, a^10b^11, No], [8, 2, 1, a^9b^11, No], [8, 3, 0, a^8b^11, No], [9, 1, 1, a^10b^11, No], [9, 2, 0, a^9b^11, No], [10, 1, 0, a^10b^11, No]], 66, 66, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

62. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=12 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(12, a^12b^12, [[0, 1, 11, a^11b^12, No], [0, 2, 10, a^10b^12, No], [0, 3, 9, a^9b^12, No], [0, 4, 8, a^8b^12, No], [0, 5, 7, a^7b^12, No], [0, 6, 6, a^6b^12, No], [0, 7, 5, a^5b^12, No], [0, 8, 4, a^4b^12, No], [0, 9, 3, a^3b^12, No], [0, 10, 2, a^2b^12, No], [0, 11, 1, a^1b^12, No], [0, 12, 0, a^0b^12, No], [1, 1, 10, a^11b^12, No], [1, 2, 9, a^10b^12, No], [1, 3, 8, a^9b^12, No], [1, 4, 7, a^8b^12, No], [1, 5, 6, a^7b^12, No], [1, 6, 5, a^6b^12, No], [1, 7, 4, a^5b^12, No], [1, 8, 3, a^4b^12, No], [1, 9, 2, a^3b^12, No], [1, 10, 1, a^2b^12, No], [1, 11, 0, a^1b^12, No], [2, 1, 9, a^11b^12, No], [2, 2, 8, a^10b^12, No], [2, 3, 7, a^9b^12, No], [2, 4, 6, a^8b^12, No], [2, 5, 5, a^7b^12, No], [2, 6, 4, a^6b^12, No], [2, 7, 3, a^5b^12, No], [2, 8, 2, a^4b^12, No], [2, 9, 1, a^3b^12, No], [2, 10, 0, a^2b^12, No], [3, 1, 8, a^11b^12, No], [3, 2, 7, a^10b^12, No], [3, 3, 6, a^9b^12, No], [3, 4, 5, a^8b^12, No], [3, 5, 4, a^7b^12, No], [3, 6, 3, a^6b^12, No], [3, 7, 2, a^5b^12, No], [3, 8, 1, a^4b^12, No], [3, 9, 0, a^3b^12, No], [4, 1, 7, a^11b^12, No], [4, 2, 6, a^10b^12, No], [4, 3, 5, a^9b^12, No], [4, 4, 4, a^8b^12, No], [4, 5, 3, a^7b^12, No], [4, 6, 2, a^6b^12, No], [4, 7, 1, a^5b^12, No], [4, 8, 0, a^4b^12, No], [5, 1, 6, a^11b^12, No], [5, 2, 5, a^10b^12, No], [5, 3, 4, a^9b^12, No], [5, 4, 3, a^8b^12, No], [5, 5, 2, a^7b^12, No], [5, 6, 1, a^6b^12, No], [5, 7, 0, a^5b^12, No], [6, 1, 5, a^11b^12, No], [6, 2, 4, a^10b^12, No], [6, 3, 3, a^9b^12, No], [6, 4, 2, a^8b^12, No], [6, 5, 1, a^7b^12, No], [6, 6, 0, a^6b^12, No], [7, 1, 4, a^11b^12, No], [7, 2, 3, a^10b^12, No], [7, 3, 2, a^9b^12, No], [7, 4, 1, a^8b^12, No], [7, 5, 0, a^7b^12, No], [8, 1, 3, a^11b^12, No], [8, 2, 2, a^10b^12, No], [8, 3, 1, a^9b^12, No], [8, 4, 0, a^8b^12, No], [9, 1, 2, a^11b^12, No], [9, 2, 1, a^10b^12, No], [9, 3, 0, a^9b^12, No], [10, 1, 1, a^11b^12, No], [10, 2, 0, a^10b^12, No], [11, 1, 0, a^11b^12, No]], 78, 78, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (12, a^12b^12, [[0, 1, 11, a^11b^12, No], [0, 2, 10, a^10b^12, No], [0, 3, 9, a^9b^12, No], [0, 4, 8, a^8b^12, No], [0, 5, 7, a^7b^12, No], [0, 6, 6, a^6b^12, No], [0, 7, 5, a^5b^12, No], [0, 8, 4, a^4b^12, No], [0, 9, 3, a^3b^12, No], [0, 10, 2, a^2b^12, No], [0, 11, 1, a^1b^12, No], [0, 12, 0, a^0b^12, No], [1, 1, 10, a^11b^12, No], [1, 2, 9, a^10b^12, No], [1, 3, 8, a^9b^12, No], [1, 4, 7, a^8b^12, No], [1, 5, 6, a^7b^12, No], [1, 6, 5, a^6b^12, No], [1, 7, 4, a^5b^12, No], [1, 8, 3, a^4b^12, No], [1, 9, 2, a^3b^12, No], [1, 10, 1, a^2b^12, No], [1, 11, 0, a^1b^12, No], [2, 1, 9, a^11b^12, No], [2, 2, 8, a^10b^12, No], [2, 3, 7, a^9b^12, No], [2, 4, 6, a^8b^12, No], [2, 5, 5, a^7b^12, No], [2, 6, 4, a^6b^12, No], [2, 7, 3, a^5b^12, No], [2, 8, 2, a^4b^12, No], [2, 9, 1, a^3b^12, No], [2, 10, 0, a^2b^12, No], [3, 1, 8, a^11b^12, No], [3, 2, 7, a^10b^12, No], [3, 3, 6, a^9b^12, No], [3, 4, 5, a^8b^12, No], [3, 5, 4, a^7b^12, No], [3, 6, 3, a^6b^12, No], [3, 7, 2, a^5b^12, No], [3, 8, 1, a^4b^12, No], [3, 9, 0, a^3b^12, No], [4, 1, 7, a^11b^12, No], [4, 2, 6, a^10b^12, No], [4, 3, 5, a^9b^12, No], [4, 4, 4, a^8b^12, No], [4, 5, 3, a^7b^12, No], [4, 6, 2, a^6b^12, No], [4, 7, 1, a^5b^12, No], [4, 8, 0, a^4b^12, No], [5, 1, 6, a^11b^12, No], [5, 2, 5, a^10b^12, No], [5, 3, 4, a^9b^12, No], [5, 4, 3, a^8b^12, No], [5, 5, 2, a^7b^12, No], [5, 6, 1, a^6b^12, No], [5, 7, 0, a^5b^12, No], [6, 1, 5, a^11b^12, No], [6, 2, 4, a^10b^12, No], [6, 3, 3, a^9b^12, No], [6, 4, 2, a^8b^12, No], [6, 5, 1, a^7b^12, No], [6, 6, 0, a^6b^12, No], [7, 1, 4, a^11b^12, No], [7, 2, 3, a^10b^12, No], [7, 3, 2, a^9b^12, No], [7, 4, 1, a^8b^12, No], [7, 5, 0, a^7b^12, No], [8, 1, 3, a^11b^12, No], [8, 2, 2, a^10b^12, No], [8, 3, 1, a^9b^12, No], [8, 4, 0, a^8b^12, No], [9, 1, 2, a^11b^12, No], [9, 2, 1, a^10b^12, No], [9, 3, 0, a^9b^12, No], [10, 1, 1, a^11b^12, No], [10, 2, 0, a^10b^12, No], [11, 1, 0, a^11b^12, No]], 78, 78, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

63. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=13 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(13, a^13b^13, [[0, 1, 12, a^12b^13, No], [0, 2, 11, a^11b^13, No], [0, 3, 10, a^10b^13, No], [0, 4, 9, a^9b^13, No], [0, 5, 8, a^8b^13, No], [0, 6, 7, a^7b^13, No], [0, 7, 6, a^6b^13, No], [0, 8, 5, a^5b^13, No], [0, 9, 4, a^4b^13, No], [0, 10, 3, a^3b^13, No], [0, 11, 2, a^2b^13, No], [0, 12, 1, a^1b^13, No], [0, 13, 0, a^0b^13, No], [1, 1, 11, a^12b^13, No], [1, 2, 10, a^11b^13, No], [1, 3, 9, a^10b^13, No], [1, 4, 8, a^9b^13, No], [1, 5, 7, a^8b^13, No], [1, 6, 6, a^7b^13, No], [1, 7, 5, a^6b^13, No], [1, 8, 4, a^5b^13, No], [1, 9, 3, a^4b^13, No], [1, 10, 2, a^3b^13, No], [1, 11, 1, a^2b^13, No], [1, 12, 0, a^1b^13, No], [2, 1, 10, a^12b^13, No], [2, 2, 9, a^11b^13, No], [2, 3, 8, a^10b^13, No], [2, 4, 7, a^9b^13, No], [2, 5, 6, a^8b^13, No], [2, 6, 5, a^7b^13, No], [2, 7, 4, a^6b^13, No], [2, 8, 3, a^5b^13, No], [2, 9, 2, a^4b^13, No], [2, 10, 1, a^3b^13, No], [2, 11, 0, a^2b^13, No], [3, 1, 9, a^12b^13, No], [3, 2, 8, a^11b^13, No], [3, 3, 7, a^10b^13, No], [3, 4, 6, a^9b^13, No], [3, 5, 5, a^8b^13, No], [3, 6, 4, a^7b^13, No], [3, 7, 3, a^6b^13, No], [3, 8, 2, a^5b^13, No], [3, 9, 1, a^4b^13, No], [3, 10, 0, a^3b^13, No], [4, 1, 8, a^12b^13, No], [4, 2, 7, a^11b^13, No], [4, 3, 6, a^10b^13, No], [4, 4, 5, a^9b^13, No], [4, 5, 4, a^8b^13, No], [4, 6, 3, a^7b^13, No], [4, 7, 2, a^6b^13, No], [4, 8, 1, a^5b^13, No], [4, 9, 0, a^4b^13, No], [5, 1, 7, a^12b^13, No], [5, 2, 6, a^11b^13, No], [5, 3, 5, a^10b^13, No], [5, 4, 4, a^9b^13, No], [5, 5, 3, a^8b^13, No], [5, 6, 2, a^7b^13, No], [5, 7, 1, a^6b^13, No], [5, 8, 0, a^5b^13, No], [6, 1, 6, a^12b^13, No], [6, 2, 5, a^11b^13, No], [6, 3, 4, a^10b^13, No], [6, 4, 3, a^9b^13, No], [6, 5, 2, a^8b^13, No], [6, 6, 1, a^7b^13, No], [6, 7, 0, a^6b^13, No], [7, 1, 5, a^12b^13, No], [7, 2, 4, a^11b^13, No], [7, 3, 3, a^10b^13, No], [7, 4, 2, a^9b^13, No], [7, 5, 1, a^8b^13, No], [7, 6, 0, a^7b^13, No], [8, 1, 4, a^12b^13, No], [8, 2, 3, a^11b^13, No], [8, 3, 2, a^10b^13, No], [8, 4, 1, a^9b^13, No], [8, 5, 0, a^8b^13, No], [9, 1, 3, a^12b^13, No], [9, 2, 2, a^11b^13, No], [9, 3, 1, a^10b^13, No], [9, 4, 0, a^9b^13, No], [10, 1, 2, a^12b^13, No], [10, 2, 1, a^11b^13, No], [10, 3, 0, a^10b^13, No], [11, 1, 1, a^12b^13, No], [11, 2, 0, a^11b^13, No], [12, 1, 0, a^12b^13, No]], 91, 91, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (13, a^13b^13, [[0, 1, 12, a^12b^13, No], [0, 2, 11, a^11b^13, No], [0, 3, 10, a^10b^13, No], [0, 4, 9, a^9b^13, No], [0, 5, 8, a^8b^13, No], [0, 6, 7, a^7b^13, No], [0, 7, 6, a^6b^13, No], [0, 8, 5, a^5b^13, No], [0, 9, 4, a^4b^13, No], [0, 10, 3, a^3b^13, No], [0, 11, 2, a^2b^13, No], [0, 12, 1, a^1b^13, No], [0, 13, 0, a^0b^13, No], [1, 1, 11, a^12b^13, No], [1, 2, 10, a^11b^13, No], [1, 3, 9, a^10b^13, No], [1, 4, 8, a^9b^13, No], [1, 5, 7, a^8b^13, No], [1, 6, 6, a^7b^13, No], [1, 7, 5, a^6b^13, No], [1, 8, 4, a^5b^13, No], [1, 9, 3, a^4b^13, No], [1, 10, 2, a^3b^13, No], [1, 11, 1, a^2b^13, No], [1, 12, 0, a^1b^13, No], [2, 1, 10, a^12b^13, No], [2, 2, 9, a^11b^13, No], [2, 3, 8, a^10b^13, No], [2, 4, 7, a^9b^13, No], [2, 5, 6, a^8b^13, No], [2, 6, 5, a^7b^13, No], [2, 7, 4, a^6b^13, No], [2, 8, 3, a^5b^13, No], [2, 9, 2, a^4b^13, No], [2, 10, 1, a^3b^13, No], [2, 11, 0, a^2b^13, No], [3, 1, 9, a^12b^13, No], [3, 2, 8, a^11b^13, No], [3, 3, 7, a^10b^13, No], [3, 4, 6, a^9b^13, No], [3, 5, 5, a^8b^13, No], [3, 6, 4, a^7b^13, No], [3, 7, 3, a^6b^13, No], [3, 8, 2, a^5b^13, No], [3, 9, 1, a^4b^13, No], [3, 10, 0, a^3b^13, No], [4, 1, 8, a^12b^13, No], [4, 2, 7, a^11b^13, No], [4, 3, 6, a^10b^13, No], [4, 4, 5, a^9b^13, No], [4, 5, 4, a^8b^13, No], [4, 6, 3, a^7b^13, No], [4, 7, 2, a^6b^13, No], [4, 8, 1, a^5b^13, No], [4, 9, 0, a^4b^13, No], [5, 1, 7, a^12b^13, No], [5, 2, 6, a^11b^13, No], [5, 3, 5, a^10b^13, No], [5, 4, 4, a^9b^13, No], [5, 5, 3, a^8b^13, No], [5, 6, 2, a^7b^13, No], [5, 7, 1, a^6b^13, No], [5, 8, 0, a^5b^13, No], [6, 1, 6, a^12b^13, No], [6, 2, 5, a^11b^13, No], [6, 3, 4, a^10b^13, No], [6, 4, 3, a^9b^13, No], [6, 5, 2, a^8b^13, No], [6, 6, 1, a^7b^13, No], [6, 7, 0, a^6b^13, No], [7, 1, 5, a^12b^13, No], [7, 2, 4, a^11b^13, No], [7, 3, 3, a^10b^13, No], [7, 4, 2, a^9b^13, No], [7, 5, 1, a^8b^13, No], [7, 6, 0, a^7b^13, No], [8, 1, 4, a^12b^13, No], [8, 2, 3, a^11b^13, No], [8, 3, 2, a^10b^13, No], [8, 4, 1, a^9b^13, No], [8, 5, 0, a^8b^13, No], [9, 1, 3, a^12b^13, No], [9, 2, 2, a^11b^13, No], [9, 3, 1, a^10b^13, No], [9, 4, 0, a^9b^13, No], [10, 1, 2, a^12b^13, No], [10, 2, 1, a^11b^13, No], [10, 3, 0, a^10b^13, No], [11, 1, 1, a^12b^13, No], [11, 2, 0, a^11b^13, No], [12, 1, 0, a^12b^13, No]], 91, 91, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

64. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=14 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(14, a^14b^14, [[0, 1, 13, a^13b^14, No], [0, 2, 12, a^12b^14, No], [0, 3, 11, a^11b^14, No], [0, 4, 10, a^10b^14, No], [0, 5, 9, a^9b^14, No], [0, 6, 8, a^8b^14, No], [0, 7, 7, a^7b^14, No], [0, 8, 6, a^6b^14, No], [0, 9, 5, a^5b^14, No], [0, 10, 4, a^4b^14, No], [0, 11, 3, a^3b^14, No], [0, 12, 2, a^2b^14, No], [0, 13, 1, a^1b^14, No], [0, 14, 0, a^0b^14, No], [1, 1, 12, a^13b^14, No], [1, 2, 11, a^12b^14, No], [1, 3, 10, a^11b^14, No], [1, 4, 9, a^10b^14, No], [1, 5, 8, a^9b^14, No], [1, 6, 7, a^8b^14, No], [1, 7, 6, a^7b^14, No], [1, 8, 5, a^6b^14, No], [1, 9, 4, a^5b^14, No], [1, 10, 3, a^4b^14, No], [1, 11, 2, a^3b^14, No], [1, 12, 1, a^2b^14, No], [1, 13, 0, a^1b^14, No], [2, 1, 11, a^13b^14, No], [2, 2, 10, a^12b^14, No], [2, 3, 9, a^11b^14, No], [2, 4, 8, a^10b^14, No], [2, 5, 7, a^9b^14, No], [2, 6, 6, a^8b^14, No], [2, 7, 5, a^7b^14, No], [2, 8, 4, a^6b^14, No], [2, 9, 3, a^5b^14, No], [2, 10, 2, a^4b^14, No], [2, 11, 1, a^3b^14, No], [2, 12, 0, a^2b^14, No], [3, 1, 10, a^13b^14, No], [3, 2, 9, a^12b^14, No], [3, 3, 8, a^11b^14, No], [3, 4, 7, a^10b^14, No], [3, 5, 6, a^9b^14, No], [3, 6, 5, a^8b^14, No], [3, 7, 4, a^7b^14, No], [3, 8, 3, a^6b^14, No], [3, 9, 2, a^5b^14, No], [3, 10, 1, a^4b^14, No], [3, 11, 0, a^3b^14, No], [4, 1, 9, a^13b^14, No], [4, 2, 8, a^12b^14, No], [4, 3, 7, a^11b^14, No], [4, 4, 6, a^10b^14, No], [4, 5, 5, a^9b^14, No], [4, 6, 4, a^8b^14, No], [4, 7, 3, a^7b^14, No], [4, 8, 2, a^6b^14, No], [4, 9, 1, a^5b^14, No], [4, 10, 0, a^4b^14, No], [5, 1, 8, a^13b^14, No], [5, 2, 7, a^12b^14, No], [5, 3, 6, a^11b^14, No], [5, 4, 5, a^10b^14, No], [5, 5, 4, a^9b^14, No], [5, 6, 3, a^8b^14, No], [5, 7, 2, a^7b^14, No], [5, 8, 1, a^6b^14, No], [5, 9, 0, a^5b^14, No], [6, 1, 7, a^13b^14, No], [6, 2, 6, a^12b^14, No], [6, 3, 5, a^11b^14, No], [6, 4, 4, a^10b^14, No], [6, 5, 3, a^9b^14, No], [6, 6, 2, a^8b^14, No], [6, 7, 1, a^7b^14, No], [6, 8, 0, a^6b^14, No], [7, 1, 6, a^13b^14, No], [7, 2, 5, a^12b^14, No], [7, 3, 4, a^11b^14, No], [7, 4, 3, a^10b^14, No], [7, 5, 2, a^9b^14, No], [7, 6, 1, a^8b^14, No], [7, 7, 0, a^7b^14, No], [8, 1, 5, a^13b^14, No], [8, 2, 4, a^12b^14, No], [8, 3, 3, a^11b^14, No], [8, 4, 2, a^10b^14, No], [8, 5, 1, a^9b^14, No], [8, 6, 0, a^8b^14, No], [9, 1, 4, a^13b^14, No], [9, 2, 3, a^12b^14, No], [9, 3, 2, a^11b^14, No], [9, 4, 1, a^10b^14, No], [9, 5, 0, a^9b^14, No], [10, 1, 3, a^13b^14, No], [10, 2, 2, a^12b^14, No], [10, 3, 1, a^11b^14, No], [10, 4, 0, a^10b^14, No], [11, 1, 2, a^13b^14, No], [11, 2, 1, a^12b^14, No], [11, 3, 0, a^11b^14, No], [12, 1, 1, a^13b^14, No], [12, 2, 0, a^12b^14, No], [13, 1, 0, a^13b^14, No]], 105, 105, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (14, a^14b^14, [[0, 1, 13, a^13b^14, No], [0, 2, 12, a^12b^14, No], [0, 3, 11, a^11b^14, No], [0, 4, 10, a^10b^14, No], [0, 5, 9, a^9b^14, No], [0, 6, 8, a^8b^14, No], [0, 7, 7, a^7b^14, No], [0, 8, 6, a^6b^14, No], [0, 9, 5, a^5b^14, No], [0, 10, 4, a^4b^14, No], [0, 11, 3, a^3b^14, No], [0, 12, 2, a^2b^14, No], [0, 13, 1, a^1b^14, No], [0, 14, 0, a^0b^14, No], [1, 1, 12, a^13b^14, No], [1, 2, 11, a^12b^14, No], [1, 3, 10, a^11b^14, No], [1, 4, 9, a^10b^14, No], [1, 5, 8, a^9b^14, No], [1, 6, 7, a^8b^14, No], [1, 7, 6, a^7b^14, No], [1, 8, 5, a^6b^14, No], [1, 9, 4, a^5b^14, No], [1, 10, 3, a^4b^14, No], [1, 11, 2, a^3b^14, No], [1, 12, 1, a^2b^14, No], [1, 13, 0, a^1b^14, No], [2, 1, 11, a^13b^14, No], [2, 2, 10, a^12b^14, No], [2, 3, 9, a^11b^14, No], [2, 4, 8, a^10b^14, No], [2, 5, 7, a^9b^14, No], [2, 6, 6, a^8b^14, No], [2, 7, 5, a^7b^14, No], [2, 8, 4, a^6b^14, No], [2, 9, 3, a^5b^14, No], [2, 10, 2, a^4b^14, No], [2, 11, 1, a^3b^14, No], [2, 12, 0, a^2b^14, No], [3, 1, 10, a^13b^14, No], [3, 2, 9, a^12b^14, No], [3, 3, 8, a^11b^14, No], [3, 4, 7, a^10b^14, No], [3, 5, 6, a^9b^14, No], [3, 6, 5, a^8b^14, No], [3, 7, 4, a^7b^14, No], [3, 8, 3, a^6b^14, No], [3, 9, 2, a^5b^14, No], [3, 10, 1, a^4b^14, No], [3, 11, 0, a^3b^14, No], [4, 1, 9, a^13b^14, No], [4, 2, 8, a^12b^14, No], [4, 3, 7, a^11b^14, No], [4, 4, 6, a^10b^14, No], [4, 5, 5, a^9b^14, No], [4, 6, 4, a^8b^14, No], [4, 7, 3, a^7b^14, No], [4, 8, 2, a^6b^14, No], [4, 9, 1, a^5b^14, No], [4, 10, 0, a^4b^14, No], [5, 1, 8, a^13b^14, No], [5, 2, 7, a^12b^14, No], [5, 3, 6, a^11b^14, No], [5, 4, 5, a^10b^14, No], [5, 5, 4, a^9b^14, No], [5, 6, 3, a^8b^14, No], [5, 7, 2, a^7b^14, No], [5, 8, 1, a^6b^14, No], [5, 9, 0, a^5b^14, No], [6, 1, 7, a^13b^14, No], [6, 2, 6, a^12b^14, No], [6, 3, 5, a^11b^14, No], [6, 4, 4, a^10b^14, No], [6, 5, 3, a^9b^14, No], [6, 6, 2, a^8b^14, No], [6, 7, 1, a^7b^14, No], [6, 8, 0, a^6b^14, No], [7, 1, 6, a^13b^14, No], [7, 2, 5, a^12b^14, No], [7, 3, 4, a^11b^14, No], [7, 4, 3, a^10b^14, No], [7, 5, 2, a^9b^14, No], [7, 6, 1, a^8b^14, No], [7, 7, 0, a^7b^14, No], [8, 1, 5, a^13b^14, No], [8, 2, 4, a^12b^14, No], [8, 3, 3, a^11b^14, No], [8, 4, 2, a^10b^14, No], [8, 5, 1, a^9b^14, No], [8, 6, 0, a^8b^14, No], [9, 1, 4, a^13b^14, No], [9, 2, 3, a^12b^14, No], [9, 3, 2, a^11b^14, No], [9, 4, 1, a^10b^14, No], [9, 5, 0, a^9b^14, No], [10, 1, 3, a^13b^14, No], [10, 2, 2, a^12b^14, No], [10, 3, 1, a^11b^14, No], [10, 4, 0, a^10b^14, No], [11, 1, 2, a^13b^14, No], [11, 2, 1, a^12b^14, No], [11, 3, 0, a^11b^14, No], [12, 1, 1, a^13b^14, No], [12, 2, 0, a^12b^14, No], [13, 1, 0, a^13b^14, No]], 105, 105, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

65. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=15 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(15, a^15b^15, [[0, 1, 14, a^14b^15, No], [0, 2, 13, a^13b^15, No], [0, 3, 12, a^12b^15, No], [0, 4, 11, a^11b^15, No], [0, 5, 10, a^10b^15, No], [0, 6, 9, a^9b^15, No], [0, 7, 8, a^8b^15, No], [0, 8, 7, a^7b^15, No], [0, 9, 6, a^6b^15, No], [0, 10, 5, a^5b^15, No], [0, 11, 4, a^4b^15, No], [0, 12, 3, a^3b^15, No], [0, 13, 2, a^2b^15, No], [0, 14, 1, a^1b^15, No], [0, 15, 0, a^0b^15, No], [1, 1, 13, a^14b^15, No], [1, 2, 12, a^13b^15, No], [1, 3, 11, a^12b^15, No], [1, 4, 10, a^11b^15, No], [1, 5, 9, a^10b^15, No], [1, 6, 8, a^9b^15, No], [1, 7, 7, a^8b^15, No], [1, 8, 6, a^7b^15, No], [1, 9, 5, a^6b^15, No], [1, 10, 4, a^5b^15, No], [1, 11, 3, a^4b^15, No], [1, 12, 2, a^3b^15, No], [1, 13, 1, a^2b^15, No], [1, 14, 0, a^1b^15, No], [2, 1, 12, a^14b^15, No], [2, 2, 11, a^13b^15, No], [2, 3, 10, a^12b^15, No], [2, 4, 9, a^11b^15, No], [2, 5, 8, a^10b^15, No], [2, 6, 7, a^9b^15, No], [2, 7, 6, a^8b^15, No], [2, 8, 5, a^7b^15, No], [2, 9, 4, a^6b^15, No], [2, 10, 3, a^5b^15, No], [2, 11, 2, a^4b^15, No], [2, 12, 1, a^3b^15, No], [2, 13, 0, a^2b^15, No], [3, 1, 11, a^14b^15, No], [3, 2, 10, a^13b^15, No], [3, 3, 9, a^12b^15, No], [3, 4, 8, a^11b^15, No], [3, 5, 7, a^10b^15, No], [3, 6, 6, a^9b^15, No], [3, 7, 5, a^8b^15, No], [3, 8, 4, a^7b^15, No], [3, 9, 3, a^6b^15, No], [3, 10, 2, a^5b^15, No], [3, 11, 1, a^4b^15, No], [3, 12, 0, a^3b^15, No], [4, 1, 10, a^14b^15, No], [4, 2, 9, a^13b^15, No], [4, 3, 8, a^12b^15, No], [4, 4, 7, a^11b^15, No], [4, 5, 6, a^10b^15, No], [4, 6, 5, a^9b^15, No], [4, 7, 4, a^8b^15, No], [4, 8, 3, a^7b^15, No], [4, 9, 2, a^6b^15, No], [4, 10, 1, a^5b^15, No], [4, 11, 0, a^4b^15, No], [5, 1, 9, a^14b^15, No], [5, 2, 8, a^13b^15, No], [5, 3, 7, a^12b^15, No], [5, 4, 6, a^11b^15, No], [5, 5, 5, a^10b^15, No], [5, 6, 4, a^9b^15, No], [5, 7, 3, a^8b^15, No], [5, 8, 2, a^7b^15, No], [5, 9, 1, a^6b^15, No], [5, 10, 0, a^5b^15, No], [6, 1, 8, a^14b^15, No], [6, 2, 7, a^13b^15, No], [6, 3, 6, a^12b^15, No], [6, 4, 5, a^11b^15, No], [6, 5, 4, a^10b^15, No], [6, 6, 3, a^9b^15, No], [6, 7, 2, a^8b^15, No], [6, 8, 1, a^7b^15, No], [6, 9, 0, a^6b^15, No], [7, 1, 7, a^14b^15, No], [7, 2, 6, a^13b^15, No], [7, 3, 5, a^12b^15, No], [7, 4, 4, a^11b^15, No], [7, 5, 3, a^10b^15, No], [7, 6, 2, a^9b^15, No], [7, 7, 1, a^8b^15, No], [7, 8, 0, a^7b^15, No], [8, 1, 6, a^14b^15, No], [8, 2, 5, a^13b^15, No], [8, 3, 4, a^12b^15, No], [8, 4, 3, a^11b^15, No], [8, 5, 2, a^10b^15, No], [8, 6, 1, a^9b^15, No], [8, 7, 0, a^8b^15, No], [9, 1, 5, a^14b^15, No], [9, 2, 4, a^13b^15, No], [9, 3, 3, a^12b^15, No], [9, 4, 2, a^11b^15, No], [9, 5, 1, a^10b^15, No], [9, 6, 0, a^9b^15, No], [10, 1, 4, a^14b^15, No], [10, 2, 3, a^13b^15, No], [10, 3, 2, a^12b^15, No], [10, 4, 1, a^11b^15, No], [10, 5, 0, a^10b^15, No], [11, 1, 3, a^14b^15, No], [11, 2, 2, a^13b^15, No], [11, 3, 1, a^12b^15, No], [11, 4, 0, a^11b^15, No], [12, 1, 2, a^14b^15, No], [12, 2, 1, a^13b^15, No], [12, 3, 0, a^12b^15, No], [13, 1, 1, a^14b^15, No], [13, 2, 0, a^13b^15, No], [14, 1, 0, a^14b^15, No]], 120, 120, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (15, a^15b^15, [[0, 1, 14, a^14b^15, No], [0, 2, 13, a^13b^15, No], [0, 3, 12, a^12b^15, No], [0, 4, 11, a^11b^15, No], [0, 5, 10, a^10b^15, No], [0, 6, 9, a^9b^15, No], [0, 7, 8, a^8b^15, No], [0, 8, 7, a^7b^15, No], [0, 9, 6, a^6b^15, No], [0, 10, 5, a^5b^15, No], [0, 11, 4, a^4b^15, No], [0, 12, 3, a^3b^15, No], [0, 13, 2, a^2b^15, No], [0, 14, 1, a^1b^15, No], [0, 15, 0, a^0b^15, No], [1, 1, 13, a^14b^15, No], [1, 2, 12, a^13b^15, No], [1, 3, 11, a^12b^15, No], [1, 4, 10, a^11b^15, No], [1, 5, 9, a^10b^15, No], [1, 6, 8, a^9b^15, No], [1, 7, 7, a^8b^15, No], [1, 8, 6, a^7b^15, No], [1, 9, 5, a^6b^15, No], [1, 10, 4, a^5b^15, No], [1, 11, 3, a^4b^15, No], [1, 12, 2, a^3b^15, No], [1, 13, 1, a^2b^15, No], [1, 14, 0, a^1b^15, No], [2, 1, 12, a^14b^15, No], [2, 2, 11, a^13b^15, No], [2, 3, 10, a^12b^15, No], [2, 4, 9, a^11b^15, No], [2, 5, 8, a^10b^15, No], [2, 6, 7, a^9b^15, No], [2, 7, 6, a^8b^15, No], [2, 8, 5, a^7b^15, No], [2, 9, 4, a^6b^15, No], [2, 10, 3, a^5b^15, No], [2, 11, 2, a^4b^15, No], [2, 12, 1, a^3b^15, No], [2, 13, 0, a^2b^15, No], [3, 1, 11, a^14b^15, No], [3, 2, 10, a^13b^15, No], [3, 3, 9, a^12b^15, No], [3, 4, 8, a^11b^15, No], [3, 5, 7, a^10b^15, No], [3, 6, 6, a^9b^15, No], [3, 7, 5, a^8b^15, No], [3, 8, 4, a^7b^15, No], [3, 9, 3, a^6b^15, No], [3, 10, 2, a^5b^15, No], [3, 11, 1, a^4b^15, No], [3, 12, 0, a^3b^15, No], [4, 1, 10, a^14b^15, No], [4, 2, 9, a^13b^15, No], [4, 3, 8, a^12b^15, No], [4, 4, 7, a^11b^15, No], [4, 5, 6, a^10b^15, No], [4, 6, 5, a^9b^15, No], [4, 7, 4, a^8b^15, No], [4, 8, 3, a^7b^15, No], [4, 9, 2, a^6b^15, No], [4, 10, 1, a^5b^15, No], [4, 11, 0, a^4b^15, No], [5, 1, 9, a^14b^15, No], [5, 2, 8, a^13b^15, No], [5, 3, 7, a^12b^15, No], [5, 4, 6, a^11b^15, No], [5, 5, 5, a^10b^15, No], [5, 6, 4, a^9b^15, No], [5, 7, 3, a^8b^15, No], [5, 8, 2, a^7b^15, No], [5, 9, 1, a^6b^15, No], [5, 10, 0, a^5b^15, No], [6, 1, 8, a^14b^15, No], [6, 2, 7, a^13b^15, No], [6, 3, 6, a^12b^15, No], [6, 4, 5, a^11b^15, No], [6, 5, 4, a^10b^15, No], [6, 6, 3, a^9b^15, No], [6, 7, 2, a^8b^15, No], [6, 8, 1, a^7b^15, No], [6, 9, 0, a^6b^15, No], [7, 1, 7, a^14b^15, No], [7, 2, 6, a^13b^15, No], [7, 3, 5, a^12b^15, No], [7, 4, 4, a^11b^15, No], [7, 5, 3, a^10b^15, No], [7, 6, 2, a^9b^15, No], [7, 7, 1, a^8b^15, No], [7, 8, 0, a^7b^15, No], [8, 1, 6, a^14b^15, No], [8, 2, 5, a^13b^15, No], [8, 3, 4, a^12b^15, No], [8, 4, 3, a^11b^15, No], [8, 5, 2, a^10b^15, No], [8, 6, 1, a^9b^15, No], [8, 7, 0, a^8b^15, No], [9, 1, 5, a^14b^15, No], [9, 2, 4, a^13b^15, No], [9, 3, 3, a^12b^15, No], [9, 4, 2, a^11b^15, No], [9, 5, 1, a^10b^15, No], [9, 6, 0, a^9b^15, No], [10, 1, 4, a^14b^15, No], [10, 2, 3, a^13b^15, No], [10, 3, 2, a^12b^15, No], [10, 4, 1, a^11b^15, No], [10, 5, 0, a^10b^15, No], [11, 1, 3, a^14b^15, No], [11, 2, 2, a^13b^15, No], [11, 3, 1, a^12b^15, No], [11, 4, 0, a^11b^15, No], [12, 1, 2, a^14b^15, No], [12, 2, 1, a^13b^15, No], [12, 3, 0, a^12b^15, No], [13, 1, 1, a^14b^15, No], [13, 2, 0, a^13b^15, No], [14, 1, 0, a^14b^15, No]], 120, 120, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

66. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=16 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(16, a^16b^16, [[0, 1, 15, a^15b^16, No], [0, 2, 14, a^14b^16, No], [0, 3, 13, a^13b^16, No], [0, 4, 12, a^12b^16, No], [0, 5, 11, a^11b^16, No], [0, 6, 10, a^10b^16, No], [0, 7, 9, a^9b^16, No], [0, 8, 8, a^8b^16, No], [0, 9, 7, a^7b^16, No], [0, 10, 6, a^6b^16, No], [0, 11, 5, a^5b^16, No], [0, 12, 4, a^4b^16, No], [0, 13, 3, a^3b^16, No], [0, 14, 2, a^2b^16, No], [0, 15, 1, a^1b^16, No], [0, 16, 0, a^0b^16, No], [1, 1, 14, a^15b^16, No], [1, 2, 13, a^14b^16, No], [1, 3, 12, a^13b^16, No], [1, 4, 11, a^12b^16, No], [1, 5, 10, a^11b^16, No], [1, 6, 9, a^10b^16, No], [1, 7, 8, a^9b^16, No], [1, 8, 7, a^8b^16, No], [1, 9, 6, a^7b^16, No], [1, 10, 5, a^6b^16, No], [1, 11, 4, a^5b^16, No], [1, 12, 3, a^4b^16, No], [1, 13, 2, a^3b^16, No], [1, 14, 1, a^2b^16, No], [1, 15, 0, a^1b^16, No], [2, 1, 13, a^15b^16, No], [2, 2, 12, a^14b^16, No], [2, 3, 11, a^13b^16, No], [2, 4, 10, a^12b^16, No], [2, 5, 9, a^11b^16, No], [2, 6, 8, a^10b^16, No], [2, 7, 7, a^9b^16, No], [2, 8, 6, a^8b^16, No], [2, 9, 5, a^7b^16, No], [2, 10, 4, a^6b^16, No], [2, 11, 3, a^5b^16, No], [2, 12, 2, a^4b^16, No], [2, 13, 1, a^3b^16, No], [2, 14, 0, a^2b^16, No], [3, 1, 12, a^15b^16, No], [3, 2, 11, a^14b^16, No], [3, 3, 10, a^13b^16, No], [3, 4, 9, a^12b^16, No], [3, 5, 8, a^11b^16, No], [3, 6, 7, a^10b^16, No], [3, 7, 6, a^9b^16, No], [3, 8, 5, a^8b^16, No], [3, 9, 4, a^7b^16, No], [3, 10, 3, a^6b^16, No], [3, 11, 2, a^5b^16, No], [3, 12, 1, a^4b^16, No], [3, 13, 0, a^3b^16, No], [4, 1, 11, a^15b^16, No], [4, 2, 10, a^14b^16, No], [4, 3, 9, a^13b^16, No], [4, 4, 8, a^12b^16, No], [4, 5, 7, a^11b^16, No], [4, 6, 6, a^10b^16, No], [4, 7, 5, a^9b^16, No], [4, 8, 4, a^8b^16, No], [4, 9, 3, a^7b^16, No], [4, 10, 2, a^6b^16, No], [4, 11, 1, a^5b^16, No], [4, 12, 0, a^4b^16, No], [5, 1, 10, a^15b^16, No], [5, 2, 9, a^14b^16, No], [5, 3, 8, a^13b^16, No], [5, 4, 7, a^12b^16, No], [5, 5, 6, a^11b^16, No], [5, 6, 5, a^10b^16, No], [5, 7, 4, a^9b^16, No], [5, 8, 3, a^8b^16, No], [5, 9, 2, a^7b^16, No], [5, 10, 1, a^6b^16, No], [5, 11, 0, a^5b^16, No], [6, 1, 9, a^15b^16, No], [6, 2, 8, a^14b^16, No], [6, 3, 7, a^13b^16, No], [6, 4, 6, a^12b^16, No], [6, 5, 5, a^11b^16, No], [6, 6, 4, a^10b^16, No], [6, 7, 3, a^9b^16, No], [6, 8, 2, a^8b^16, No], [6, 9, 1, a^7b^16, No], [6, 10, 0, a^6b^16, No], [7, 1, 8, a^15b^16, No], [7, 2, 7, a^14b^16, No], [7, 3, 6, a^13b^16, No], [7, 4, 5, a^12b^16, No], [7, 5, 4, a^11b^16, No], [7, 6, 3, a^10b^16, No], [7, 7, 2, a^9b^16, No], [7, 8, 1, a^8b^16, No], [7, 9, 0, a^7b^16, No], [8, 1, 7, a^15b^16, No], [8, 2, 6, a^14b^16, No], [8, 3, 5, a^13b^16, No], [8, 4, 4, a^12b^16, No], [8, 5, 3, a^11b^16, No], [8, 6, 2, a^10b^16, No], [8, 7, 1, a^9b^16, No], [8, 8, 0, a^8b^16, No], [9, 1, 6, a^15b^16, No], [9, 2, 5, a^14b^16, No], [9, 3, 4, a^13b^16, No], [9, 4, 3, a^12b^16, No], [9, 5, 2, a^11b^16, No], [9, 6, 1, a^10b^16, No], [9, 7, 0, a^9b^16, No], [10, 1, 5, a^15b^16, No], [10, 2, 4, a^14b^16, No], [10, 3, 3, a^13b^16, No], [10, 4, 2, a^12b^16, No], [10, 5, 1, a^11b^16, No], [10, 6, 0, a^10b^16, No], [11, 1, 4, a^15b^16, No], [11, 2, 3, a^14b^16, No], [11, 3, 2, a^13b^16, No], [11, 4, 1, a^12b^16, No], [11, 5, 0, a^11b^16, No], [12, 1, 3, a^15b^16, No], [12, 2, 2, a^14b^16, No], [12, 3, 1, a^13b^16, No], [12, 4, 0, a^12b^16, No], [13, 1, 2, a^15b^16, No], [13, 2, 1, a^14b^16, No], [13, 3, 0, a^13b^16, No], [14, 1, 1, a^15b^16, No], [14, 2, 0, a^14b^16, No], [15, 1, 0, a^15b^16, No]], 136, 136, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (16, a^16b^16, [[0, 1, 15, a^15b^16, No], [0, 2, 14, a^14b^16, No], [0, 3, 13, a^13b^16, No], [0, 4, 12, a^12b^16, No], [0, 5, 11, a^11b^16, No], [0, 6, 10, a^10b^16, No], [0, 7, 9, a^9b^16, No], [0, 8, 8, a^8b^16, No], [0, 9, 7, a^7b^16, No], [0, 10, 6, a^6b^16, No], [0, 11, 5, a^5b^16, No], [0, 12, 4, a^4b^16, No], [0, 13, 3, a^3b^16, No], [0, 14, 2, a^2b^16, No], [0, 15, 1, a^1b^16, No], [0, 16, 0, a^0b^16, No], [1, 1, 14, a^15b^16, No], [1, 2, 13, a^14b^16, No], [1, 3, 12, a^13b^16, No], [1, 4, 11, a^12b^16, No], [1, 5, 10, a^11b^16, No], [1, 6, 9, a^10b^16, No], [1, 7, 8, a^9b^16, No], [1, 8, 7, a^8b^16, No], [1, 9, 6, a^7b^16, No], [1, 10, 5, a^6b^16, No], [1, 11, 4, a^5b^16, No], [1, 12, 3, a^4b^16, No], [1, 13, 2, a^3b^16, No], [1, 14, 1, a^2b^16, No], [1, 15, 0, a^1b^16, No], [2, 1, 13, a^15b^16, No], [2, 2, 12, a^14b^16, No], [2, 3, 11, a^13b^16, No], [2, 4, 10, a^12b^16, No], [2, 5, 9, a^11b^16, No], [2, 6, 8, a^10b^16, No], [2, 7, 7, a^9b^16, No], [2, 8, 6, a^8b^16, No], [2, 9, 5, a^7b^16, No], [2, 10, 4, a^6b^16, No], [2, 11, 3, a^5b^16, No], [2, 12, 2, a^4b^16, No], [2, 13, 1, a^3b^16, No], [2, 14, 0, a^2b^16, No], [3, 1, 12, a^15b^16, No], [3, 2, 11, a^14b^16, No], [3, 3, 10, a^13b^16, No], [3, 4, 9, a^12b^16, No], [3, 5, 8, a^11b^16, No], [3, 6, 7, a^10b^16, No], [3, 7, 6, a^9b^16, No], [3, 8, 5, a^8b^16, No], [3, 9, 4, a^7b^16, No], [3, 10, 3, a^6b^16, No], [3, 11, 2, a^5b^16, No], [3, 12, 1, a^4b^16, No], [3, 13, 0, a^3b^16, No], [4, 1, 11, a^15b^16, No], [4, 2, 10, a^14b^16, No], [4, 3, 9, a^13b^16, No], [4, 4, 8, a^12b^16, No], [4, 5, 7, a^11b^16, No], [4, 6, 6, a^10b^16, No], [4, 7, 5, a^9b^16, No], [4, 8, 4, a^8b^16, No], [4, 9, 3, a^7b^16, No], [4, 10, 2, a^6b^16, No], [4, 11, 1, a^5b^16, No], [4, 12, 0, a^4b^16, No], [5, 1, 10, a^15b^16, No], [5, 2, 9, a^14b^16, No], [5, 3, 8, a^13b^16, No], [5, 4, 7, a^12b^16, No], [5, 5, 6, a^11b^16, No], [5, 6, 5, a^10b^16, No], [5, 7, 4, a^9b^16, No], [5, 8, 3, a^8b^16, No], [5, 9, 2, a^7b^16, No], [5, 10, 1, a^6b^16, No], [5, 11, 0, a^5b^16, No], [6, 1, 9, a^15b^16, No], [6, 2, 8, a^14b^16, No], [6, 3, 7, a^13b^16, No], [6, 4, 6, a^12b^16, No], [6, 5, 5, a^11b^16, No], [6, 6, 4, a^10b^16, No], [6, 7, 3, a^9b^16, No], [6, 8, 2, a^8b^16, No], [6, 9, 1, a^7b^16, No], [6, 10, 0, a^6b^16, No], [7, 1, 8, a^15b^16, No], [7, 2, 7, a^14b^16, No], [7, 3, 6, a^13b^16, No], [7, 4, 5, a^12b^16, No], [7, 5, 4, a^11b^16, No], [7, 6, 3, a^10b^16, No], [7, 7, 2, a^9b^16, No], [7, 8, 1, a^8b^16, No], [7, 9, 0, a^7b^16, No], [8, 1, 7, a^15b^16, No], [8, 2, 6, a^14b^16, No], [8, 3, 5, a^13b^16, No], [8, 4, 4, a^12b^16, No], [8, 5, 3, a^11b^16, No], [8, 6, 2, a^10b^16, No], [8, 7, 1, a^9b^16, No], [8, 8, 0, a^8b^16, No], [9, 1, 6, a^15b^16, No], [9, 2, 5, a^14b^16, No], [9, 3, 4, a^13b^16, No], [9, 4, 3, a^12b^16, No], [9, 5, 2, a^11b^16, No], [9, 6, 1, a^10b^16, No], [9, 7, 0, a^9b^16, No], [10, 1, 5, a^15b^16, No], [10, 2, 4, a^14b^16, No], [10, 3, 3, a^13b^16, No], [10, 4, 2, a^12b^16, No], [10, 5, 1, a^11b^16, No], [10, 6, 0, a^10b^16, No], [11, 1, 4, a^15b^16, No], [11, 2, 3, a^14b^16, No], [11, 3, 2, a^13b^16, No], [11, 4, 1, a^12b^16, No], [11, 5, 0, a^11b^16, No], [12, 1, 3, a^15b^16, No], [12, 2, 2, a^14b^16, No], [12, 3, 1, a^13b^16, No], [12, 4, 0, a^12b^16, No], [13, 1, 2, a^15b^16, No], [13, 2, 1, a^14b^16, No], [13, 3, 0, a^13b^16, No], [14, 1, 1, a^15b^16, No], [14, 2, 0, a^14b^16, No], [15, 1, 0, a^15b^16, No]], 136, 136, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

67. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=17 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(17, a^17b^17, [[0, 1, 16, a^16b^17, No], [0, 2, 15, a^15b^17, No], [0, 3, 14, a^14b^17, No], [0, 4, 13, a^13b^17, No], [0, 5, 12, a^12b^17, No], [0, 6, 11, a^11b^17, No], [0, 7, 10, a^10b^17, No], [0, 8, 9, a^9b^17, No], [0, 9, 8, a^8b^17, No], [0, 10, 7, a^7b^17, No], [0, 11, 6, a^6b^17, No], [0, 12, 5, a^5b^17, No], [0, 13, 4, a^4b^17, No], [0, 14, 3, a^3b^17, No], [0, 15, 2, a^2b^17, No], [0, 16, 1, a^1b^17, No], [0, 17, 0, a^0b^17, No], [1, 1, 15, a^16b^17, No], [1, 2, 14, a^15b^17, No], [1, 3, 13, a^14b^17, No], [1, 4, 12, a^13b^17, No], [1, 5, 11, a^12b^17, No], [1, 6, 10, a^11b^17, No], [1, 7, 9, a^10b^17, No], [1, 8, 8, a^9b^17, No], [1, 9, 7, a^8b^17, No], [1, 10, 6, a^7b^17, No], [1, 11, 5, a^6b^17, No], [1, 12, 4, a^5b^17, No], [1, 13, 3, a^4b^17, No], [1, 14, 2, a^3b^17, No], [1, 15, 1, a^2b^17, No], [1, 16, 0, a^1b^17, No], [2, 1, 14, a^16b^17, No], [2, 2, 13, a^15b^17, No], [2, 3, 12, a^14b^17, No], [2, 4, 11, a^13b^17, No], [2, 5, 10, a^12b^17, No], [2, 6, 9, a^11b^17, No], [2, 7, 8, a^10b^17, No], [2, 8, 7, a^9b^17, No], [2, 9, 6, a^8b^17, No], [2, 10, 5, a^7b^17, No], [2, 11, 4, a^6b^17, No], [2, 12, 3, a^5b^17, No], [2, 13, 2, a^4b^17, No], [2, 14, 1, a^3b^17, No], [2, 15, 0, a^2b^17, No], [3, 1, 13, a^16b^17, No], [3, 2, 12, a^15b^17, No], [3, 3, 11, a^14b^17, No], [3, 4, 10, a^13b^17, No], [3, 5, 9, a^12b^17, No], [3, 6, 8, a^11b^17, No], [3, 7, 7, a^10b^17, No], [3, 8, 6, a^9b^17, No], [3, 9, 5, a^8b^17, No], [3, 10, 4, a^7b^17, No], [3, 11, 3, a^6b^17, No], [3, 12, 2, a^5b^17, No], [3, 13, 1, a^4b^17, No], [3, 14, 0, a^3b^17, No], [4, 1, 12, a^16b^17, No], [4, 2, 11, a^15b^17, No], [4, 3, 10, a^14b^17, No], [4, 4, 9, a^13b^17, No], [4, 5, 8, a^12b^17, No], [4, 6, 7, a^11b^17, No], [4, 7, 6, a^10b^17, No], [4, 8, 5, a^9b^17, No], [4, 9, 4, a^8b^17, No], [4, 10, 3, a^7b^17, No], [4, 11, 2, a^6b^17, No], [4, 12, 1, a^5b^17, No], [4, 13, 0, a^4b^17, No], [5, 1, 11, a^16b^17, No], [5, 2, 10, a^15b^17, No], [5, 3, 9, a^14b^17, No], [5, 4, 8, a^13b^17, No], [5, 5, 7, a^12b^17, No], [5, 6, 6, a^11b^17, No], [5, 7, 5, a^10b^17, No], [5, 8, 4, a^9b^17, No], [5, 9, 3, a^8b^17, No], [5, 10, 2, a^7b^17, No], [5, 11, 1, a^6b^17, No], [5, 12, 0, a^5b^17, No], [6, 1, 10, a^16b^17, No], [6, 2, 9, a^15b^17, No], [6, 3, 8, a^14b^17, No], [6, 4, 7, a^13b^17, No], [6, 5, 6, a^12b^17, No], [6, 6, 5, a^11b^17, No], [6, 7, 4, a^10b^17, No], [6, 8, 3, a^9b^17, No], [6, 9, 2, a^8b^17, No], [6, 10, 1, a^7b^17, No], [6, 11, 0, a^6b^17, No], [7, 1, 9, a^16b^17, No], [7, 2, 8, a^15b^17, No], [7, 3, 7, a^14b^17, No], [7, 4, 6, a^13b^17, No], [7, 5, 5, a^12b^17, No], [7, 6, 4, a^11b^17, No], [7, 7, 3, a^10b^17, No], [7, 8, 2, a^9b^17, No], [7, 9, 1, a^8b^17, No], [7, 10, 0, a^7b^17, No], [8, 1, 8, a^16b^17, No], [8, 2, 7, a^15b^17, No], [8, 3, 6, a^14b^17, No], [8, 4, 5, a^13b^17, No], [8, 5, 4, a^12b^17, No], [8, 6, 3, a^11b^17, No], [8, 7, 2, a^10b^17, No], [8, 8, 1, a^9b^17, No], [8, 9, 0, a^8b^17, No], [9, 1, 7, a^16b^17, No], [9, 2, 6, a^15b^17, No], [9, 3, 5, a^14b^17, No], [9, 4, 4, a^13b^17, No], [9, 5, 3, a^12b^17, No], [9, 6, 2, a^11b^17, No], [9, 7, 1, a^10b^17, No], [9, 8, 0, a^9b^17, No], [10, 1, 6, a^16b^17, No], [10, 2, 5, a^15b^17, No], [10, 3, 4, a^14b^17, No], [10, 4, 3, a^13b^17, No], [10, 5, 2, a^12b^17, No], [10, 6, 1, a^11b^17, No], [10, 7, 0, a^10b^17, No], [11, 1, 5, a^16b^17, No], [11, 2, 4, a^15b^17, No], [11, 3, 3, a^14b^17, No], [11, 4, 2, a^13b^17, No], [11, 5, 1, a^12b^17, No], [11, 6, 0, a^11b^17, No], [12, 1, 4, a^16b^17, No], [12, 2, 3, a^15b^17, No], [12, 3, 2, a^14b^17, No], [12, 4, 1, a^13b^17, No], [12, 5, 0, a^12b^17, No], [13, 1, 3, a^16b^17, No], [13, 2, 2, a^15b^17, No], [13, 3, 1, a^14b^17, No], [13, 4, 0, a^13b^17, No], [14, 1, 2, a^16b^17, No], [14, 2, 1, a^15b^17, No], [14, 3, 0, a^14b^17, No], [15, 1, 1, a^16b^17, No], [15, 2, 0, a^15b^17, No], [16, 1, 0, a^16b^17, No]], 153, 153, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (17, a^17b^17, [[0, 1, 16, a^16b^17, No], [0, 2, 15, a^15b^17, No], [0, 3, 14, a^14b^17, No], [0, 4, 13, a^13b^17, No], [0, 5, 12, a^12b^17, No], [0, 6, 11, a^11b^17, No], [0, 7, 10, a^10b^17, No], [0, 8, 9, a^9b^17, No], [0, 9, 8, a^8b^17, No], [0, 10, 7, a^7b^17, No], [0, 11, 6, a^6b^17, No], [0, 12, 5, a^5b^17, No], [0, 13, 4, a^4b^17, No], [0, 14, 3, a^3b^17, No], [0, 15, 2, a^2b^17, No], [0, 16, 1, a^1b^17, No], [0, 17, 0, a^0b^17, No], [1, 1, 15, a^16b^17, No], [1, 2, 14, a^15b^17, No], [1, 3, 13, a^14b^17, No], [1, 4, 12, a^13b^17, No], [1, 5, 11, a^12b^17, No], [1, 6, 10, a^11b^17, No], [1, 7, 9, a^10b^17, No], [1, 8, 8, a^9b^17, No], [1, 9, 7, a^8b^17, No], [1, 10, 6, a^7b^17, No], [1, 11, 5, a^6b^17, No], [1, 12, 4, a^5b^17, No], [1, 13, 3, a^4b^17, No], [1, 14, 2, a^3b^17, No], [1, 15, 1, a^2b^17, No], [1, 16, 0, a^1b^17, No], [2, 1, 14, a^16b^17, No], [2, 2, 13, a^15b^17, No], [2, 3, 12, a^14b^17, No], [2, 4, 11, a^13b^17, No], [2, 5, 10, a^12b^17, No], [2, 6, 9, a^11b^17, No], [2, 7, 8, a^10b^17, No], [2, 8, 7, a^9b^17, No], [2, 9, 6, a^8b^17, No], [2, 10, 5, a^7b^17, No], [2, 11, 4, a^6b^17, No], [2, 12, 3, a^5b^17, No], [2, 13, 2, a^4b^17, No], [2, 14, 1, a^3b^17, No], [2, 15, 0, a^2b^17, No], [3, 1, 13, a^16b^17, No], [3, 2, 12, a^15b^17, No], [3, 3, 11, a^14b^17, No], [3, 4, 10, a^13b^17, No], [3, 5, 9, a^12b^17, No], [3, 6, 8, a^11b^17, No], [3, 7, 7, a^10b^17, No], [3, 8, 6, a^9b^17, No], [3, 9, 5, a^8b^17, No], [3, 10, 4, a^7b^17, No], [3, 11, 3, a^6b^17, No], [3, 12, 2, a^5b^17, No], [3, 13, 1, a^4b^17, No], [3, 14, 0, a^3b^17, No], [4, 1, 12, a^16b^17, No], [4, 2, 11, a^15b^17, No], [4, 3, 10, a^14b^17, No], [4, 4, 9, a^13b^17, No], [4, 5, 8, a^12b^17, No], [4, 6, 7, a^11b^17, No], [4, 7, 6, a^10b^17, No], [4, 8, 5, a^9b^17, No], [4, 9, 4, a^8b^17, No], [4, 10, 3, a^7b^17, No], [4, 11, 2, a^6b^17, No], [4, 12, 1, a^5b^17, No], [4, 13, 0, a^4b^17, No], [5, 1, 11, a^16b^17, No], [5, 2, 10, a^15b^17, No], [5, 3, 9, a^14b^17, No], [5, 4, 8, a^13b^17, No], [5, 5, 7, a^12b^17, No], [5, 6, 6, a^11b^17, No], [5, 7, 5, a^10b^17, No], [5, 8, 4, a^9b^17, No], [5, 9, 3, a^8b^17, No], [5, 10, 2, a^7b^17, No], [5, 11, 1, a^6b^17, No], [5, 12, 0, a^5b^17, No], [6, 1, 10, a^16b^17, No], [6, 2, 9, a^15b^17, No], [6, 3, 8, a^14b^17, No], [6, 4, 7, a^13b^17, No], [6, 5, 6, a^12b^17, No], [6, 6, 5, a^11b^17, No], [6, 7, 4, a^10b^17, No], [6, 8, 3, a^9b^17, No], [6, 9, 2, a^8b^17, No], [6, 10, 1, a^7b^17, No], [6, 11, 0, a^6b^17, No], [7, 1, 9, a^16b^17, No], [7, 2, 8, a^15b^17, No], [7, 3, 7, a^14b^17, No], [7, 4, 6, a^13b^17, No], [7, 5, 5, a^12b^17, No], [7, 6, 4, a^11b^17, No], [7, 7, 3, a^10b^17, No], [7, 8, 2, a^9b^17, No], [7, 9, 1, a^8b^17, No], [7, 10, 0, a^7b^17, No], [8, 1, 8, a^16b^17, No], [8, 2, 7, a^15b^17, No], [8, 3, 6, a^14b^17, No], [8, 4, 5, a^13b^17, No], [8, 5, 4, a^12b^17, No], [8, 6, 3, a^11b^17, No], [8, 7, 2, a^10b^17, No], [8, 8, 1, a^9b^17, No], [8, 9, 0, a^8b^17, No], [9, 1, 7, a^16b^17, No], [9, 2, 6, a^15b^17, No], [9, 3, 5, a^14b^17, No], [9, 4, 4, a^13b^17, No], [9, 5, 3, a^12b^17, No], [9, 6, 2, a^11b^17, No], [9, 7, 1, a^10b^17, No], [9, 8, 0, a^9b^17, No], [10, 1, 6, a^16b^17, No], [10, 2, 5, a^15b^17, No], [10, 3, 4, a^14b^17, No], [10, 4, 3, a^13b^17, No], [10, 5, 2, a^12b^17, No], [10, 6, 1, a^11b^17, No], [10, 7, 0, a^10b^17, No], [11, 1, 5, a^16b^17, No], [11, 2, 4, a^15b^17, No], [11, 3, 3, a^14b^17, No], [11, 4, 2, a^13b^17, No], [11, 5, 1, a^12b^17, No], [11, 6, 0, a^11b^17, No], [12, 1, 4, a^16b^17, No], [12, 2, 3, a^15b^17, No], [12, 3, 2, a^14b^17, No], [12, 4, 1, a^13b^17, No], [12, 5, 0, a^12b^17, No], [13, 1, 3, a^16b^17, No], [13, 2, 2, a^15b^17, No], [13, 3, 1, a^14b^17, No], [13, 4, 0, a^13b^17, No], [14, 1, 2, a^16b^17, No], [14, 2, 1, a^15b^17, No], [14, 3, 0, a^14b^17, No], [15, 1, 1, a^16b^17, No], [15, 2, 0, a^15b^17, No], [16, 1, 0, a^16b^17, No]], 153, 153, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

68. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=18 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

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(18, a^18b^18, [[0, 1, 17, a^17b^18, No], [0, 2, 16, a^16b^18, No], [0, 3, 15, a^15b^18, No], [0, 4, 14, a^14b^18, No], [0, 5, 13, a^13b^18, No], [0, 6, 12, a^12b^18, No], [0, 7, 11, a^11b^18, No], [0, 8, 10, a^10b^18, No], [0, 9, 9, a^9b^18, No], [0, 10, 8, a^8b^18, No], [0, 11, 7, a^7b^18, No], [0, 12, 6, a^6b^18, No], [0, 13, 5, a^5b^18, No], [0, 14, 4, a^4b^18, No], [0, 15, 3, a^3b^18, No], [0, 16, 2, a^2b^18, No], [0, 17, 1, a^1b^18, No], [0, 18, 0, a^0b^18, No], [1, 1, 16, a^17b^18, No], [1, 2, 15, a^16b^18, No], [1, 3, 14, a^15b^18, No], [1, 4, 13, a^14b^18, No], [1, 5, 12, a^13b^18, No], [1, 6, 11, a^12b^18, No], [1, 7, 10, a^11b^18, No], [1, 8, 9, a^10b^18, No], [1, 9, 8, a^9b^18, No], [1, 10, 7, a^8b^18, No], [1, 11, 6, a^7b^18, No], [1, 12, 5, a^6b^18, No], [1, 13, 4, a^5b^18, No], [1, 14, 3, a^4b^18, No], [1, 15, 2, a^3b^18, No], [1, 16, 1, a^2b^18, No], [1, 17, 0, a^1b^18, No], [2, 1, 15, a^17b^18, No], [2, 2, 14, a^16b^18, No], [2, 3, 13, a^15b^18, No], [2, 4, 12, a^14b^18, No], [2, 5, 11, a^13b^18, No], [2, 6, 10, a^12b^18, No], [2, 7, 9, a^11b^18, No], [2, 8, 8, a^10b^18, No], [2, 9, 7, a^9b^18, No], [2, 10, 6, a^8b^18, No], [2, 11, 5, a^7b^18, No], [2, 12, 4, a^6b^18, No], [2, 13, 3, a^5b^18, No], [2, 14, 2, a^4b^18, No], [2, 15, 1, a^3b^18, No], [2, 16, 0, a^2b^18, No], [3, 1, 14, a^17b^18, No], [3, 2, 13, a^16b^18, No], [3, 3, 12, a^15b^18, No], [3, 4, 11, a^14b^18, No], [3, 5, 10, a^13b^18, No], [3, 6, 9, a^12b^18, No], [3, 7, 8, a^11b^18, No], [3, 8, 7, a^10b^18, No], [3, 9, 6, a^9b^18, No], [3, 10, 5, a^8b^18, No], [3, 11, 4, a^7b^18, No], [3, 12, 3, a^6b^18, No], [3, 13, 2, a^5b^18, No], [3, 14, 1, a^4b^18, No], [3, 15, 0, a^3b^18, No], [4, 1, 13, a^17b^18, No], [4, 2, 12, a^16b^18, No], [4, 3, 11, a^15b^18, No], [4, 4, 10, a^14b^18, No], [4, 5, 9, a^13b^18, No], [4, 6, 8, a^12b^18, No], [4, 7, 7, a^11b^18, No], [4, 8, 6, a^10b^18, No], [4, 9, 5, a^9b^18, No], [4, 10, 4, a^8b^18, No], [4, 11, 3, a^7b^18, No], [4, 12, 2, a^6b^18, No], [4, 13, 1, a^5b^18, No], [4, 14, 0, a^4b^18, No], [5, 1, 12, a^17b^18, No], [5, 2, 11, a^16b^18, No], [5, 3, 10, a^15b^18, No], [5, 4, 9, a^14b^18, No], [5, 5, 8, a^13b^18, No], [5, 6, 7, a^12b^18, No], [5, 7, 6, a^11b^18, No], [5, 8, 5, a^10b^18, No], [5, 9, 4, a^9b^18, No], [5, 10, 3, a^8b^18, No], [5, 11, 2, a^7b^18, No], [5, 12, 1, a^6b^18, No], [5, 13, 0, a^5b^18, No], [6, 1, 11, a^17b^18, No], [6, 2, 10, a^16b^18, No], [6, 3, 9, a^15b^18, No], [6, 4, 8, a^14b^18, No], [6, 5, 7, a^13b^18, No], [6, 6, 6, a^12b^18, No], [6, 7, 5, a^11b^18, No], [6, 8, 4, a^10b^18, No], [6, 9, 3, a^9b^18, No], [6, 10, 2, a^8b^18, No], [6, 11, 1, a^7b^18, No], [6, 12, 0, a^6b^18, No], [7, 1, 10, a^17b^18, No], [7, 2, 9, a^16b^18, No], [7, 3, 8, a^15b^18, No], [7, 4, 7, a^14b^18, No], [7, 5, 6, a^13b^18, No], [7, 6, 5, a^12b^18, No], [7, 7, 4, a^11b^18, No], [7, 8, 3, a^10b^18, No], [7, 9, 2, a^9b^18, No], [7, 10, 1, a^8b^18, No], [7, 11, 0, a^7b^18, No], [8, 1, 9, a^17b^18, No], [8, 2, 8, a^16b^18, No], [8, 3, 7, a^15b^18, No], [8, 4, 6, a^14b^18, No], [8, 5, 5, a^13b^18, No], [8, 6, 4, a^12b^18, No], [8, 7, 3, a^11b^18, No], [8, 8, 2, a^10b^18, No], [8, 9, 1, a^9b^18, No], [8, 10, 0, a^8b^18, No], [9, 1, 8, a^17b^18, No], [9, 2, 7, a^16b^18, No], [9, 3, 6, a^15b^18, No], [9, 4, 5, a^14b^18, No], [9, 5, 4, a^13b^18, No], [9, 6, 3, a^12b^18, No], [9, 7, 2, a^11b^18, No], [9, 8, 1, a^10b^18, No], [9, 9, 0, a^9b^18, No], [10, 1, 7, a^17b^18, No], [10, 2, 6, a^16b^18, No], [10, 3, 5, a^15b^18, No], [10, 4, 4, a^14b^18, No], [10, 5, 3, a^13b^18, No], [10, 6, 2, a^12b^18, No], [10, 7, 1, a^11b^18, No], [10, 8, 0, a^10b^18, No], [11, 1, 6, a^17b^18, No], [11, 2, 5, a^16b^18, No], [11, 3, 4, a^15b^18, No], [11, 4, 3, a^14b^18, No], [11, 5, 2, a^13b^18, No], [11, 6, 1, a^12b^18, No], [11, 7, 0, a^11b^18, No], [12, 1, 5, a^17b^18, No], [12, 2, 4, a^16b^18, No], [12, 3, 3, a^15b^18, No], [12, 4, 2, a^14b^18, No], [12, 5, 1, a^13b^18, No], [12, 6, 0, a^12b^18, No], [13, 1, 4, a^17b^18, No], [13, 2, 3, a^16b^18, No], [13, 3, 2, a^15b^18, No], [13, 4, 1, a^14b^18, No], [13, 5, 0, a^13b^18, No], [14, 1, 3, a^17b^18, No], [14, 2, 2, a^16b^18, No], [14, 3, 1, a^15b^18, No], [14, 4, 0, a^14b^18, No], [15, 1, 2, a^17b^18, No], [15, 2, 1, a^16b^18, No], [15, 3, 0, a^15b^18, No], [16, 1, 1, a^17b^18, No], [16, 2, 0, a^16b^18, No], [17, 1, 0, a^17b^18, No]], 171, 171, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (18, a^18b^18, [[0, 1, 17, a^17b^18, No], [0, 2, 16, a^16b^18, No], [0, 3, 15, a^15b^18, No], [0, 4, 14, a^14b^18, No], [0, 5, 13, a^13b^18, No], [0, 6, 12, a^12b^18, No], [0, 7, 11, a^11b^18, No], [0, 8, 10, a^10b^18, No], [0, 9, 9, a^9b^18, No], [0, 10, 8, a^8b^18, No], [0, 11, 7, a^7b^18, No], [0, 12, 6, a^6b^18, No], [0, 13, 5, a^5b^18, No], [0, 14, 4, a^4b^18, No], [0, 15, 3, a^3b^18, No], [0, 16, 2, a^2b^18, No], [0, 17, 1, a^1b^18, No], [0, 18, 0, a^0b^18, No], [1, 1, 16, a^17b^18, No], [1, 2, 15, a^16b^18, No], [1, 3, 14, a^15b^18, No], [1, 4, 13, a^14b^18, No], [1, 5, 12, a^13b^18, No], [1, 6, 11, a^12b^18, No], [1, 7, 10, a^11b^18, No], [1, 8, 9, a^10b^18, No], [1, 9, 8, a^9b^18, No], [1, 10, 7, a^8b^18, No], [1, 11, 6, a^7b^18, No], [1, 12, 5, a^6b^18, No], [1, 13, 4, a^5b^18, No], [1, 14, 3, a^4b^18, No], [1, 15, 2, a^3b^18, No], [1, 16, 1, a^2b^18, No], [1, 17, 0, a^1b^18, No], [2, 1, 15, a^17b^18, No], [2, 2, 14, a^16b^18, No], [2, 3, 13, a^15b^18, No], [2, 4, 12, a^14b^18, No], [2, 5, 11, a^13b^18, No], [2, 6, 10, a^12b^18, No], [2, 7, 9, a^11b^18, No], [2, 8, 8, a^10b^18, No], [2, 9, 7, a^9b^18, No], [2, 10, 6, a^8b^18, No], [2, 11, 5, a^7b^18, No], [2, 12, 4, a^6b^18, No], [2, 13, 3, a^5b^18, No], [2, 14, 2, a^4b^18, No], [2, 15, 1, a^3b^18, No], [2, 16, 0, a^2b^18, No], [3, 1, 14, a^17b^18, No], [3, 2, 13, a^16b^18, No], [3, 3, 12, a^15b^18, No], [3, 4, 11, a^14b^18, No], [3, 5, 10, a^13b^18, No], [3, 6, 9, a^12b^18, No], [3, 7, 8, a^11b^18, No], [3, 8, 7, a^10b^18, No], [3, 9, 6, a^9b^18, No], [3, 10, 5, a^8b^18, No], [3, 11, 4, a^7b^18, No], [3, 12, 3, a^6b^18, No], [3, 13, 2, a^5b^18, No], [3, 14, 1, a^4b^18, No], [3, 15, 0, a^3b^18, No], [4, 1, 13, a^17b^18, No], [4, 2, 12, a^16b^18, No], [4, 3, 11, a^15b^18, No], [4, 4, 10, a^14b^18, No], [4, 5, 9, a^13b^18, No], [4, 6, 8, a^12b^18, No], [4, 7, 7, a^11b^18, No], [4, 8, 6, a^10b^18, No], [4, 9, 5, a^9b^18, No], [4, 10, 4, a^8b^18, No], [4, 11, 3, a^7b^18, No], [4, 12, 2, a^6b^18, No], [4, 13, 1, a^5b^18, No], [4, 14, 0, a^4b^18, No], [5, 1, 12, a^17b^18, No], [5, 2, 11, a^16b^18, No], [5, 3, 10, a^15b^18, No], [5, 4, 9, a^14b^18, No], [5, 5, 8, a^13b^18, No], [5, 6, 7, a^12b^18, No], [5, 7, 6, a^11b^18, No], [5, 8, 5, a^10b^18, No], [5, 9, 4, a^9b^18, No], [5, 10, 3, a^8b^18, No], [5, 11, 2, a^7b^18, No], [5, 12, 1, a^6b^18, No], [5, 13, 0, a^5b^18, No], [6, 1, 11, a^17b^18, No], [6, 2, 10, a^16b^18, No], [6, 3, 9, a^15b^18, No], [6, 4, 8, a^14b^18, No], [6, 5, 7, a^13b^18, No], [6, 6, 6, a^12b^18, No], [6, 7, 5, a^11b^18, No], [6, 8, 4, a^10b^18, No], [6, 9, 3, a^9b^18, No], [6, 10, 2, a^8b^18, No], [6, 11, 1, a^7b^18, No], [6, 12, 0, a^6b^18, No], [7, 1, 10, a^17b^18, No], [7, 2, 9, a^16b^18, No], [7, 3, 8, a^15b^18, No], [7, 4, 7, a^14b^18, No], [7, 5, 6, a^13b^18, No], [7, 6, 5, a^12b^18, No], [7, 7, 4, a^11b^18, No], [7, 8, 3, a^10b^18, No], [7, 9, 2, a^9b^18, No], [7, 10, 1, a^8b^18, No], [7, 11, 0, a^7b^18, No], [8, 1, 9, a^17b^18, No], [8, 2, 8, a^16b^18, No], [8, 3, 7, a^15b^18, No], [8, 4, 6, a^14b^18, No], [8, 5, 5, a^13b^18, No], [8, 6, 4, a^12b^18, No], [8, 7, 3, a^11b^18, No], [8, 8, 2, a^10b^18, No], [8, 9, 1, a^9b^18, No], [8, 10, 0, a^8b^18, No], [9, 1, 8, a^17b^18, No], [9, 2, 7, a^16b^18, No], [9, 3, 6, a^15b^18, No], [9, 4, 5, a^14b^18, No], [9, 5, 4, a^13b^18, No], [9, 6, 3, a^12b^18, No], [9, 7, 2, a^11b^18, No], [9, 8, 1, a^10b^18, No], [9, 9, 0, a^9b^18, No], [10, 1, 7, a^17b^18, No], [10, 2, 6, a^16b^18, No], [10, 3, 5, a^15b^18, No], [10, 4, 4, a^14b^18, No], [10, 5, 3, a^13b^18, No], [10, 6, 2, a^12b^18, No], [10, 7, 1, a^11b^18, No], [10, 8, 0, a^10b^18, No], [11, 1, 6, a^17b^18, No], [11, 2, 5, a^16b^18, No], [11, 3, 4, a^15b^18, No], [11, 4, 3, a^14b^18, No], [11, 5, 2, a^13b^18, No], [11, 6, 1, a^12b^18, No], [11, 7, 0, a^11b^18, No], [12, 1, 5, a^17b^18, No], [12, 2, 4, a^16b^18, No], [12, 3, 3, a^15b^18, No], [12, 4, 2, a^14b^18, No], [12, 5, 1, a^13b^18, No], [12, 6, 0, a^12b^18, No], [13, 1, 4, a^17b^18, No], [13, 2, 3, a^16b^18, No], [13, 3, 2, a^15b^18, No], [13, 4, 1, a^14b^18, No], [13, 5, 0, a^13b^18, No], [14, 1, 3, a^17b^18, No], [14, 2, 2, a^16b^18, No], [14, 3, 1, a^15b^18, No], [14, 4, 0, a^14b^18, No], [15, 1, 2, a^17b^18, No], [15, 2, 1, a^16b^18, No], [15, 3, 0, a^15b^18, No], [16, 1, 1, a^17b^18, No], [16, 2, 0, a^16b^18, No], [17, 1, 0, a^17b^18, No]], 171, 171, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

69. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=19 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

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(19, a^19b^19, [[0, 1, 18, a^18b^19, No], [0, 2, 17, a^17b^19, No], [0, 3, 16, a^16b^19, No], [0, 4, 15, a^15b^19, No], [0, 5, 14, a^14b^19, No], [0, 6, 13, a^13b^19, No], [0, 7, 12, a^12b^19, No], [0, 8, 11, a^11b^19, No], [0, 9, 10, a^10b^19, No], [0, 10, 9, a^9b^19, No], [0, 11, 8, a^8b^19, No], [0, 12, 7, a^7b^19, No], [0, 13, 6, a^6b^19, No], [0, 14, 5, a^5b^19, No], [0, 15, 4, a^4b^19, No], [0, 16, 3, a^3b^19, No], [0, 17, 2, a^2b^19, No], [0, 18, 1, a^1b^19, No], [0, 19, 0, a^0b^19, No], [1, 1, 17, a^18b^19, No], [1, 2, 16, a^17b^19, No], [1, 3, 15, a^16b^19, No], [1, 4, 14, a^15b^19, No], [1, 5, 13, a^14b^19, No], [1, 6, 12, a^13b^19, No], [1, 7, 11, a^12b^19, No], [1, 8, 10, a^11b^19, No], [1, 9, 9, a^10b^19, No], [1, 10, 8, a^9b^19, No], [1, 11, 7, a^8b^19, No], [1, 12, 6, a^7b^19, No], [1, 13, 5, a^6b^19, No], [1, 14, 4, a^5b^19, No], [1, 15, 3, a^4b^19, No], [1, 16, 2, a^3b^19, No], [1, 17, 1, a^2b^19, No], [1, 18, 0, a^1b^19, No], [2, 1, 16, a^18b^19, No], [2, 2, 15, a^17b^19, No], [2, 3, 14, a^16b^19, No], [2, 4, 13, a^15b^19, No], [2, 5, 12, a^14b^19, No], [2, 6, 11, a^13b^19, No], [2, 7, 10, a^12b^19, No], [2, 8, 9, a^11b^19, No], [2, 9, 8, a^10b^19, No], [2, 10, 7, a^9b^19, No], [2, 11, 6, a^8b^19, No], [2, 12, 5, a^7b^19, No], [2, 13, 4, a^6b^19, No], [2, 14, 3, a^5b^19, No], [2, 15, 2, a^4b^19, No], [2, 16, 1, a^3b^19, No], [2, 17, 0, a^2b^19, No], [3, 1, 15, a^18b^19, No], [3, 2, 14, a^17b^19, No], [3, 3, 13, a^16b^19, No], [3, 4, 12, a^15b^19, No], [3, 5, 11, a^14b^19, No], [3, 6, 10, a^13b^19, No], [3, 7, 9, a^12b^19, No], [3, 8, 8, a^11b^19, No], [3, 9, 7, a^10b^19, No], [3, 10, 6, a^9b^19, No], [3, 11, 5, a^8b^19, No], [3, 12, 4, a^7b^19, No], [3, 13, 3, a^6b^19, No], [3, 14, 2, a^5b^19, No], [3, 15, 1, a^4b^19, No], [3, 16, 0, a^3b^19, No], [4, 1, 14, a^18b^19, No], [4, 2, 13, a^17b^19, No], [4, 3, 12, a^16b^19, No], [4, 4, 11, a^15b^19, No], [4, 5, 10, a^14b^19, No], [4, 6, 9, a^13b^19, No], [4, 7, 8, a^12b^19, No], [4, 8, 7, a^11b^19, No], [4, 9, 6, a^10b^19, No], [4, 10, 5, a^9b^19, No], [4, 11, 4, a^8b^19, No], [4, 12, 3, a^7b^19, No], [4, 13, 2, a^6b^19, No], [4, 14, 1, a^5b^19, No], [4, 15, 0, a^4b^19, No], [5, 1, 13, a^18b^19, No], [5, 2, 12, a^17b^19, No], [5, 3, 11, a^16b^19, No], [5, 4, 10, a^15b^19, No], [5, 5, 9, a^14b^19, No], [5, 6, 8, a^13b^19, No], [5, 7, 7, a^12b^19, No], [5, 8, 6, a^11b^19, No], [5, 9, 5, a^10b^19, No], [5, 10, 4, a^9b^19, No], [5, 11, 3, a^8b^19, No], [5, 12, 2, a^7b^19, No], [5, 13, 1, a^6b^19, No], [5, 14, 0, a^5b^19, No], [6, 1, 12, a^18b^19, No], [6, 2, 11, a^17b^19, No], [6, 3, 10, a^16b^19, No], [6, 4, 9, a^15b^19, No], [6, 5, 8, a^14b^19, No], [6, 6, 7, a^13b^19, No], [6, 7, 6, a^12b^19, No], [6, 8, 5, a^11b^19, No], [6, 9, 4, a^10b^19, No], [6, 10, 3, a^9b^19, No], [6, 11, 2, a^8b^19, No], [6, 12, 1, a^7b^19, No], [6, 13, 0, a^6b^19, No], [7, 1, 11, a^18b^19, No], [7, 2, 10, a^17b^19, No], [7, 3, 9, a^16b^19, No], [7, 4, 8, a^15b^19, No], [7, 5, 7, a^14b^19, No], [7, 6, 6, a^13b^19, No], [7, 7, 5, a^12b^19, No], [7, 8, 4, a^11b^19, No], [7, 9, 3, a^10b^19, No], [7, 10, 2, a^9b^19, No], [7, 11, 1, a^8b^19, No], [7, 12, 0, a^7b^19, No], [8, 1, 10, a^18b^19, No], [8, 2, 9, a^17b^19, No], [8, 3, 8, a^16b^19, No], [8, 4, 7, a^15b^19, No], [8, 5, 6, a^14b^19, No], [8, 6, 5, a^13b^19, No], [8, 7, 4, a^12b^19, No], [8, 8, 3, a^11b^19, No], [8, 9, 2, a^10b^19, No], [8, 10, 1, a^9b^19, No], [8, 11, 0, a^8b^19, No], [9, 1, 9, a^18b^19, No], [9, 2, 8, a^17b^19, No], [9, 3, 7, a^16b^19, No], [9, 4, 6, a^15b^19, No], [9, 5, 5, a^14b^19, No], [9, 6, 4, a^13b^19, No], [9, 7, 3, a^12b^19, No], [9, 8, 2, a^11b^19, No], [9, 9, 1, a^10b^19, No], [9, 10, 0, a^9b^19, No], [10, 1, 8, a^18b^19, No], [10, 2, 7, a^17b^19, No], [10, 3, 6, a^16b^19, No], [10, 4, 5, a^15b^19, No], [10, 5, 4, a^14b^19, No], [10, 6, 3, a^13b^19, No], [10, 7, 2, a^12b^19, No], [10, 8, 1, a^11b^19, No], [10, 9, 0, a^10b^19, No], [11, 1, 7, a^18b^19, No], [11, 2, 6, a^17b^19, No], [11, 3, 5, a^16b^19, No], [11, 4, 4, a^15b^19, No], [11, 5, 3, a^14b^19, No], [11, 6, 2, a^13b^19, No], [11, 7, 1, a^12b^19, No], [11, 8, 0, a^11b^19, No], [12, 1, 6, a^18b^19, No], [12, 2, 5, a^17b^19, No], [12, 3, 4, a^16b^19, No], [12, 4, 3, a^15b^19, No], [12, 5, 2, a^14b^19, No], [12, 6, 1, a^13b^19, No], [12, 7, 0, a^12b^19, No], [13, 1, 5, a^18b^19, No], [13, 2, 4, a^17b^19, No], [13, 3, 3, a^16b^19, No], [13, 4, 2, a^15b^19, No], [13, 5, 1, a^14b^19, No], [13, 6, 0, a^13b^19, No], [14, 1, 4, a^18b^19, No], [14, 2, 3, a^17b^19, No], [14, 3, 2, a^16b^19, No], [14, 4, 1, a^15b^19, No], [14, 5, 0, a^14b^19, No], [15, 1, 3, a^18b^19, No], [15, 2, 2, a^17b^19, No], [15, 3, 1, a^16b^19, No], [15, 4, 0, a^15b^19, No], [16, 1, 2, a^18b^19, No], [16, 2, 1, a^17b^19, No], [16, 3, 0, a^16b^19, No], [17, 1, 1, a^18b^19, No], [17, 2, 0, a^17b^19, No], [18, 1, 0, a^18b^19, No]], 190, 190, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (19, a^19b^19, [[0, 1, 18, a^18b^19, No], [0, 2, 17, a^17b^19, No], [0, 3, 16, a^16b^19, No], [0, 4, 15, a^15b^19, No], [0, 5, 14, a^14b^19, No], [0, 6, 13, a^13b^19, No], [0, 7, 12, a^12b^19, No], [0, 8, 11, a^11b^19, No], [0, 9, 10, a^10b^19, No], [0, 10, 9, a^9b^19, No], [0, 11, 8, a^8b^19, No], [0, 12, 7, a^7b^19, No], [0, 13, 6, a^6b^19, No], [0, 14, 5, a^5b^19, No], [0, 15, 4, a^4b^19, No], [0, 16, 3, a^3b^19, No], [0, 17, 2, a^2b^19, No], [0, 18, 1, a^1b^19, No], [0, 19, 0, a^0b^19, No], [1, 1, 17, a^18b^19, No], [1, 2, 16, a^17b^19, No], [1, 3, 15, a^16b^19, No], [1, 4, 14, a^15b^19, No], [1, 5, 13, a^14b^19, No], [1, 6, 12, a^13b^19, No], [1, 7, 11, a^12b^19, No], [1, 8, 10, a^11b^19, No], [1, 9, 9, a^10b^19, No], [1, 10, 8, a^9b^19, No], [1, 11, 7, a^8b^19, No], [1, 12, 6, a^7b^19, No], [1, 13, 5, a^6b^19, No], [1, 14, 4, a^5b^19, No], [1, 15, 3, a^4b^19, No], [1, 16, 2, a^3b^19, No], [1, 17, 1, a^2b^19, No], [1, 18, 0, a^1b^19, No], [2, 1, 16, a^18b^19, No], [2, 2, 15, a^17b^19, No], [2, 3, 14, a^16b^19, No], [2, 4, 13, a^15b^19, No], [2, 5, 12, a^14b^19, No], [2, 6, 11, a^13b^19, No], [2, 7, 10, a^12b^19, No], [2, 8, 9, a^11b^19, No], [2, 9, 8, a^10b^19, No], [2, 10, 7, a^9b^19, No], [2, 11, 6, a^8b^19, No], [2, 12, 5, a^7b^19, No], [2, 13, 4, a^6b^19, No], [2, 14, 3, a^5b^19, No], [2, 15, 2, a^4b^19, No], [2, 16, 1, a^3b^19, No], [2, 17, 0, a^2b^19, No], [3, 1, 15, a^18b^19, No], [3, 2, 14, a^17b^19, No], [3, 3, 13, a^16b^19, No], [3, 4, 12, a^15b^19, No], [3, 5, 11, a^14b^19, No], [3, 6, 10, a^13b^19, No], [3, 7, 9, a^12b^19, No], [3, 8, 8, a^11b^19, No], [3, 9, 7, a^10b^19, No], [3, 10, 6, a^9b^19, No], [3, 11, 5, a^8b^19, No], [3, 12, 4, a^7b^19, No], [3, 13, 3, a^6b^19, No], [3, 14, 2, a^5b^19, No], [3, 15, 1, a^4b^19, No], [3, 16, 0, a^3b^19, No], [4, 1, 14, a^18b^19, No], [4, 2, 13, a^17b^19, No], [4, 3, 12, a^16b^19, No], [4, 4, 11, a^15b^19, No], [4, 5, 10, a^14b^19, No], [4, 6, 9, a^13b^19, No], [4, 7, 8, a^12b^19, No], [4, 8, 7, a^11b^19, No], [4, 9, 6, a^10b^19, No], [4, 10, 5, a^9b^19, No], [4, 11, 4, a^8b^19, No], [4, 12, 3, a^7b^19, No], [4, 13, 2, a^6b^19, No], [4, 14, 1, a^5b^19, No], [4, 15, 0, a^4b^19, No], [5, 1, 13, a^18b^19, No], [5, 2, 12, a^17b^19, No], [5, 3, 11, a^16b^19, No], [5, 4, 10, a^15b^19, No], [5, 5, 9, a^14b^19, No], [5, 6, 8, a^13b^19, No], [5, 7, 7, a^12b^19, No], [5, 8, 6, a^11b^19, No], [5, 9, 5, a^10b^19, No], [5, 10, 4, a^9b^19, No], [5, 11, 3, a^8b^19, No], [5, 12, 2, a^7b^19, No], [5, 13, 1, a^6b^19, No], [5, 14, 0, a^5b^19, No], [6, 1, 12, a^18b^19, No], [6, 2, 11, a^17b^19, No], [6, 3, 10, a^16b^19, No], [6, 4, 9, a^15b^19, No], [6, 5, 8, a^14b^19, No], [6, 6, 7, a^13b^19, No], [6, 7, 6, a^12b^19, No], [6, 8, 5, a^11b^19, No], [6, 9, 4, a^10b^19, No], [6, 10, 3, a^9b^19, No], [6, 11, 2, a^8b^19, No], [6, 12, 1, a^7b^19, No], [6, 13, 0, a^6b^19, No], [7, 1, 11, a^18b^19, No], [7, 2, 10, a^17b^19, No], [7, 3, 9, a^16b^19, No], [7, 4, 8, a^15b^19, No], [7, 5, 7, a^14b^19, No], [7, 6, 6, a^13b^19, No], [7, 7, 5, a^12b^19, No], [7, 8, 4, a^11b^19, No], [7, 9, 3, a^10b^19, No], [7, 10, 2, a^9b^19, No], [7, 11, 1, a^8b^19, No], [7, 12, 0, a^7b^19, No], [8, 1, 10, a^18b^19, No], [8, 2, 9, a^17b^19, No], [8, 3, 8, a^16b^19, No], [8, 4, 7, a^15b^19, No], [8, 5, 6, a^14b^19, No], [8, 6, 5, a^13b^19, No], [8, 7, 4, a^12b^19, No], [8, 8, 3, a^11b^19, No], [8, 9, 2, a^10b^19, No], [8, 10, 1, a^9b^19, No], [8, 11, 0, a^8b^19, No], [9, 1, 9, a^18b^19, No], [9, 2, 8, a^17b^19, No], [9, 3, 7, a^16b^19, No], [9, 4, 6, a^15b^19, No], [9, 5, 5, a^14b^19, No], [9, 6, 4, a^13b^19, No], [9, 7, 3, a^12b^19, No], [9, 8, 2, a^11b^19, No], [9, 9, 1, a^10b^19, No], [9, 10, 0, a^9b^19, No], [10, 1, 8, a^18b^19, No], [10, 2, 7, a^17b^19, No], [10, 3, 6, a^16b^19, No], [10, 4, 5, a^15b^19, No], [10, 5, 4, a^14b^19, No], [10, 6, 3, a^13b^19, No], [10, 7, 2, a^12b^19, No], [10, 8, 1, a^11b^19, No], [10, 9, 0, a^10b^19, No], [11, 1, 7, a^18b^19, No], [11, 2, 6, a^17b^19, No], [11, 3, 5, a^16b^19, No], [11, 4, 4, a^15b^19, No], [11, 5, 3, a^14b^19, No], [11, 6, 2, a^13b^19, No], [11, 7, 1, a^12b^19, No], [11, 8, 0, a^11b^19, No], [12, 1, 6, a^18b^19, No], [12, 2, 5, a^17b^19, No], [12, 3, 4, a^16b^19, No], [12, 4, 3, a^15b^19, No], [12, 5, 2, a^14b^19, No], [12, 6, 1, a^13b^19, No], [12, 7, 0, a^12b^19, No], [13, 1, 5, a^18b^19, No], [13, 2, 4, a^17b^19, No], [13, 3, 3, a^16b^19, No], [13, 4, 2, a^15b^19, No], [13, 5, 1, a^14b^19, No], [13, 6, 0, a^13b^19, No], [14, 1, 4, a^18b^19, No], [14, 2, 3, a^17b^19, No], [14, 3, 2, a^16b^19, No], [14, 4, 1, a^15b^19, No], [14, 5, 0, a^14b^19, No], [15, 1, 3, a^18b^19, No], [15, 2, 2, a^17b^19, No], [15, 3, 1, a^16b^19, No], [15, 4, 0, a^15b^19, No], [16, 1, 2, a^18b^19, No], [16, 2, 1, a^17b^19, No], [16, 3, 0, a^16b^19, No], [17, 1, 1, a^18b^19, No], [17, 2, 0, a^17b^19, No], [18, 1, 0, a^18b^19, No]], 190, 190, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

70. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=20 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(20, a^20b^20, [[0, 1, 19, a^19b^20, No], [0, 2, 18, a^18b^20, No], [0, 3, 17, a^17b^20, No], [0, 4, 16, a^16b^20, No], [0, 5, 15, a^15b^20, No], [0, 6, 14, a^14b^20, No], [0, 7, 13, a^13b^20, No], [0, 8, 12, a^12b^20, No], [0, 9, 11, a^11b^20, No], [0, 10, 10, a^10b^20, No], [0, 11, 9, a^9b^20, No], [0, 12, 8, a^8b^20, No], [0, 13, 7, a^7b^20, No], [0, 14, 6, a^6b^20, No], [0, 15, 5, a^5b^20, No], [0, 16, 4, a^4b^20, No], [0, 17, 3, a^3b^20, No], [0, 18, 2, a^2b^20, No], [0, 19, 1, a^1b^20, No], [0, 20, 0, a^0b^20, No], [1, 1, 18, a^19b^20, No], [1, 2, 17, a^18b^20, No], [1, 3, 16, a^17b^20, No], [1, 4, 15, a^16b^20, No], [1, 5, 14, a^15b^20, No], [1, 6, 13, a^14b^20, No], [1, 7, 12, a^13b^20, No], [1, 8, 11, a^12b^20, No], [1, 9, 10, a^11b^20, No], [1, 10, 9, a^10b^20, No], [1, 11, 8, a^9b^20, No], [1, 12, 7, a^8b^20, No], [1, 13, 6, a^7b^20, No], [1, 14, 5, a^6b^20, No], [1, 15, 4, a^5b^20, No], [1, 16, 3, a^4b^20, No], [1, 17, 2, a^3b^20, No], [1, 18, 1, a^2b^20, No], [1, 19, 0, a^1b^20, No], [2, 1, 17, a^19b^20, No], [2, 2, 16, a^18b^20, No], [2, 3, 15, a^17b^20, No], [2, 4, 14, a^16b^20, No], [2, 5, 13, a^15b^20, No], [2, 6, 12, a^14b^20, No], [2, 7, 11, a^13b^20, No], [2, 8, 10, a^12b^20, No], [2, 9, 9, a^11b^20, No], [2, 10, 8, a^10b^20, No], [2, 11, 7, a^9b^20, No], [2, 12, 6, a^8b^20, No], [2, 13, 5, a^7b^20, No], [2, 14, 4, a^6b^20, No], [2, 15, 3, a^5b^20, No], [2, 16, 2, a^4b^20, No], [2, 17, 1, a^3b^20, No], [2, 18, 0, a^2b^20, No], [3, 1, 16, a^19b^20, No], [3, 2, 15, a^18b^20, No], [3, 3, 14, a^17b^20, No], [3, 4, 13, a^16b^20, No], [3, 5, 12, a^15b^20, No], [3, 6, 11, a^14b^20, No], [3, 7, 10, a^13b^20, No], [3, 8, 9, a^12b^20, No], [3, 9, 8, a^11b^20, No], [3, 10, 7, a^10b^20, No], [3, 11, 6, a^9b^20, No], [3, 12, 5, a^8b^20, No], [3, 13, 4, a^7b^20, No], [3, 14, 3, a^6b^20, No], [3, 15, 2, a^5b^20, No], [3, 16, 1, a^4b^20, No], [3, 17, 0, a^3b^20, No], [4, 1, 15, a^19b^20, No], [4, 2, 14, a^18b^20, No], [4, 3, 13, a^17b^20, No], [4, 4, 12, a^16b^20, No], [4, 5, 11, a^15b^20, No], [4, 6, 10, a^14b^20, No], [4, 7, 9, a^13b^20, No], [4, 8, 8, a^12b^20, No], [4, 9, 7, a^11b^20, No], [4, 10, 6, a^10b^20, No], [4, 11, 5, a^9b^20, No], [4, 12, 4, a^8b^20, No], [4, 13, 3, a^7b^20, No], [4, 14, 2, a^6b^20, No], [4, 15, 1, a^5b^20, No], [4, 16, 0, a^4b^20, No], [5, 1, 14, a^19b^20, No], [5, 2, 13, a^18b^20, No], [5, 3, 12, a^17b^20, No], [5, 4, 11, a^16b^20, No], [5, 5, 10, a^15b^20, No], [5, 6, 9, a^14b^20, No], [5, 7, 8, a^13b^20, No], [5, 8, 7, a^12b^20, No], [5, 9, 6, a^11b^20, No], [5, 10, 5, a^10b^20, No], [5, 11, 4, a^9b^20, No], [5, 12, 3, a^8b^20, No], [5, 13, 2, a^7b^20, No], [5, 14, 1, a^6b^20, No], [5, 15, 0, a^5b^20, No], [6, 1, 13, a^19b^20, No], [6, 2, 12, a^18b^20, No], [6, 3, 11, a^17b^20, No], [6, 4, 10, a^16b^20, No], [6, 5, 9, a^15b^20, No], [6, 6, 8, a^14b^20, No], [6, 7, 7, a^13b^20, No], [6, 8, 6, a^12b^20, No], [6, 9, 5, a^11b^20, No], [6, 10, 4, a^10b^20, No], [6, 11, 3, a^9b^20, No], [6, 12, 2, a^8b^20, No], [6, 13, 1, a^7b^20, No], [6, 14, 0, a^6b^20, No], [7, 1, 12, a^19b^20, No], [7, 2, 11, a^18b^20, No], [7, 3, 10, a^17b^20, No], [7, 4, 9, a^16b^20, No], [7, 5, 8, a^15b^20, No], [7, 6, 7, a^14b^20, No], [7, 7, 6, a^13b^20, No], [7, 8, 5, a^12b^20, No], [7, 9, 4, a^11b^20, No], [7, 10, 3, a^10b^20, No], [7, 11, 2, a^9b^20, No], [7, 12, 1, a^8b^20, No], [7, 13, 0, a^7b^20, No], [8, 1, 11, a^19b^20, No], [8, 2, 10, a^18b^20, No], [8, 3, 9, a^17b^20, No], [8, 4, 8, a^16b^20, No], [8, 5, 7, a^15b^20, No], [8, 6, 6, a^14b^20, No], [8, 7, 5, a^13b^20, No], [8, 8, 4, a^12b^20, No], [8, 9, 3, a^11b^20, No], [8, 10, 2, a^10b^20, No], [8, 11, 1, a^9b^20, No], [8, 12, 0, a^8b^20, No], [9, 1, 10, a^19b^20, No], [9, 2, 9, a^18b^20, No], [9, 3, 8, a^17b^20, No], [9, 4, 7, a^16b^20, No], [9, 5, 6, a^15b^20, No], [9, 6, 5, a^14b^20, No], [9, 7, 4, a^13b^20, No], [9, 8, 3, a^12b^20, No], [9, 9, 2, a^11b^20, No], [9, 10, 1, a^10b^20, No], [9, 11, 0, a^9b^20, No], [10, 1, 9, a^19b^20, No], [10, 2, 8, a^18b^20, No], [10, 3, 7, a^17b^20, No], [10, 4, 6, a^16b^20, No], [10, 5, 5, a^15b^20, No], [10, 6, 4, a^14b^20, No], [10, 7, 3, a^13b^20, No], [10, 8, 2, a^12b^20, No], [10, 9, 1, a^11b^20, No], [10, 10, 0, a^10b^20, No], [11, 1, 8, a^19b^20, No], [11, 2, 7, a^18b^20, No], [11, 3, 6, a^17b^20, No], [11, 4, 5, a^16b^20, No], [11, 5, 4, a^15b^20, No], [11, 6, 3, a^14b^20, No], [11, 7, 2, a^13b^20, No], [11, 8, 1, a^12b^20, No], [11, 9, 0, a^11b^20, No], [12, 1, 7, a^19b^20, No], [12, 2, 6, a^18b^20, No], [12, 3, 5, a^17b^20, No], [12, 4, 4, a^16b^20, No], [12, 5, 3, a^15b^20, No], [12, 6, 2, a^14b^20, No], [12, 7, 1, a^13b^20, No], [12, 8, 0, a^12b^20, No], [13, 1, 6, a^19b^20, No], [13, 2, 5, a^18b^20, No], [13, 3, 4, a^17b^20, No], [13, 4, 3, a^16b^20, No], [13, 5, 2, a^15b^20, No], [13, 6, 1, a^14b^20, No], [13, 7, 0, a^13b^20, No], [14, 1, 5, a^19b^20, No], [14, 2, 4, a^18b^20, No], [14, 3, 3, a^17b^20, No], [14, 4, 2, a^16b^20, No], [14, 5, 1, a^15b^20, No], [14, 6, 0, a^14b^20, No], [15, 1, 4, a^19b^20, No], [15, 2, 3, a^18b^20, No], [15, 3, 2, a^17b^20, No], [15, 4, 1, a^16b^20, No], [15, 5, 0, a^15b^20, No], [16, 1, 3, a^19b^20, No], [16, 2, 2, a^18b^20, No], [16, 3, 1, a^17b^20, No], [16, 4, 0, a^16b^20, No], [17, 1, 2, a^19b^20, No], [17, 2, 1, a^18b^20, No], [17, 3, 0, a^17b^20, No], [18, 1, 1, a^19b^20, No], [18, 2, 0, a^18b^20, No], [19, 1, 0, a^19b^20, No]], 210, 210, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (20, a^20b^20, [[0, 1, 19, a^19b^20, No], [0, 2, 18, a^18b^20, No], [0, 3, 17, a^17b^20, No], [0, 4, 16, a^16b^20, No], [0, 5, 15, a^15b^20, No], [0, 6, 14, a^14b^20, No], [0, 7, 13, a^13b^20, No], [0, 8, 12, a^12b^20, No], [0, 9, 11, a^11b^20, No], [0, 10, 10, a^10b^20, No], [0, 11, 9, a^9b^20, No], [0, 12, 8, a^8b^20, No], [0, 13, 7, a^7b^20, No], [0, 14, 6, a^6b^20, No], [0, 15, 5, a^5b^20, No], [0, 16, 4, a^4b^20, No], [0, 17, 3, a^3b^20, No], [0, 18, 2, a^2b^20, No], [0, 19, 1, a^1b^20, No], [0, 20, 0, a^0b^20, No], [1, 1, 18, a^19b^20, No], [1, 2, 17, a^18b^20, No], [1, 3, 16, a^17b^20, No], [1, 4, 15, a^16b^20, No], [1, 5, 14, a^15b^20, No], [1, 6, 13, a^14b^20, No], [1, 7, 12, a^13b^20, No], [1, 8, 11, a^12b^20, No], [1, 9, 10, a^11b^20, No], [1, 10, 9, a^10b^20, No], [1, 11, 8, a^9b^20, No], [1, 12, 7, a^8b^20, No], [1, 13, 6, a^7b^20, No], [1, 14, 5, a^6b^20, No], [1, 15, 4, a^5b^20, No], [1, 16, 3, a^4b^20, No], [1, 17, 2, a^3b^20, No], [1, 18, 1, a^2b^20, No], [1, 19, 0, a^1b^20, No], [2, 1, 17, a^19b^20, No], [2, 2, 16, a^18b^20, No], [2, 3, 15, a^17b^20, No], [2, 4, 14, a^16b^20, No], [2, 5, 13, a^15b^20, No], [2, 6, 12, a^14b^20, No], [2, 7, 11, a^13b^20, No], [2, 8, 10, a^12b^20, No], [2, 9, 9, a^11b^20, No], [2, 10, 8, a^10b^20, No], [2, 11, 7, a^9b^20, No], [2, 12, 6, a^8b^20, No], [2, 13, 5, a^7b^20, No], [2, 14, 4, a^6b^20, No], [2, 15, 3, a^5b^20, No], [2, 16, 2, a^4b^20, No], [2, 17, 1, a^3b^20, No], [2, 18, 0, a^2b^20, No], [3, 1, 16, a^19b^20, No], [3, 2, 15, a^18b^20, No], [3, 3, 14, a^17b^20, No], [3, 4, 13, a^16b^20, No], [3, 5, 12, a^15b^20, No], [3, 6, 11, a^14b^20, No], [3, 7, 10, a^13b^20, No], [3, 8, 9, a^12b^20, No], [3, 9, 8, a^11b^20, No], [3, 10, 7, a^10b^20, No], [3, 11, 6, a^9b^20, No], [3, 12, 5, a^8b^20, No], [3, 13, 4, a^7b^20, No], [3, 14, 3, a^6b^20, No], [3, 15, 2, a^5b^20, No], [3, 16, 1, a^4b^20, No], [3, 17, 0, a^3b^20, No], [4, 1, 15, a^19b^20, No], [4, 2, 14, a^18b^20, No], [4, 3, 13, a^17b^20, No], [4, 4, 12, a^16b^20, No], [4, 5, 11, a^15b^20, No], [4, 6, 10, a^14b^20, No], [4, 7, 9, a^13b^20, No], [4, 8, 8, a^12b^20, No], [4, 9, 7, a^11b^20, No], [4, 10, 6, a^10b^20, No], [4, 11, 5, a^9b^20, No], [4, 12, 4, a^8b^20, No], [4, 13, 3, a^7b^20, No], [4, 14, 2, a^6b^20, No], [4, 15, 1, a^5b^20, No], [4, 16, 0, a^4b^20, No], [5, 1, 14, a^19b^20, No], [5, 2, 13, a^18b^20, No], [5, 3, 12, a^17b^20, No], [5, 4, 11, a^16b^20, No], [5, 5, 10, a^15b^20, No], [5, 6, 9, a^14b^20, No], [5, 7, 8, a^13b^20, No], [5, 8, 7, a^12b^20, No], [5, 9, 6, a^11b^20, No], [5, 10, 5, a^10b^20, No], [5, 11, 4, a^9b^20, No], [5, 12, 3, a^8b^20, No], [5, 13, 2, a^7b^20, No], [5, 14, 1, a^6b^20, No], [5, 15, 0, a^5b^20, No], [6, 1, 13, a^19b^20, No], [6, 2, 12, a^18b^20, No], [6, 3, 11, a^17b^20, No], [6, 4, 10, a^16b^20, No], [6, 5, 9, a^15b^20, No], [6, 6, 8, a^14b^20, No], [6, 7, 7, a^13b^20, No], [6, 8, 6, a^12b^20, No], [6, 9, 5, a^11b^20, No], [6, 10, 4, a^10b^20, No], [6, 11, 3, a^9b^20, No], [6, 12, 2, a^8b^20, No], [6, 13, 1, a^7b^20, No], [6, 14, 0, a^6b^20, No], [7, 1, 12, a^19b^20, No], [7, 2, 11, a^18b^20, No], [7, 3, 10, a^17b^20, No], [7, 4, 9, a^16b^20, No], [7, 5, 8, a^15b^20, No], [7, 6, 7, a^14b^20, No], [7, 7, 6, a^13b^20, No], [7, 8, 5, a^12b^20, No], [7, 9, 4, a^11b^20, No], [7, 10, 3, a^10b^20, No], [7, 11, 2, a^9b^20, No], [7, 12, 1, a^8b^20, No], [7, 13, 0, a^7b^20, No], [8, 1, 11, a^19b^20, No], [8, 2, 10, a^18b^20, No], [8, 3, 9, a^17b^20, No], [8, 4, 8, a^16b^20, No], [8, 5, 7, a^15b^20, No], [8, 6, 6, a^14b^20, No], [8, 7, 5, a^13b^20, No], [8, 8, 4, a^12b^20, No], [8, 9, 3, a^11b^20, No], [8, 10, 2, a^10b^20, No], [8, 11, 1, a^9b^20, No], [8, 12, 0, a^8b^20, No], [9, 1, 10, a^19b^20, No], [9, 2, 9, a^18b^20, No], [9, 3, 8, a^17b^20, No], [9, 4, 7, a^16b^20, No], [9, 5, 6, a^15b^20, No], [9, 6, 5, a^14b^20, No], [9, 7, 4, a^13b^20, No], [9, 8, 3, a^12b^20, No], [9, 9, 2, a^11b^20, No], [9, 10, 1, a^10b^20, No], [9, 11, 0, a^9b^20, No], [10, 1, 9, a^19b^20, No], [10, 2, 8, a^18b^20, No], [10, 3, 7, a^17b^20, No], [10, 4, 6, a^16b^20, No], [10, 5, 5, a^15b^20, No], [10, 6, 4, a^14b^20, No], [10, 7, 3, a^13b^20, No], [10, 8, 2, a^12b^20, No], [10, 9, 1, a^11b^20, No], [10, 10, 0, a^10b^20, No], [11, 1, 8, a^19b^20, No], [11, 2, 7, a^18b^20, No], [11, 3, 6, a^17b^20, No], [11, 4, 5, a^16b^20, No], [11, 5, 4, a^15b^20, No], [11, 6, 3, a^14b^20, No], [11, 7, 2, a^13b^20, No], [11, 8, 1, a^12b^20, No], [11, 9, 0, a^11b^20, No], [12, 1, 7, a^19b^20, No], [12, 2, 6, a^18b^20, No], [12, 3, 5, a^17b^20, No], [12, 4, 4, a^16b^20, No], [12, 5, 3, a^15b^20, No], [12, 6, 2, a^14b^20, No], [12, 7, 1, a^13b^20, No], [12, 8, 0, a^12b^20, No], [13, 1, 6, a^19b^20, No], [13, 2, 5, a^18b^20, No], [13, 3, 4, a^17b^20, No], [13, 4, 3, a^16b^20, No], [13, 5, 2, a^15b^20, No], [13, 6, 1, a^14b^20, No], [13, 7, 0, a^13b^20, No], [14, 1, 5, a^19b^20, No], [14, 2, 4, a^18b^20, No], [14, 3, 3, a^17b^20, No], [14, 4, 2, a^16b^20, No], [14, 5, 1, a^15b^20, No], [14, 6, 0, a^14b^20, No], [15, 1, 4, a^19b^20, No], [15, 2, 3, a^18b^20, No], [15, 3, 2, a^17b^20, No], [15, 4, 1, a^16b^20, No], [15, 5, 0, a^15b^20, No], [16, 1, 3, a^19b^20, No], [16, 2, 2, a^18b^20, No], [16, 3, 1, a^17b^20, No], [16, 4, 0, a^16b^20, No], [17, 1, 2, a^19b^20, No], [17, 2, 1, a^18b^20, No], [17, 3, 0, a^17b^20, No], [18, 1, 1, a^19b^20, No], [18, 2, 0, a^18b^20, No], [19, 1, 0, a^19b^20, No]], 210, 210, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

71. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=21 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(21, a^21b^21, [[0, 1, 20, a^20b^21, No], [0, 2, 19, a^19b^21, No], [0, 3, 18, a^18b^21, No], [0, 4, 17, a^17b^21, No], [0, 5, 16, a^16b^21, No], [0, 6, 15, a^15b^21, No], [0, 7, 14, a^14b^21, No], [0, 8, 13, a^13b^21, No], [0, 9, 12, a^12b^21, No], [0, 10, 11, a^11b^21, No], [0, 11, 10, a^10b^21, No], [0, 12, 9, a^9b^21, No], [0, 13, 8, a^8b^21, No], [0, 14, 7, a^7b^21, No], [0, 15, 6, a^6b^21, No], [0, 16, 5, a^5b^21, No], [0, 17, 4, a^4b^21, No], [0, 18, 3, a^3b^21, No], [0, 19, 2, a^2b^21, No], [0, 20, 1, a^1b^21, No], [0, 21, 0, a^0b^21, No], [1, 1, 19, a^20b^21, No], [1, 2, 18, a^19b^21, No], [1, 3, 17, a^18b^21, No], [1, 4, 16, a^17b^21, No], [1, 5, 15, a^16b^21, No], [1, 6, 14, a^15b^21, No], [1, 7, 13, a^14b^21, No], [1, 8, 12, a^13b^21, No], [1, 9, 11, a^12b^21, No], [1, 10, 10, a^11b^21, No], [1, 11, 9, a^10b^21, No], [1, 12, 8, a^9b^21, No], [1, 13, 7, a^8b^21, No], [1, 14, 6, a^7b^21, No], [1, 15, 5, a^6b^21, No], [1, 16, 4, a^5b^21, No], [1, 17, 3, a^4b^21, No], [1, 18, 2, a^3b^21, No], [1, 19, 1, a^2b^21, No], [1, 20, 0, a^1b^21, No], [2, 1, 18, a^20b^21, No], [2, 2, 17, a^19b^21, No], [2, 3, 16, a^18b^21, No], [2, 4, 15, a^17b^21, No], [2, 5, 14, a^16b^21, No], [2, 6, 13, a^15b^21, No], [2, 7, 12, a^14b^21, No], [2, 8, 11, a^13b^21, No], [2, 9, 10, a^12b^21, No], [2, 10, 9, a^11b^21, No], [2, 11, 8, a^10b^21, No], [2, 12, 7, a^9b^21, No], [2, 13, 6, a^8b^21, No], [2, 14, 5, a^7b^21, No], [2, 15, 4, a^6b^21, No], [2, 16, 3, a^5b^21, No], [2, 17, 2, a^4b^21, No], [2, 18, 1, a^3b^21, No], [2, 19, 0, a^2b^21, No], [3, 1, 17, a^20b^21, No], [3, 2, 16, a^19b^21, No], [3, 3, 15, a^18b^21, No], [3, 4, 14, a^17b^21, No], [3, 5, 13, a^16b^21, No], [3, 6, 12, a^15b^21, No], [3, 7, 11, a^14b^21, No], [3, 8, 10, a^13b^21, No], [3, 9, 9, a^12b^21, No], [3, 10, 8, a^11b^21, No], [3, 11, 7, a^10b^21, No], [3, 12, 6, a^9b^21, No], [3, 13, 5, a^8b^21, No], [3, 14, 4, a^7b^21, No], [3, 15, 3, a^6b^21, No], [3, 16, 2, a^5b^21, No], [3, 17, 1, a^4b^21, No], [3, 18, 0, a^3b^21, No], [4, 1, 16, a^20b^21, No], [4, 2, 15, a^19b^21, No], [4, 3, 14, a^18b^21, No], [4, 4, 13, a^17b^21, No], [4, 5, 12, a^16b^21, No], [4, 6, 11, a^15b^21, No], [4, 7, 10, a^14b^21, No], [4, 8, 9, a^13b^21, No], [4, 9, 8, a^12b^21, No], [4, 10, 7, a^11b^21, No], [4, 11, 6, a^10b^21, No], [4, 12, 5, a^9b^21, No], [4, 13, 4, a^8b^21, No], [4, 14, 3, a^7b^21, No], [4, 15, 2, a^6b^21, No], [4, 16, 1, a^5b^21, No], [4, 17, 0, a^4b^21, No], [5, 1, 15, a^20b^21, No], [5, 2, 14, a^19b^21, No], [5, 3, 13, a^18b^21, No], [5, 4, 12, a^17b^21, No], [5, 5, 11, a^16b^21, No], [5, 6, 10, a^15b^21, No], [5, 7, 9, a^14b^21, No], [5, 8, 8, a^13b^21, No], [5, 9, 7, a^12b^21, No], [5, 10, 6, a^11b^21, No], [5, 11, 5, a^10b^21, No], [5, 12, 4, a^9b^21, No], [5, 13, 3, a^8b^21, No], [5, 14, 2, a^7b^21, No], [5, 15, 1, a^6b^21, No], [5, 16, 0, a^5b^21, No], [6, 1, 14, a^20b^21, No], [6, 2, 13, a^19b^21, No], [6, 3, 12, a^18b^21, No], [6, 4, 11, a^17b^21, No], [6, 5, 10, a^16b^21, No], [6, 6, 9, a^15b^21, No], [6, 7, 8, a^14b^21, No], [6, 8, 7, a^13b^21, No], [6, 9, 6, a^12b^21, No], [6, 10, 5, a^11b^21, No], [6, 11, 4, a^10b^21, No], [6, 12, 3, a^9b^21, No], [6, 13, 2, a^8b^21, No], [6, 14, 1, a^7b^21, No], [6, 15, 0, a^6b^21, No], [7, 1, 13, a^20b^21, No], [7, 2, 12, a^19b^21, No], [7, 3, 11, a^18b^21, No], [7, 4, 10, a^17b^21, No], [7, 5, 9, a^16b^21, No], [7, 6, 8, a^15b^21, No], [7, 7, 7, a^14b^21, No], [7, 8, 6, a^13b^21, No], [7, 9, 5, a^12b^21, No], [7, 10, 4, a^11b^21, No], [7, 11, 3, a^10b^21, No], [7, 12, 2, a^9b^21, No], [7, 13, 1, a^8b^21, No], [7, 14, 0, a^7b^21, No], [8, 1, 12, a^20b^21, No], [8, 2, 11, a^19b^21, No], [8, 3, 10, a^18b^21, No], [8, 4, 9, a^17b^21, No], [8, 5, 8, a^16b^21, No], [8, 6, 7, a^15b^21, No], [8, 7, 6, a^14b^21, No], [8, 8, 5, a^13b^21, No], [8, 9, 4, a^12b^21, No], [8, 10, 3, a^11b^21, No], [8, 11, 2, a^10b^21, No], [8, 12, 1, a^9b^21, No], [8, 13, 0, a^8b^21, No], [9, 1, 11, a^20b^21, No], [9, 2, 10, a^19b^21, No], [9, 3, 9, a^18b^21, No], [9, 4, 8, a^17b^21, No], [9, 5, 7, a^16b^21, No], [9, 6, 6, a^15b^21, No], [9, 7, 5, a^14b^21, No], [9, 8, 4, a^13b^21, No], [9, 9, 3, a^12b^21, No], [9, 10, 2, a^11b^21, No], [9, 11, 1, a^10b^21, No], [9, 12, 0, a^9b^21, No], [10, 1, 10, a^20b^21, No], [10, 2, 9, a^19b^21, No], [10, 3, 8, a^18b^21, No], [10, 4, 7, a^17b^21, No], [10, 5, 6, a^16b^21, No], [10, 6, 5, a^15b^21, No], [10, 7, 4, a^14b^21, No], [10, 8, 3, a^13b^21, No], [10, 9, 2, a^12b^21, No], [10, 10, 1, a^11b^21, No], [10, 11, 0, a^10b^21, No], [11, 1, 9, a^20b^21, No], [11, 2, 8, a^19b^21, No], [11, 3, 7, a^18b^21, No], [11, 4, 6, a^17b^21, No], [11, 5, 5, a^16b^21, No], [11, 6, 4, a^15b^21, No], [11, 7, 3, a^14b^21, No], [11, 8, 2, a^13b^21, No], [11, 9, 1, a^12b^21, No], [11, 10, 0, a^11b^21, No], [12, 1, 8, a^20b^21, No], [12, 2, 7, a^19b^21, No], [12, 3, 6, a^18b^21, No], [12, 4, 5, a^17b^21, No], [12, 5, 4, a^16b^21, No], [12, 6, 3, a^15b^21, No], [12, 7, 2, a^14b^21, No], [12, 8, 1, a^13b^21, No], [12, 9, 0, a^12b^21, No], [13, 1, 7, a^20b^21, No], [13, 2, 6, a^19b^21, No], [13, 3, 5, a^18b^21, No], [13, 4, 4, a^17b^21, No], [13, 5, 3, a^16b^21, No], [13, 6, 2, a^15b^21, No], [13, 7, 1, a^14b^21, No], [13, 8, 0, a^13b^21, No], [14, 1, 6, a^20b^21, No], [14, 2, 5, a^19b^21, No], [14, 3, 4, a^18b^21, No], [14, 4, 3, a^17b^21, No], [14, 5, 2, a^16b^21, No], [14, 6, 1, a^15b^21, No], [14, 7, 0, a^14b^21, No], [15, 1, 5, a^20b^21, No], [15, 2, 4, a^19b^21, No], [15, 3, 3, a^18b^21, No], [15, 4, 2, a^17b^21, No], [15, 5, 1, a^16b^21, No], [15, 6, 0, a^15b^21, No], [16, 1, 4, a^20b^21, No], [16, 2, 3, a^19b^21, No], [16, 3, 2, a^18b^21, No], [16, 4, 1, a^17b^21, No], [16, 5, 0, a^16b^21, No], [17, 1, 3, a^20b^21, No], [17, 2, 2, a^19b^21, No], [17, 3, 1, a^18b^21, No], [17, 4, 0, a^17b^21, No], [18, 1, 2, a^20b^21, No], [18, 2, 1, a^19b^21, No], [18, 3, 0, a^18b^21, No], [19, 1, 1, a^20b^21, No], [19, 2, 0, a^19b^21, No], [20, 1, 0, a^20b^21, No]], 231, 231, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (21, a^21b^21, [[0, 1, 20, a^20b^21, No], [0, 2, 19, a^19b^21, No], [0, 3, 18, a^18b^21, No], [0, 4, 17, a^17b^21, No], [0, 5, 16, a^16b^21, No], [0, 6, 15, a^15b^21, No], [0, 7, 14, a^14b^21, No], [0, 8, 13, a^13b^21, No], [0, 9, 12, a^12b^21, No], [0, 10, 11, a^11b^21, No], [0, 11, 10, a^10b^21, No], [0, 12, 9, a^9b^21, No], [0, 13, 8, a^8b^21, No], [0, 14, 7, a^7b^21, No], [0, 15, 6, a^6b^21, No], [0, 16, 5, a^5b^21, No], [0, 17, 4, a^4b^21, No], [0, 18, 3, a^3b^21, No], [0, 19, 2, a^2b^21, No], [0, 20, 1, a^1b^21, No], [0, 21, 0, a^0b^21, No], [1, 1, 19, a^20b^21, No], [1, 2, 18, a^19b^21, No], [1, 3, 17, a^18b^21, No], [1, 4, 16, a^17b^21, No], [1, 5, 15, a^16b^21, No], [1, 6, 14, a^15b^21, No], [1, 7, 13, a^14b^21, No], [1, 8, 12, a^13b^21, No], [1, 9, 11, a^12b^21, No], [1, 10, 10, a^11b^21, No], [1, 11, 9, a^10b^21, No], [1, 12, 8, a^9b^21, No], [1, 13, 7, a^8b^21, No], [1, 14, 6, a^7b^21, No], [1, 15, 5, a^6b^21, No], [1, 16, 4, a^5b^21, No], [1, 17, 3, a^4b^21, No], [1, 18, 2, a^3b^21, No], [1, 19, 1, a^2b^21, No], [1, 20, 0, a^1b^21, No], [2, 1, 18, a^20b^21, No], [2, 2, 17, a^19b^21, No], [2, 3, 16, a^18b^21, No], [2, 4, 15, a^17b^21, No], [2, 5, 14, a^16b^21, No], [2, 6, 13, a^15b^21, No], [2, 7, 12, a^14b^21, No], [2, 8, 11, a^13b^21, No], [2, 9, 10, a^12b^21, No], [2, 10, 9, a^11b^21, No], [2, 11, 8, a^10b^21, No], [2, 12, 7, a^9b^21, No], [2, 13, 6, a^8b^21, No], [2, 14, 5, a^7b^21, No], [2, 15, 4, a^6b^21, No], [2, 16, 3, a^5b^21, No], [2, 17, 2, a^4b^21, No], [2, 18, 1, a^3b^21, No], [2, 19, 0, a^2b^21, No], [3, 1, 17, a^20b^21, No], [3, 2, 16, a^19b^21, No], [3, 3, 15, a^18b^21, No], [3, 4, 14, a^17b^21, No], [3, 5, 13, a^16b^21, No], [3, 6, 12, a^15b^21, No], [3, 7, 11, a^14b^21, No], [3, 8, 10, a^13b^21, No], [3, 9, 9, a^12b^21, No], [3, 10, 8, a^11b^21, No], [3, 11, 7, a^10b^21, No], [3, 12, 6, a^9b^21, No], [3, 13, 5, a^8b^21, No], [3, 14, 4, a^7b^21, No], [3, 15, 3, a^6b^21, No], [3, 16, 2, a^5b^21, No], [3, 17, 1, a^4b^21, No], [3, 18, 0, a^3b^21, No], [4, 1, 16, a^20b^21, No], [4, 2, 15, a^19b^21, No], [4, 3, 14, a^18b^21, No], [4, 4, 13, a^17b^21, No], [4, 5, 12, a^16b^21, No], [4, 6, 11, a^15b^21, No], [4, 7, 10, a^14b^21, No], [4, 8, 9, a^13b^21, No], [4, 9, 8, a^12b^21, No], [4, 10, 7, a^11b^21, No], [4, 11, 6, a^10b^21, No], [4, 12, 5, a^9b^21, No], [4, 13, 4, a^8b^21, No], [4, 14, 3, a^7b^21, No], [4, 15, 2, a^6b^21, No], [4, 16, 1, a^5b^21, No], [4, 17, 0, a^4b^21, No], [5, 1, 15, a^20b^21, No], [5, 2, 14, a^19b^21, No], [5, 3, 13, a^18b^21, No], [5, 4, 12, a^17b^21, No], [5, 5, 11, a^16b^21, No], [5, 6, 10, a^15b^21, No], [5, 7, 9, a^14b^21, No], [5, 8, 8, a^13b^21, No], [5, 9, 7, a^12b^21, No], [5, 10, 6, a^11b^21, No], [5, 11, 5, a^10b^21, No], [5, 12, 4, a^9b^21, No], [5, 13, 3, a^8b^21, No], [5, 14, 2, a^7b^21, No], [5, 15, 1, a^6b^21, No], [5, 16, 0, a^5b^21, No], [6, 1, 14, a^20b^21, No], [6, 2, 13, a^19b^21, No], [6, 3, 12, a^18b^21, No], [6, 4, 11, a^17b^21, No], [6, 5, 10, a^16b^21, No], [6, 6, 9, a^15b^21, No], [6, 7, 8, a^14b^21, No], [6, 8, 7, a^13b^21, No], [6, 9, 6, a^12b^21, No], [6, 10, 5, a^11b^21, No], [6, 11, 4, a^10b^21, No], [6, 12, 3, a^9b^21, No], [6, 13, 2, a^8b^21, No], [6, 14, 1, a^7b^21, No], [6, 15, 0, a^6b^21, No], [7, 1, 13, a^20b^21, No], [7, 2, 12, a^19b^21, No], [7, 3, 11, a^18b^21, No], [7, 4, 10, a^17b^21, No], [7, 5, 9, a^16b^21, No], [7, 6, 8, a^15b^21, No], [7, 7, 7, a^14b^21, No], [7, 8, 6, a^13b^21, No], [7, 9, 5, a^12b^21, No], [7, 10, 4, a^11b^21, No], [7, 11, 3, a^10b^21, No], [7, 12, 2, a^9b^21, No], [7, 13, 1, a^8b^21, No], [7, 14, 0, a^7b^21, No], [8, 1, 12, a^20b^21, No], [8, 2, 11, a^19b^21, No], [8, 3, 10, a^18b^21, No], [8, 4, 9, a^17b^21, No], [8, 5, 8, a^16b^21, No], [8, 6, 7, a^15b^21, No], [8, 7, 6, a^14b^21, No], [8, 8, 5, a^13b^21, No], [8, 9, 4, a^12b^21, No], [8, 10, 3, a^11b^21, No], [8, 11, 2, a^10b^21, No], [8, 12, 1, a^9b^21, No], [8, 13, 0, a^8b^21, No], [9, 1, 11, a^20b^21, No], [9, 2, 10, a^19b^21, No], [9, 3, 9, a^18b^21, No], [9, 4, 8, a^17b^21, No], [9, 5, 7, a^16b^21, No], [9, 6, 6, a^15b^21, No], [9, 7, 5, a^14b^21, No], [9, 8, 4, a^13b^21, No], [9, 9, 3, a^12b^21, No], [9, 10, 2, a^11b^21, No], [9, 11, 1, a^10b^21, No], [9, 12, 0, a^9b^21, No], [10, 1, 10, a^20b^21, No], [10, 2, 9, a^19b^21, No], [10, 3, 8, a^18b^21, No], [10, 4, 7, a^17b^21, No], [10, 5, 6, a^16b^21, No], [10, 6, 5, a^15b^21, No], [10, 7, 4, a^14b^21, No], [10, 8, 3, a^13b^21, No], [10, 9, 2, a^12b^21, No], [10, 10, 1, a^11b^21, No], [10, 11, 0, a^10b^21, No], [11, 1, 9, a^20b^21, No], [11, 2, 8, a^19b^21, No], [11, 3, 7, a^18b^21, No], [11, 4, 6, a^17b^21, No], [11, 5, 5, a^16b^21, No], [11, 6, 4, a^15b^21, No], [11, 7, 3, a^14b^21, No], [11, 8, 2, a^13b^21, No], [11, 9, 1, a^12b^21, No], [11, 10, 0, a^11b^21, No], [12, 1, 8, a^20b^21, No], [12, 2, 7, a^19b^21, No], [12, 3, 6, a^18b^21, No], [12, 4, 5, a^17b^21, No], [12, 5, 4, a^16b^21, No], [12, 6, 3, a^15b^21, No], [12, 7, 2, a^14b^21, No], [12, 8, 1, a^13b^21, No], [12, 9, 0, a^12b^21, No], [13, 1, 7, a^20b^21, No], [13, 2, 6, a^19b^21, No], [13, 3, 5, a^18b^21, No], [13, 4, 4, a^17b^21, No], [13, 5, 3, a^16b^21, No], [13, 6, 2, a^15b^21, No], [13, 7, 1, a^14b^21, No], [13, 8, 0, a^13b^21, No], [14, 1, 6, a^20b^21, No], [14, 2, 5, a^19b^21, No], [14, 3, 4, a^18b^21, No], [14, 4, 3, a^17b^21, No], [14, 5, 2, a^16b^21, No], [14, 6, 1, a^15b^21, No], [14, 7, 0, a^14b^21, No], [15, 1, 5, a^20b^21, No], [15, 2, 4, a^19b^21, No], [15, 3, 3, a^18b^21, No], [15, 4, 2, a^17b^21, No], [15, 5, 1, a^16b^21, No], [15, 6, 0, a^15b^21, No], [16, 1, 4, a^20b^21, No], [16, 2, 3, a^19b^21, No], [16, 3, 2, a^18b^21, No], [16, 4, 1, a^17b^21, No], [16, 5, 0, a^16b^21, No], [17, 1, 3, a^20b^21, No], [17, 2, 2, a^19b^21, No], [17, 3, 1, a^18b^21, No], [17, 4, 0, a^17b^21, No], [18, 1, 2, a^20b^21, No], [18, 2, 1, a^19b^21, No], [18, 3, 0, a^18b^21, No], [19, 1, 1, a^20b^21, No], [19, 2, 0, a^19b^21, No], [20, 1, 0, a^20b^21, No]], 231, 231, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

72. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=22 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(22, a^22b^22, [[0, 1, 21, a^21b^22, No], [0, 2, 20, a^20b^22, No], [0, 3, 19, a^19b^22, No], [0, 4, 18, a^18b^22, No], [0, 5, 17, a^17b^22, No], [0, 6, 16, a^16b^22, No], [0, 7, 15, a^15b^22, No], [0, 8, 14, a^14b^22, No], [0, 9, 13, a^13b^22, No], [0, 10, 12, a^12b^22, No], [0, 11, 11, a^11b^22, No], [0, 12, 10, a^10b^22, No], [0, 13, 9, a^9b^22, No], [0, 14, 8, a^8b^22, No], [0, 15, 7, a^7b^22, No], [0, 16, 6, a^6b^22, No], [0, 17, 5, a^5b^22, No], [0, 18, 4, a^4b^22, No], [0, 19, 3, a^3b^22, No], [0, 20, 2, a^2b^22, No], [0, 21, 1, a^1b^22, No], [0, 22, 0, a^0b^22, No], [1, 1, 20, a^21b^22, No], [1, 2, 19, a^20b^22, No], [1, 3, 18, a^19b^22, No], [1, 4, 17, a^18b^22, No], [1, 5, 16, a^17b^22, No], [1, 6, 15, a^16b^22, No], [1, 7, 14, a^15b^22, No], [1, 8, 13, a^14b^22, No], [1, 9, 12, a^13b^22, No], [1, 10, 11, a^12b^22, No], [1, 11, 10, a^11b^22, No], [1, 12, 9, a^10b^22, No], [1, 13, 8, a^9b^22, No], [1, 14, 7, a^8b^22, No], [1, 15, 6, a^7b^22, No], [1, 16, 5, a^6b^22, No], [1, 17, 4, a^5b^22, No], [1, 18, 3, a^4b^22, No], [1, 19, 2, a^3b^22, No], [1, 20, 1, a^2b^22, No], [1, 21, 0, a^1b^22, No], [2, 1, 19, a^21b^22, No], [2, 2, 18, a^20b^22, No], [2, 3, 17, a^19b^22, No], [2, 4, 16, a^18b^22, No], [2, 5, 15, a^17b^22, No], [2, 6, 14, a^16b^22, No], [2, 7, 13, a^15b^22, No], [2, 8, 12, a^14b^22, No], [2, 9, 11, a^13b^22, No], [2, 10, 10, a^12b^22, No], [2, 11, 9, a^11b^22, No], [2, 12, 8, a^10b^22, No], [2, 13, 7, a^9b^22, No], [2, 14, 6, a^8b^22, No], [2, 15, 5, a^7b^22, No], [2, 16, 4, a^6b^22, No], [2, 17, 3, a^5b^22, No], [2, 18, 2, a^4b^22, No], [2, 19, 1, a^3b^22, No], [2, 20, 0, a^2b^22, No], [3, 1, 18, a^21b^22, No], [3, 2, 17, a^20b^22, No], [3, 3, 16, a^19b^22, No], [3, 4, 15, a^18b^22, No], [3, 5, 14, a^17b^22, No], [3, 6, 13, a^16b^22, No], [3, 7, 12, a^15b^22, No], [3, 8, 11, a^14b^22, No], [3, 9, 10, a^13b^22, No], [3, 10, 9, a^12b^22, No], [3, 11, 8, a^11b^22, No], [3, 12, 7, a^10b^22, No], [3, 13, 6, a^9b^22, No], [3, 14, 5, a^8b^22, No], [3, 15, 4, a^7b^22, No], [3, 16, 3, a^6b^22, No], [3, 17, 2, a^5b^22, No], [3, 18, 1, a^4b^22, No], [3, 19, 0, a^3b^22, No], [4, 1, 17, a^21b^22, No], [4, 2, 16, a^20b^22, No], [4, 3, 15, a^19b^22, No], [4, 4, 14, a^18b^22, No], [4, 5, 13, a^17b^22, No], [4, 6, 12, a^16b^22, No], [4, 7, 11, a^15b^22, No], [4, 8, 10, a^14b^22, No], [4, 9, 9, a^13b^22, No], [4, 10, 8, a^12b^22, No], [4, 11, 7, a^11b^22, No], [4, 12, 6, a^10b^22, No], [4, 13, 5, a^9b^22, No], [4, 14, 4, a^8b^22, No], [4, 15, 3, a^7b^22, No], [4, 16, 2, a^6b^22, No], [4, 17, 1, a^5b^22, No], [4, 18, 0, a^4b^22, No], [5, 1, 16, a^21b^22, No], [5, 2, 15, a^20b^22, No], [5, 3, 14, a^19b^22, No], [5, 4, 13, a^18b^22, No], [5, 5, 12, a^17b^22, No], [5, 6, 11, a^16b^22, No], [5, 7, 10, a^15b^22, No], [5, 8, 9, a^14b^22, No], [5, 9, 8, a^13b^22, No], [5, 10, 7, a^12b^22, No], [5, 11, 6, a^11b^22, No], [5, 12, 5, a^10b^22, No], [5, 13, 4, a^9b^22, No], [5, 14, 3, a^8b^22, No], [5, 15, 2, a^7b^22, No], [5, 16, 1, a^6b^22, No], [5, 17, 0, a^5b^22, No], [6, 1, 15, a^21b^22, No], [6, 2, 14, a^20b^22, No], [6, 3, 13, a^19b^22, No], [6, 4, 12, a^18b^22, No], [6, 5, 11, a^17b^22, No], [6, 6, 10, a^16b^22, No], [6, 7, 9, a^15b^22, No], [6, 8, 8, a^14b^22, No], [6, 9, 7, a^13b^22, No], [6, 10, 6, a^12b^22, No], [6, 11, 5, a^11b^22, No], [6, 12, 4, a^10b^22, No], [6, 13, 3, a^9b^22, No], [6, 14, 2, a^8b^22, No], [6, 15, 1, a^7b^22, No], [6, 16, 0, a^6b^22, No], [7, 1, 14, a^21b^22, No], [7, 2, 13, a^20b^22, No], [7, 3, 12, a^19b^22, No], [7, 4, 11, a^18b^22, No], [7, 5, 10, a^17b^22, No], [7, 6, 9, a^16b^22, No], [7, 7, 8, a^15b^22, No], [7, 8, 7, a^14b^22, No], [7, 9, 6, a^13b^22, No], [7, 10, 5, a^12b^22, No], [7, 11, 4, a^11b^22, No], [7, 12, 3, a^10b^22, No], [7, 13, 2, a^9b^22, No], [7, 14, 1, a^8b^22, No], [7, 15, 0, a^7b^22, No], [8, 1, 13, a^21b^22, No], [8, 2, 12, a^20b^22, No], [8, 3, 11, a^19b^22, No], [8, 4, 10, a^18b^22, No], [8, 5, 9, a^17b^22, No], [8, 6, 8, a^16b^22, No], [8, 7, 7, a^15b^22, No], [8, 8, 6, a^14b^22, No], [8, 9, 5, a^13b^22, No], [8, 10, 4, a^12b^22, No], [8, 11, 3, a^11b^22, No], [8, 12, 2, a^10b^22, No], [8, 13, 1, a^9b^22, No], [8, 14, 0, a^8b^22, No], [9, 1, 12, a^21b^22, No], [9, 2, 11, a^20b^22, No], [9, 3, 10, a^19b^22, No], [9, 4, 9, a^18b^22, No], [9, 5, 8, a^17b^22, No], [9, 6, 7, a^16b^22, No], [9, 7, 6, a^15b^22, No], [9, 8, 5, a^14b^22, No], [9, 9, 4, a^13b^22, No], [9, 10, 3, a^12b^22, No], [9, 11, 2, a^11b^22, No], [9, 12, 1, a^10b^22, No], [9, 13, 0, a^9b^22, No], [10, 1, 11, a^21b^22, No], [10, 2, 10, a^20b^22, No], [10, 3, 9, a^19b^22, No], [10, 4, 8, a^18b^22, No], [10, 5, 7, a^17b^22, No], [10, 6, 6, a^16b^22, No], [10, 7, 5, a^15b^22, No], [10, 8, 4, a^14b^22, No], [10, 9, 3, a^13b^22, No], [10, 10, 2, a^12b^22, No], [10, 11, 1, a^11b^22, No], [10, 12, 0, a^10b^22, No], [11, 1, 10, a^21b^22, No], [11, 2, 9, a^20b^22, No], [11, 3, 8, a^19b^22, No], [11, 4, 7, a^18b^22, No], [11, 5, 6, a^17b^22, No], [11, 6, 5, a^16b^22, No], [11, 7, 4, a^15b^22, No], [11, 8, 3, a^14b^22, No], [11, 9, 2, a^13b^22, No], [11, 10, 1, a^12b^22, No], [11, 11, 0, a^11b^22, No], [12, 1, 9, a^21b^22, No], [12, 2, 8, a^20b^22, No], [12, 3, 7, a^19b^22, No], [12, 4, 6, a^18b^22, No], [12, 5, 5, a^17b^22, No], [12, 6, 4, a^16b^22, No], [12, 7, 3, a^15b^22, No], [12, 8, 2, a^14b^22, No], [12, 9, 1, a^13b^22, No], [12, 10, 0, a^12b^22, No], [13, 1, 8, a^21b^22, No], [13, 2, 7, a^20b^22, No], [13, 3, 6, a^19b^22, No], [13, 4, 5, a^18b^22, No], [13, 5, 4, a^17b^22, No], [13, 6, 3, a^16b^22, No], [13, 7, 2, a^15b^22, No], [13, 8, 1, a^14b^22, No], [13, 9, 0, a^13b^22, No], [14, 1, 7, a^21b^22, No], [14, 2, 6, a^20b^22, No], [14, 3, 5, a^19b^22, No], [14, 4, 4, a^18b^22, No], [14, 5, 3, a^17b^22, No], [14, 6, 2, a^16b^22, No], [14, 7, 1, a^15b^22, No], [14, 8, 0, a^14b^22, No], [15, 1, 6, a^21b^22, No], [15, 2, 5, a^20b^22, No], [15, 3, 4, a^19b^22, No], [15, 4, 3, a^18b^22, No], [15, 5, 2, a^17b^22, No], [15, 6, 1, a^16b^22, No], [15, 7, 0, a^15b^22, No], [16, 1, 5, a^21b^22, No], [16, 2, 4, a^20b^22, No], [16, 3, 3, a^19b^22, No], [16, 4, 2, a^18b^22, No], [16, 5, 1, a^17b^22, No], [16, 6, 0, a^16b^22, No], [17, 1, 4, a^21b^22, No], [17, 2, 3, a^20b^22, No], [17, 3, 2, a^19b^22, No], [17, 4, 1, a^18b^22, No], [17, 5, 0, a^17b^22, No], [18, 1, 3, a^21b^22, No], [18, 2, 2, a^20b^22, No], [18, 3, 1, a^19b^22, No], [18, 4, 0, a^18b^22, No], [19, 1, 2, a^21b^22, No], [19, 2, 1, a^20b^22, No], [19, 3, 0, a^19b^22, No], [20, 1, 1, a^21b^22, No], [20, 2, 0, a^20b^22, No], [21, 1, 0, a^21b^22, No]], 253, 253, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (22, a^22b^22, [[0, 1, 21, a^21b^22, No], [0, 2, 20, a^20b^22, No], [0, 3, 19, a^19b^22, No], [0, 4, 18, a^18b^22, No], [0, 5, 17, a^17b^22, No], [0, 6, 16, a^16b^22, No], [0, 7, 15, a^15b^22, No], [0, 8, 14, a^14b^22, No], [0, 9, 13, a^13b^22, No], [0, 10, 12, a^12b^22, No], [0, 11, 11, a^11b^22, No], [0, 12, 10, a^10b^22, No], [0, 13, 9, a^9b^22, No], [0, 14, 8, a^8b^22, No], [0, 15, 7, a^7b^22, No], [0, 16, 6, a^6b^22, No], [0, 17, 5, a^5b^22, No], [0, 18, 4, a^4b^22, No], [0, 19, 3, a^3b^22, No], [0, 20, 2, a^2b^22, No], [0, 21, 1, a^1b^22, No], [0, 22, 0, a^0b^22, No], [1, 1, 20, a^21b^22, No], [1, 2, 19, a^20b^22, No], [1, 3, 18, a^19b^22, No], [1, 4, 17, a^18b^22, No], [1, 5, 16, a^17b^22, No], [1, 6, 15, a^16b^22, No], [1, 7, 14, a^15b^22, No], [1, 8, 13, a^14b^22, No], [1, 9, 12, a^13b^22, No], [1, 10, 11, a^12b^22, No], [1, 11, 10, a^11b^22, No], [1, 12, 9, a^10b^22, No], [1, 13, 8, a^9b^22, No], [1, 14, 7, a^8b^22, No], [1, 15, 6, a^7b^22, No], [1, 16, 5, a^6b^22, No], [1, 17, 4, a^5b^22, No], [1, 18, 3, a^4b^22, No], [1, 19, 2, a^3b^22, No], [1, 20, 1, a^2b^22, No], [1, 21, 0, a^1b^22, No], [2, 1, 19, a^21b^22, No], [2, 2, 18, a^20b^22, No], [2, 3, 17, a^19b^22, No], [2, 4, 16, a^18b^22, No], [2, 5, 15, a^17b^22, No], [2, 6, 14, a^16b^22, No], [2, 7, 13, a^15b^22, No], [2, 8, 12, a^14b^22, No], [2, 9, 11, a^13b^22, No], [2, 10, 10, a^12b^22, No], [2, 11, 9, a^11b^22, No], [2, 12, 8, a^10b^22, No], [2, 13, 7, a^9b^22, No], [2, 14, 6, a^8b^22, No], [2, 15, 5, a^7b^22, No], [2, 16, 4, a^6b^22, No], [2, 17, 3, a^5b^22, No], [2, 18, 2, a^4b^22, No], [2, 19, 1, a^3b^22, No], [2, 20, 0, a^2b^22, No], [3, 1, 18, a^21b^22, No], [3, 2, 17, a^20b^22, No], [3, 3, 16, a^19b^22, No], [3, 4, 15, a^18b^22, No], [3, 5, 14, a^17b^22, No], [3, 6, 13, a^16b^22, No], [3, 7, 12, a^15b^22, No], [3, 8, 11, a^14b^22, No], [3, 9, 10, a^13b^22, No], [3, 10, 9, a^12b^22, No], [3, 11, 8, a^11b^22, No], [3, 12, 7, a^10b^22, No], [3, 13, 6, a^9b^22, No], [3, 14, 5, a^8b^22, No], [3, 15, 4, a^7b^22, No], [3, 16, 3, a^6b^22, No], [3, 17, 2, a^5b^22, No], [3, 18, 1, a^4b^22, No], [3, 19, 0, a^3b^22, No], [4, 1, 17, a^21b^22, No], [4, 2, 16, a^20b^22, No], [4, 3, 15, a^19b^22, No], [4, 4, 14, a^18b^22, No], [4, 5, 13, a^17b^22, No], [4, 6, 12, a^16b^22, No], [4, 7, 11, a^15b^22, No], [4, 8, 10, a^14b^22, No], [4, 9, 9, a^13b^22, No], [4, 10, 8, a^12b^22, No], [4, 11, 7, a^11b^22, No], [4, 12, 6, a^10b^22, No], [4, 13, 5, a^9b^22, No], [4, 14, 4, a^8b^22, No], [4, 15, 3, a^7b^22, No], [4, 16, 2, a^6b^22, No], [4, 17, 1, a^5b^22, No], [4, 18, 0, a^4b^22, No], [5, 1, 16, a^21b^22, No], [5, 2, 15, a^20b^22, No], [5, 3, 14, a^19b^22, No], [5, 4, 13, a^18b^22, No], [5, 5, 12, a^17b^22, No], [5, 6, 11, a^16b^22, No], [5, 7, 10, a^15b^22, No], [5, 8, 9, a^14b^22, No], [5, 9, 8, a^13b^22, No], [5, 10, 7, a^12b^22, No], [5, 11, 6, a^11b^22, No], [5, 12, 5, a^10b^22, No], [5, 13, 4, a^9b^22, No], [5, 14, 3, a^8b^22, No], [5, 15, 2, a^7b^22, No], [5, 16, 1, a^6b^22, No], [5, 17, 0, a^5b^22, No], [6, 1, 15, a^21b^22, No], [6, 2, 14, a^20b^22, No], [6, 3, 13, a^19b^22, No], [6, 4, 12, a^18b^22, No], [6, 5, 11, a^17b^22, No], [6, 6, 10, a^16b^22, No], [6, 7, 9, a^15b^22, No], [6, 8, 8, a^14b^22, No], [6, 9, 7, a^13b^22, No], [6, 10, 6, a^12b^22, No], [6, 11, 5, a^11b^22, No], [6, 12, 4, a^10b^22, No], [6, 13, 3, a^9b^22, No], [6, 14, 2, a^8b^22, No], [6, 15, 1, a^7b^22, No], [6, 16, 0, a^6b^22, No], [7, 1, 14, a^21b^22, No], [7, 2, 13, a^20b^22, No], [7, 3, 12, a^19b^22, No], [7, 4, 11, a^18b^22, No], [7, 5, 10, a^17b^22, No], [7, 6, 9, a^16b^22, No], [7, 7, 8, a^15b^22, No], [7, 8, 7, a^14b^22, No], [7, 9, 6, a^13b^22, No], [7, 10, 5, a^12b^22, No], [7, 11, 4, a^11b^22, No], [7, 12, 3, a^10b^22, No], [7, 13, 2, a^9b^22, No], [7, 14, 1, a^8b^22, No], [7, 15, 0, a^7b^22, No], [8, 1, 13, a^21b^22, No], [8, 2, 12, a^20b^22, No], [8, 3, 11, a^19b^22, No], [8, 4, 10, a^18b^22, No], [8, 5, 9, a^17b^22, No], [8, 6, 8, a^16b^22, No], [8, 7, 7, a^15b^22, No], [8, 8, 6, a^14b^22, No], [8, 9, 5, a^13b^22, No], [8, 10, 4, a^12b^22, No], [8, 11, 3, a^11b^22, No], [8, 12, 2, a^10b^22, No], [8, 13, 1, a^9b^22, No], [8, 14, 0, a^8b^22, No], [9, 1, 12, a^21b^22, No], [9, 2, 11, a^20b^22, No], [9, 3, 10, a^19b^22, No], [9, 4, 9, a^18b^22, No], [9, 5, 8, a^17b^22, No], [9, 6, 7, a^16b^22, No], [9, 7, 6, a^15b^22, No], [9, 8, 5, a^14b^22, No], [9, 9, 4, a^13b^22, No], [9, 10, 3, a^12b^22, No], [9, 11, 2, a^11b^22, No], [9, 12, 1, a^10b^22, No], [9, 13, 0, a^9b^22, No], [10, 1, 11, a^21b^22, No], [10, 2, 10, a^20b^22, No], [10, 3, 9, a^19b^22, No], [10, 4, 8, a^18b^22, No], [10, 5, 7, a^17b^22, No], [10, 6, 6, a^16b^22, No], [10, 7, 5, a^15b^22, No], [10, 8, 4, a^14b^22, No], [10, 9, 3, a^13b^22, No], [10, 10, 2, a^12b^22, No], [10, 11, 1, a^11b^22, No], [10, 12, 0, a^10b^22, No], [11, 1, 10, a^21b^22, No], [11, 2, 9, a^20b^22, No], [11, 3, 8, a^19b^22, No], [11, 4, 7, a^18b^22, No], [11, 5, 6, a^17b^22, No], [11, 6, 5, a^16b^22, No], [11, 7, 4, a^15b^22, No], [11, 8, 3, a^14b^22, No], [11, 9, 2, a^13b^22, No], [11, 10, 1, a^12b^22, No], [11, 11, 0, a^11b^22, No], [12, 1, 9, a^21b^22, No], [12, 2, 8, a^20b^22, No], [12, 3, 7, a^19b^22, No], [12, 4, 6, a^18b^22, No], [12, 5, 5, a^17b^22, No], [12, 6, 4, a^16b^22, No], [12, 7, 3, a^15b^22, No], [12, 8, 2, a^14b^22, No], [12, 9, 1, a^13b^22, No], [12, 10, 0, a^12b^22, No], [13, 1, 8, a^21b^22, No], [13, 2, 7, a^20b^22, No], [13, 3, 6, a^19b^22, No], [13, 4, 5, a^18b^22, No], [13, 5, 4, a^17b^22, No], [13, 6, 3, a^16b^22, No], [13, 7, 2, a^15b^22, No], [13, 8, 1, a^14b^22, No], [13, 9, 0, a^13b^22, No], [14, 1, 7, a^21b^22, No], [14, 2, 6, a^20b^22, No], [14, 3, 5, a^19b^22, No], [14, 4, 4, a^18b^22, No], [14, 5, 3, a^17b^22, No], [14, 6, 2, a^16b^22, No], [14, 7, 1, a^15b^22, No], [14, 8, 0, a^14b^22, No], [15, 1, 6, a^21b^22, No], [15, 2, 5, a^20b^22, No], [15, 3, 4, a^19b^22, No], [15, 4, 3, a^18b^22, No], [15, 5, 2, a^17b^22, No], [15, 6, 1, a^16b^22, No], [15, 7, 0, a^15b^22, No], [16, 1, 5, a^21b^22, No], [16, 2, 4, a^20b^22, No], [16, 3, 3, a^19b^22, No], [16, 4, 2, a^18b^22, No], [16, 5, 1, a^17b^22, No], [16, 6, 0, a^16b^22, No], [17, 1, 4, a^21b^22, No], [17, 2, 3, a^20b^22, No], [17, 3, 2, a^19b^22, No], [17, 4, 1, a^18b^22, No], [17, 5, 0, a^17b^22, No], [18, 1, 3, a^21b^22, No], [18, 2, 2, a^20b^22, No], [18, 3, 1, a^19b^22, No], [18, 4, 0, a^18b^22, No], [19, 1, 2, a^21b^22, No], [19, 2, 1, a^20b^22, No], [19, 3, 0, a^19b^22, No], [20, 1, 1, a^21b^22, No], [20, 2, 0, a^20b^22, No], [21, 1, 0, a^21b^22, No]], 253, 253, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

73. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=23 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(23, a^23b^23, [[0, 1, 22, a^22b^23, No], [0, 2, 21, a^21b^23, No], [0, 3, 20, a^20b^23, No], [0, 4, 19, a^19b^23, No], [0, 5, 18, a^18b^23, No], [0, 6, 17, a^17b^23, No], [0, 7, 16, a^16b^23, No], [0, 8, 15, a^15b^23, No], [0, 9, 14, a^14b^23, No], [0, 10, 13, a^13b^23, No], [0, 11, 12, a^12b^23, No], [0, 12, 11, a^11b^23, No], [0, 13, 10, a^10b^23, No], [0, 14, 9, a^9b^23, No], [0, 15, 8, a^8b^23, No], [0, 16, 7, a^7b^23, No], [0, 17, 6, a^6b^23, No], [0, 18, 5, a^5b^23, No], [0, 19, 4, a^4b^23, No], [0, 20, 3, a^3b^23, No], [0, 21, 2, a^2b^23, No], [0, 22, 1, a^1b^23, No], [0, 23, 0, a^0b^23, No], [1, 1, 21, a^22b^23, No], [1, 2, 20, a^21b^23, No], [1, 3, 19, a^20b^23, No], [1, 4, 18, a^19b^23, No], [1, 5, 17, a^18b^23, No], [1, 6, 16, a^17b^23, No], [1, 7, 15, a^16b^23, No], [1, 8, 14, a^15b^23, No], [1, 9, 13, a^14b^23, No], [1, 10, 12, a^13b^23, No], [1, 11, 11, a^12b^23, No], [1, 12, 10, a^11b^23, No], [1, 13, 9, a^10b^23, No], [1, 14, 8, a^9b^23, No], [1, 15, 7, a^8b^23, No], [1, 16, 6, a^7b^23, No], [1, 17, 5, a^6b^23, No], [1, 18, 4, a^5b^23, No], [1, 19, 3, a^4b^23, No], [1, 20, 2, a^3b^23, No], [1, 21, 1, a^2b^23, No], [1, 22, 0, a^1b^23, No], [2, 1, 20, a^22b^23, No], [2, 2, 19, a^21b^23, No], [2, 3, 18, a^20b^23, No], [2, 4, 17, a^19b^23, No], [2, 5, 16, a^18b^23, No], [2, 6, 15, a^17b^23, No], [2, 7, 14, a^16b^23, No], [2, 8, 13, a^15b^23, No], [2, 9, 12, a^14b^23, No], [2, 10, 11, a^13b^23, No], [2, 11, 10, a^12b^23, No], [2, 12, 9, a^11b^23, No], [2, 13, 8, a^10b^23, No], [2, 14, 7, a^9b^23, No], [2, 15, 6, a^8b^23, No], [2, 16, 5, a^7b^23, No], [2, 17, 4, a^6b^23, No], [2, 18, 3, a^5b^23, No], [2, 19, 2, a^4b^23, No], [2, 20, 1, a^3b^23, No], [2, 21, 0, a^2b^23, No], [3, 1, 19, a^22b^23, No], [3, 2, 18, a^21b^23, No], [3, 3, 17, a^20b^23, No], [3, 4, 16, a^19b^23, No], [3, 5, 15, a^18b^23, No], [3, 6, 14, a^17b^23, No], [3, 7, 13, a^16b^23, No], [3, 8, 12, a^15b^23, No], [3, 9, 11, a^14b^23, No], [3, 10, 10, a^13b^23, No], [3, 11, 9, a^12b^23, No], [3, 12, 8, a^11b^23, No], [3, 13, 7, a^10b^23, No], [3, 14, 6, a^9b^23, No], [3, 15, 5, a^8b^23, No], [3, 16, 4, a^7b^23, No], [3, 17, 3, a^6b^23, No], [3, 18, 2, a^5b^23, No], [3, 19, 1, a^4b^23, No], [3, 20, 0, a^3b^23, No], [4, 1, 18, a^22b^23, No], [4, 2, 17, a^21b^23, No], [4, 3, 16, a^20b^23, No], [4, 4, 15, a^19b^23, No], [4, 5, 14, a^18b^23, No], [4, 6, 13, a^17b^23, No], [4, 7, 12, a^16b^23, No], [4, 8, 11, a^15b^23, No], [4, 9, 10, a^14b^23, No], [4, 10, 9, a^13b^23, No], [4, 11, 8, a^12b^23, No], [4, 12, 7, a^11b^23, No], [4, 13, 6, a^10b^23, No], [4, 14, 5, a^9b^23, No], [4, 15, 4, a^8b^23, No], [4, 16, 3, a^7b^23, No], [4, 17, 2, a^6b^23, No], [4, 18, 1, a^5b^23, No], [4, 19, 0, a^4b^23, No], [5, 1, 17, a^22b^23, No], [5, 2, 16, a^21b^23, No], [5, 3, 15, a^20b^23, No], [5, 4, 14, a^19b^23, No], [5, 5, 13, a^18b^23, No], [5, 6, 12, a^17b^23, No], [5, 7, 11, a^16b^23, No], [5, 8, 10, a^15b^23, No], [5, 9, 9, a^14b^23, No], [5, 10, 8, a^13b^23, No], [5, 11, 7, a^12b^23, No], [5, 12, 6, a^11b^23, No], [5, 13, 5, a^10b^23, No], [5, 14, 4, a^9b^23, No], [5, 15, 3, a^8b^23, No], [5, 16, 2, a^7b^23, No], [5, 17, 1, a^6b^23, No], [5, 18, 0, a^5b^23, No], [6, 1, 16, a^22b^23, No], [6, 2, 15, a^21b^23, No], [6, 3, 14, a^20b^23, No], [6, 4, 13, a^19b^23, No], [6, 5, 12, a^18b^23, No], [6, 6, 11, a^17b^23, No], [6, 7, 10, a^16b^23, No], [6, 8, 9, a^15b^23, No], [6, 9, 8, a^14b^23, No], [6, 10, 7, a^13b^23, No], [6, 11, 6, a^12b^23, No], [6, 12, 5, a^11b^23, No], [6, 13, 4, a^10b^23, No], [6, 14, 3, a^9b^23, No], [6, 15, 2, a^8b^23, No], [6, 16, 1, a^7b^23, No], [6, 17, 0, a^6b^23, No], [7, 1, 15, a^22b^23, No], [7, 2, 14, a^21b^23, No], [7, 3, 13, a^20b^23, No], [7, 4, 12, a^19b^23, No], [7, 5, 11, a^18b^23, No], [7, 6, 10, a^17b^23, No], [7, 7, 9, a^16b^23, No], [7, 8, 8, a^15b^23, No], [7, 9, 7, a^14b^23, No], [7, 10, 6, a^13b^23, No], [7, 11, 5, a^12b^23, No], [7, 12, 4, a^11b^23, No], [7, 13, 3, a^10b^23, No], [7, 14, 2, a^9b^23, No], [7, 15, 1, a^8b^23, No], [7, 16, 0, a^7b^23, No], [8, 1, 14, a^22b^23, No], [8, 2, 13, a^21b^23, No], [8, 3, 12, a^20b^23, No], [8, 4, 11, a^19b^23, No], [8, 5, 10, a^18b^23, No], [8, 6, 9, a^17b^23, No], [8, 7, 8, a^16b^23, No], [8, 8, 7, a^15b^23, No], [8, 9, 6, a^14b^23, No], [8, 10, 5, a^13b^23, No], [8, 11, 4, a^12b^23, No], [8, 12, 3, a^11b^23, No], [8, 13, 2, a^10b^23, No], [8, 14, 1, a^9b^23, No], [8, 15, 0, a^8b^23, No], [9, 1, 13, a^22b^23, No], [9, 2, 12, a^21b^23, No], [9, 3, 11, a^20b^23, No], [9, 4, 10, a^19b^23, No], [9, 5, 9, a^18b^23, No], [9, 6, 8, a^17b^23, No], [9, 7, 7, a^16b^23, No], [9, 8, 6, a^15b^23, No], [9, 9, 5, a^14b^23, No], [9, 10, 4, a^13b^23, No], [9, 11, 3, a^12b^23, No], [9, 12, 2, a^11b^23, No], [9, 13, 1, a^10b^23, No], [9, 14, 0, a^9b^23, No], [10, 1, 12, a^22b^23, No], [10, 2, 11, a^21b^23, No], [10, 3, 10, a^20b^23, No], [10, 4, 9, a^19b^23, No], [10, 5, 8, a^18b^23, No], [10, 6, 7, a^17b^23, No], [10, 7, 6, a^16b^23, No], [10, 8, 5, a^15b^23, No], [10, 9, 4, a^14b^23, No], [10, 10, 3, a^13b^23, No], [10, 11, 2, a^12b^23, No], [10, 12, 1, a^11b^23, No], [10, 13, 0, a^10b^23, No], [11, 1, 11, a^22b^23, No], [11, 2, 10, a^21b^23, No], [11, 3, 9, a^20b^23, No], [11, 4, 8, a^19b^23, No], [11, 5, 7, a^18b^23, No], [11, 6, 6, a^17b^23, No], [11, 7, 5, a^16b^23, No], [11, 8, 4, a^15b^23, No], [11, 9, 3, a^14b^23, No], [11, 10, 2, a^13b^23, No], [11, 11, 1, a^12b^23, No], [11, 12, 0, a^11b^23, No], [12, 1, 10, a^22b^23, No], [12, 2, 9, a^21b^23, No], [12, 3, 8, a^20b^23, No], [12, 4, 7, a^19b^23, No], [12, 5, 6, a^18b^23, No], [12, 6, 5, a^17b^23, No], [12, 7, 4, a^16b^23, No], [12, 8, 3, a^15b^23, No], [12, 9, 2, a^14b^23, No], [12, 10, 1, a^13b^23, No], [12, 11, 0, a^12b^23, No], [13, 1, 9, a^22b^23, No], [13, 2, 8, a^21b^23, No], [13, 3, 7, a^20b^23, No], [13, 4, 6, a^19b^23, No], [13, 5, 5, a^18b^23, No], [13, 6, 4, a^17b^23, No], [13, 7, 3, a^16b^23, No], [13, 8, 2, a^15b^23, No], [13, 9, 1, a^14b^23, No], [13, 10, 0, a^13b^23, No], [14, 1, 8, a^22b^23, No], [14, 2, 7, a^21b^23, No], [14, 3, 6, a^20b^23, No], [14, 4, 5, a^19b^23, No], [14, 5, 4, a^18b^23, No], [14, 6, 3, a^17b^23, No], [14, 7, 2, a^16b^23, No], [14, 8, 1, a^15b^23, No], [14, 9, 0, a^14b^23, No], [15, 1, 7, a^22b^23, No], [15, 2, 6, a^21b^23, No], [15, 3, 5, a^20b^23, No], [15, 4, 4, a^19b^23, No], [15, 5, 3, a^18b^23, No], [15, 6, 2, a^17b^23, No], [15, 7, 1, a^16b^23, No], [15, 8, 0, a^15b^23, No], [16, 1, 6, a^22b^23, No], [16, 2, 5, a^21b^23, No], [16, 3, 4, a^20b^23, No], [16, 4, 3, a^19b^23, No], [16, 5, 2, a^18b^23, No], [16, 6, 1, a^17b^23, No], [16, 7, 0, a^16b^23, No], [17, 1, 5, a^22b^23, No], [17, 2, 4, a^21b^23, No], [17, 3, 3, a^20b^23, No], [17, 4, 2, a^19b^23, No], [17, 5, 1, a^18b^23, No], [17, 6, 0, a^17b^23, No], [18, 1, 4, a^22b^23, No], [18, 2, 3, a^21b^23, No], [18, 3, 2, a^20b^23, No], [18, 4, 1, a^19b^23, No], [18, 5, 0, a^18b^23, No], [19, 1, 3, a^22b^23, No], [19, 2, 2, a^21b^23, No], [19, 3, 1, a^20b^23, No], [19, 4, 0, a^19b^23, No], [20, 1, 2, a^22b^23, No], [20, 2, 1, a^21b^23, No], [20, 3, 0, a^20b^23, No], [21, 1, 1, a^22b^23, No], [21, 2, 0, a^21b^23, No], [22, 1, 0, a^22b^23, No]], 276, 276, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (23, a^23b^23, [[0, 1, 22, a^22b^23, No], [0, 2, 21, a^21b^23, No], [0, 3, 20, a^20b^23, No], [0, 4, 19, a^19b^23, No], [0, 5, 18, a^18b^23, No], [0, 6, 17, a^17b^23, No], [0, 7, 16, a^16b^23, No], [0, 8, 15, a^15b^23, No], [0, 9, 14, a^14b^23, No], [0, 10, 13, a^13b^23, No], [0, 11, 12, a^12b^23, No], [0, 12, 11, a^11b^23, No], [0, 13, 10, a^10b^23, No], [0, 14, 9, a^9b^23, No], [0, 15, 8, a^8b^23, No], [0, 16, 7, a^7b^23, No], [0, 17, 6, a^6b^23, No], [0, 18, 5, a^5b^23, No], [0, 19, 4, a^4b^23, No], [0, 20, 3, a^3b^23, No], [0, 21, 2, a^2b^23, No], [0, 22, 1, a^1b^23, No], [0, 23, 0, a^0b^23, No], [1, 1, 21, a^22b^23, No], [1, 2, 20, a^21b^23, No], [1, 3, 19, a^20b^23, No], [1, 4, 18, a^19b^23, No], [1, 5, 17, a^18b^23, No], [1, 6, 16, a^17b^23, No], [1, 7, 15, a^16b^23, No], [1, 8, 14, a^15b^23, No], [1, 9, 13, a^14b^23, No], [1, 10, 12, a^13b^23, No], [1, 11, 11, a^12b^23, No], [1, 12, 10, a^11b^23, No], [1, 13, 9, a^10b^23, No], [1, 14, 8, a^9b^23, No], [1, 15, 7, a^8b^23, No], [1, 16, 6, a^7b^23, No], [1, 17, 5, a^6b^23, No], [1, 18, 4, a^5b^23, No], [1, 19, 3, a^4b^23, No], [1, 20, 2, a^3b^23, No], [1, 21, 1, a^2b^23, No], [1, 22, 0, a^1b^23, No], [2, 1, 20, a^22b^23, No], [2, 2, 19, a^21b^23, No], [2, 3, 18, a^20b^23, No], [2, 4, 17, a^19b^23, No], [2, 5, 16, a^18b^23, No], [2, 6, 15, a^17b^23, No], [2, 7, 14, a^16b^23, No], [2, 8, 13, a^15b^23, No], [2, 9, 12, a^14b^23, No], [2, 10, 11, a^13b^23, No], [2, 11, 10, a^12b^23, No], [2, 12, 9, a^11b^23, No], [2, 13, 8, a^10b^23, No], [2, 14, 7, a^9b^23, No], [2, 15, 6, a^8b^23, No], [2, 16, 5, a^7b^23, No], [2, 17, 4, a^6b^23, No], [2, 18, 3, a^5b^23, No], [2, 19, 2, a^4b^23, No], [2, 20, 1, a^3b^23, No], [2, 21, 0, a^2b^23, No], [3, 1, 19, a^22b^23, No], [3, 2, 18, a^21b^23, No], [3, 3, 17, a^20b^23, No], [3, 4, 16, a^19b^23, No], [3, 5, 15, a^18b^23, No], [3, 6, 14, a^17b^23, No], [3, 7, 13, a^16b^23, No], [3, 8, 12, a^15b^23, No], [3, 9, 11, a^14b^23, No], [3, 10, 10, a^13b^23, No], [3, 11, 9, a^12b^23, No], [3, 12, 8, a^11b^23, No], [3, 13, 7, a^10b^23, No], [3, 14, 6, a^9b^23, No], [3, 15, 5, a^8b^23, No], [3, 16, 4, a^7b^23, No], [3, 17, 3, a^6b^23, No], [3, 18, 2, a^5b^23, No], [3, 19, 1, a^4b^23, No], [3, 20, 0, a^3b^23, No], [4, 1, 18, a^22b^23, No], [4, 2, 17, a^21b^23, No], [4, 3, 16, a^20b^23, No], [4, 4, 15, a^19b^23, No], [4, 5, 14, a^18b^23, No], [4, 6, 13, a^17b^23, No], [4, 7, 12, a^16b^23, No], [4, 8, 11, a^15b^23, No], [4, 9, 10, a^14b^23, No], [4, 10, 9, a^13b^23, No], [4, 11, 8, a^12b^23, No], [4, 12, 7, a^11b^23, No], [4, 13, 6, a^10b^23, No], [4, 14, 5, a^9b^23, No], [4, 15, 4, a^8b^23, No], [4, 16, 3, a^7b^23, No], [4, 17, 2, a^6b^23, No], [4, 18, 1, a^5b^23, No], [4, 19, 0, a^4b^23, No], [5, 1, 17, a^22b^23, No], [5, 2, 16, a^21b^23, No], [5, 3, 15, a^20b^23, No], [5, 4, 14, a^19b^23, No], [5, 5, 13, a^18b^23, No], [5, 6, 12, a^17b^23, No], [5, 7, 11, a^16b^23, No], [5, 8, 10, a^15b^23, No], [5, 9, 9, a^14b^23, No], [5, 10, 8, a^13b^23, No], [5, 11, 7, a^12b^23, No], [5, 12, 6, a^11b^23, No], [5, 13, 5, a^10b^23, No], [5, 14, 4, a^9b^23, No], [5, 15, 3, a^8b^23, No], [5, 16, 2, a^7b^23, No], [5, 17, 1, a^6b^23, No], [5, 18, 0, a^5b^23, No], [6, 1, 16, a^22b^23, No], [6, 2, 15, a^21b^23, No], [6, 3, 14, a^20b^23, No], [6, 4, 13, a^19b^23, No], [6, 5, 12, a^18b^23, No], [6, 6, 11, a^17b^23, No], [6, 7, 10, a^16b^23, No], [6, 8, 9, a^15b^23, No], [6, 9, 8, a^14b^23, No], [6, 10, 7, a^13b^23, No], [6, 11, 6, a^12b^23, No], [6, 12, 5, a^11b^23, No], [6, 13, 4, a^10b^23, No], [6, 14, 3, a^9b^23, No], [6, 15, 2, a^8b^23, No], [6, 16, 1, a^7b^23, No], [6, 17, 0, a^6b^23, No], [7, 1, 15, a^22b^23, No], [7, 2, 14, a^21b^23, No], [7, 3, 13, a^20b^23, No], [7, 4, 12, a^19b^23, No], [7, 5, 11, a^18b^23, No], [7, 6, 10, a^17b^23, No], [7, 7, 9, a^16b^23, No], [7, 8, 8, a^15b^23, No], [7, 9, 7, a^14b^23, No], [7, 10, 6, a^13b^23, No], [7, 11, 5, a^12b^23, No], [7, 12, 4, a^11b^23, No], [7, 13, 3, a^10b^23, No], [7, 14, 2, a^9b^23, No], [7, 15, 1, a^8b^23, No], [7, 16, 0, a^7b^23, No], [8, 1, 14, a^22b^23, No], [8, 2, 13, a^21b^23, No], [8, 3, 12, a^20b^23, No], [8, 4, 11, a^19b^23, No], [8, 5, 10, a^18b^23, No], [8, 6, 9, a^17b^23, No], [8, 7, 8, a^16b^23, No], [8, 8, 7, a^15b^23, No], [8, 9, 6, a^14b^23, No], [8, 10, 5, a^13b^23, No], [8, 11, 4, a^12b^23, No], [8, 12, 3, a^11b^23, No], [8, 13, 2, a^10b^23, No], [8, 14, 1, a^9b^23, No], [8, 15, 0, a^8b^23, No], [9, 1, 13, a^22b^23, No], [9, 2, 12, a^21b^23, No], [9, 3, 11, a^20b^23, No], [9, 4, 10, a^19b^23, No], [9, 5, 9, a^18b^23, No], [9, 6, 8, a^17b^23, No], [9, 7, 7, a^16b^23, No], [9, 8, 6, a^15b^23, No], [9, 9, 5, a^14b^23, No], [9, 10, 4, a^13b^23, No], [9, 11, 3, a^12b^23, No], [9, 12, 2, a^11b^23, No], [9, 13, 1, a^10b^23, No], [9, 14, 0, a^9b^23, No], [10, 1, 12, a^22b^23, No], [10, 2, 11, a^21b^23, No], [10, 3, 10, a^20b^23, No], [10, 4, 9, a^19b^23, No], [10, 5, 8, a^18b^23, No], [10, 6, 7, a^17b^23, No], [10, 7, 6, a^16b^23, No], [10, 8, 5, a^15b^23, No], [10, 9, 4, a^14b^23, No], [10, 10, 3, a^13b^23, No], [10, 11, 2, a^12b^23, No], [10, 12, 1, a^11b^23, No], [10, 13, 0, a^10b^23, No], [11, 1, 11, a^22b^23, No], [11, 2, 10, a^21b^23, No], [11, 3, 9, a^20b^23, No], [11, 4, 8, a^19b^23, No], [11, 5, 7, a^18b^23, No], [11, 6, 6, a^17b^23, No], [11, 7, 5, a^16b^23, No], [11, 8, 4, a^15b^23, No], [11, 9, 3, a^14b^23, No], [11, 10, 2, a^13b^23, No], [11, 11, 1, a^12b^23, No], [11, 12, 0, a^11b^23, No], [12, 1, 10, a^22b^23, No], [12, 2, 9, a^21b^23, No], [12, 3, 8, a^20b^23, No], [12, 4, 7, a^19b^23, No], [12, 5, 6, a^18b^23, No], [12, 6, 5, a^17b^23, No], [12, 7, 4, a^16b^23, No], [12, 8, 3, a^15b^23, No], [12, 9, 2, a^14b^23, No], [12, 10, 1, a^13b^23, No], [12, 11, 0, a^12b^23, No], [13, 1, 9, a^22b^23, No], [13, 2, 8, a^21b^23, No], [13, 3, 7, a^20b^23, No], [13, 4, 6, a^19b^23, No], [13, 5, 5, a^18b^23, No], [13, 6, 4, a^17b^23, No], [13, 7, 3, a^16b^23, No], [13, 8, 2, a^15b^23, No], [13, 9, 1, a^14b^23, No], [13, 10, 0, a^13b^23, No], [14, 1, 8, a^22b^23, No], [14, 2, 7, a^21b^23, No], [14, 3, 6, a^20b^23, No], [14, 4, 5, a^19b^23, No], [14, 5, 4, a^18b^23, No], [14, 6, 3, a^17b^23, No], [14, 7, 2, a^16b^23, No], [14, 8, 1, a^15b^23, No], [14, 9, 0, a^14b^23, No], [15, 1, 7, a^22b^23, No], [15, 2, 6, a^21b^23, No], [15, 3, 5, a^20b^23, No], [15, 4, 4, a^19b^23, No], [15, 5, 3, a^18b^23, No], [15, 6, 2, a^17b^23, No], [15, 7, 1, a^16b^23, No], [15, 8, 0, a^15b^23, No], [16, 1, 6, a^22b^23, No], [16, 2, 5, a^21b^23, No], [16, 3, 4, a^20b^23, No], [16, 4, 3, a^19b^23, No], [16, 5, 2, a^18b^23, No], [16, 6, 1, a^17b^23, No], [16, 7, 0, a^16b^23, No], [17, 1, 5, a^22b^23, No], [17, 2, 4, a^21b^23, No], [17, 3, 3, a^20b^23, No], [17, 4, 2, a^19b^23, No], [17, 5, 1, a^18b^23, No], [17, 6, 0, a^17b^23, No], [18, 1, 4, a^22b^23, No], [18, 2, 3, a^21b^23, No], [18, 3, 2, a^20b^23, No], [18, 4, 1, a^19b^23, No], [18, 5, 0, a^18b^23, No], [19, 1, 3, a^22b^23, No], [19, 2, 2, a^21b^23, No], [19, 3, 1, a^20b^23, No], [19, 4, 0, a^19b^23, No], [20, 1, 2, a^22b^23, No], [20, 2, 1, a^21b^23, No], [20, 3, 0, a^20b^23, No], [21, 1, 1, a^22b^23, No], [21, 2, 0, a^21b^23, No], [22, 1, 0, a^22b^23, No]], 276, 276, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

74. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=24 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(24, a^24b^24, [[0, 1, 23, a^23b^24, No], [0, 2, 22, a^22b^24, No], [0, 3, 21, a^21b^24, No], [0, 4, 20, a^20b^24, No], [0, 5, 19, a^19b^24, No], [0, 6, 18, a^18b^24, No], [0, 7, 17, a^17b^24, No], [0, 8, 16, a^16b^24, No], [0, 9, 15, a^15b^24, No], [0, 10, 14, a^14b^24, No], [0, 11, 13, a^13b^24, No], [0, 12, 12, a^12b^24, No], [0, 13, 11, a^11b^24, No], [0, 14, 10, a^10b^24, No], [0, 15, 9, a^9b^24, No], [0, 16, 8, a^8b^24, No], [0, 17, 7, a^7b^24, No], [0, 18, 6, a^6b^24, No], [0, 19, 5, a^5b^24, No], [0, 20, 4, a^4b^24, No], [0, 21, 3, a^3b^24, No], [0, 22, 2, a^2b^24, No], [0, 23, 1, a^1b^24, No], [0, 24, 0, a^0b^24, No], [1, 1, 22, a^23b^24, No], [1, 2, 21, a^22b^24, No], [1, 3, 20, a^21b^24, No], [1, 4, 19, a^20b^24, No], [1, 5, 18, a^19b^24, No], [1, 6, 17, a^18b^24, No], [1, 7, 16, a^17b^24, No], [1, 8, 15, a^16b^24, No], [1, 9, 14, a^15b^24, No], [1, 10, 13, a^14b^24, No], [1, 11, 12, a^13b^24, No], [1, 12, 11, a^12b^24, No], [1, 13, 10, a^11b^24, No], [1, 14, 9, a^10b^24, No], [1, 15, 8, a^9b^24, No], [1, 16, 7, a^8b^24, No], [1, 17, 6, a^7b^24, No], [1, 18, 5, a^6b^24, No], [1, 19, 4, a^5b^24, No], [1, 20, 3, a^4b^24, No], [1, 21, 2, a^3b^24, No], [1, 22, 1, a^2b^24, No], [1, 23, 0, a^1b^24, No], [2, 1, 21, a^23b^24, No], [2, 2, 20, a^22b^24, No], [2, 3, 19, a^21b^24, No], [2, 4, 18, a^20b^24, No], [2, 5, 17, a^19b^24, No], [2, 6, 16, a^18b^24, No], [2, 7, 15, a^17b^24, No], [2, 8, 14, a^16b^24, No], [2, 9, 13, a^15b^24, No], [2, 10, 12, a^14b^24, No], [2, 11, 11, a^13b^24, No], [2, 12, 10, a^12b^24, No], [2, 13, 9, a^11b^24, No], [2, 14, 8, a^10b^24, No], [2, 15, 7, a^9b^24, No], [2, 16, 6, a^8b^24, No], [2, 17, 5, a^7b^24, No], [2, 18, 4, a^6b^24, No], [2, 19, 3, a^5b^24, No], [2, 20, 2, a^4b^24, No], [2, 21, 1, a^3b^24, No], [2, 22, 0, a^2b^24, No], [3, 1, 20, a^23b^24, No], [3, 2, 19, a^22b^24, No], [3, 3, 18, a^21b^24, No], [3, 4, 17, a^20b^24, No], [3, 5, 16, a^19b^24, No], [3, 6, 15, a^18b^24, No], [3, 7, 14, a^17b^24, No], [3, 8, 13, a^16b^24, No], [3, 9, 12, a^15b^24, No], [3, 10, 11, a^14b^24, No], [3, 11, 10, a^13b^24, No], [3, 12, 9, a^12b^24, No], [3, 13, 8, a^11b^24, No], [3, 14, 7, a^10b^24, No], [3, 15, 6, a^9b^24, No], [3, 16, 5, a^8b^24, No], [3, 17, 4, a^7b^24, No], [3, 18, 3, a^6b^24, No], [3, 19, 2, a^5b^24, No], [3, 20, 1, a^4b^24, No], [3, 21, 0, a^3b^24, No], [4, 1, 19, a^23b^24, No], [4, 2, 18, a^22b^24, No], [4, 3, 17, a^21b^24, No], [4, 4, 16, a^20b^24, No], [4, 5, 15, a^19b^24, No], [4, 6, 14, a^18b^24, No], [4, 7, 13, a^17b^24, No], [4, 8, 12, a^16b^24, No], [4, 9, 11, a^15b^24, No], [4, 10, 10, a^14b^24, No], [4, 11, 9, a^13b^24, No], [4, 12, 8, a^12b^24, No], [4, 13, 7, a^11b^24, No], [4, 14, 6, a^10b^24, No], [4, 15, 5, a^9b^24, No], [4, 16, 4, a^8b^24, No], [4, 17, 3, a^7b^24, No], [4, 18, 2, a^6b^24, No], [4, 19, 1, a^5b^24, No], [4, 20, 0, a^4b^24, No], [5, 1, 18, a^23b^24, No], [5, 2, 17, a^22b^24, No], [5, 3, 16, a^21b^24, No], [5, 4, 15, a^20b^24, No], [5, 5, 14, a^19b^24, No], [5, 6, 13, a^18b^24, No], [5, 7, 12, a^17b^24, No], [5, 8, 11, a^16b^24, No], [5, 9, 10, a^15b^24, No], [5, 10, 9, a^14b^24, No], [5, 11, 8, a^13b^24, No], [5, 12, 7, a^12b^24, No], [5, 13, 6, a^11b^24, No], [5, 14, 5, a^10b^24, No], [5, 15, 4, a^9b^24, No], [5, 16, 3, a^8b^24, No], [5, 17, 2, a^7b^24, No], [5, 18, 1, a^6b^24, No], [5, 19, 0, a^5b^24, No], [6, 1, 17, a^23b^24, No], [6, 2, 16, a^22b^24, No], [6, 3, 15, a^21b^24, No], [6, 4, 14, a^20b^24, No], [6, 5, 13, a^19b^24, No], [6, 6, 12, a^18b^24, No], [6, 7, 11, a^17b^24, No], [6, 8, 10, a^16b^24, No], [6, 9, 9, a^15b^24, No], [6, 10, 8, a^14b^24, No], [6, 11, 7, a^13b^24, No], [6, 12, 6, a^12b^24, No], [6, 13, 5, a^11b^24, No], [6, 14, 4, a^10b^24, No], [6, 15, 3, a^9b^24, No], [6, 16, 2, a^8b^24, No], [6, 17, 1, a^7b^24, No], [6, 18, 0, a^6b^24, No], [7, 1, 16, a^23b^24, No], [7, 2, 15, a^22b^24, No], [7, 3, 14, a^21b^24, No], [7, 4, 13, a^20b^24, No], [7, 5, 12, a^19b^24, No], [7, 6, 11, a^18b^24, No], [7, 7, 10, a^17b^24, No], [7, 8, 9, a^16b^24, No], [7, 9, 8, a^15b^24, No], [7, 10, 7, a^14b^24, No], [7, 11, 6, a^13b^24, No], [7, 12, 5, a^12b^24, No], [7, 13, 4, a^11b^24, No], [7, 14, 3, a^10b^24, No], [7, 15, 2, a^9b^24, No], [7, 16, 1, a^8b^24, No], [7, 17, 0, a^7b^24, No], [8, 1, 15, a^23b^24, No], [8, 2, 14, a^22b^24, No], [8, 3, 13, a^21b^24, No], [8, 4, 12, a^20b^24, No], [8, 5, 11, a^19b^24, No], [8, 6, 10, a^18b^24, No], [8, 7, 9, a^17b^24, No], [8, 8, 8, a^16b^24, No], [8, 9, 7, a^15b^24, No], [8, 10, 6, a^14b^24, No], [8, 11, 5, a^13b^24, No], [8, 12, 4, a^12b^24, No], [8, 13, 3, a^11b^24, No], [8, 14, 2, a^10b^24, No], [8, 15, 1, a^9b^24, No], [8, 16, 0, a^8b^24, No], [9, 1, 14, a^23b^24, No], [9, 2, 13, a^22b^24, No], [9, 3, 12, a^21b^24, No], [9, 4, 11, a^20b^24, No], [9, 5, 10, a^19b^24, No], [9, 6, 9, a^18b^24, No], [9, 7, 8, a^17b^24, No], [9, 8, 7, a^16b^24, No], [9, 9, 6, a^15b^24, No], [9, 10, 5, a^14b^24, No], [9, 11, 4, a^13b^24, No], [9, 12, 3, a^12b^24, No], [9, 13, 2, a^11b^24, No], [9, 14, 1, a^10b^24, No], [9, 15, 0, a^9b^24, No], [10, 1, 13, a^23b^24, No], [10, 2, 12, a^22b^24, No], [10, 3, 11, a^21b^24, No], [10, 4, 10, a^20b^24, No], [10, 5, 9, a^19b^24, No], [10, 6, 8, a^18b^24, No], [10, 7, 7, a^17b^24, No], [10, 8, 6, a^16b^24, No], [10, 9, 5, a^15b^24, No], [10, 10, 4, a^14b^24, No], [10, 11, 3, a^13b^24, No], [10, 12, 2, a^12b^24, No], [10, 13, 1, a^11b^24, No], [10, 14, 0, a^10b^24, No], [11, 1, 12, a^23b^24, No], [11, 2, 11, a^22b^24, No], [11, 3, 10, a^21b^24, No], [11, 4, 9, a^20b^24, No], [11, 5, 8, a^19b^24, No], [11, 6, 7, a^18b^24, No], [11, 7, 6, a^17b^24, No], [11, 8, 5, a^16b^24, No], [11, 9, 4, a^15b^24, No], [11, 10, 3, a^14b^24, No], [11, 11, 2, a^13b^24, No], [11, 12, 1, a^12b^24, No], [11, 13, 0, a^11b^24, No], [12, 1, 11, a^23b^24, No], [12, 2, 10, a^22b^24, No], [12, 3, 9, a^21b^24, No], [12, 4, 8, a^20b^24, No], [12, 5, 7, a^19b^24, No], [12, 6, 6, a^18b^24, No], [12, 7, 5, a^17b^24, No], [12, 8, 4, a^16b^24, No], [12, 9, 3, a^15b^24, No], [12, 10, 2, a^14b^24, No], [12, 11, 1, a^13b^24, No], [12, 12, 0, a^12b^24, No], [13, 1, 10, a^23b^24, No], [13, 2, 9, a^22b^24, No], [13, 3, 8, a^21b^24, No], [13, 4, 7, a^20b^24, No], [13, 5, 6, a^19b^24, No], [13, 6, 5, a^18b^24, No], [13, 7, 4, a^17b^24, No], [13, 8, 3, a^16b^24, No], [13, 9, 2, a^15b^24, No], [13, 10, 1, a^14b^24, No], [13, 11, 0, a^13b^24, No], [14, 1, 9, a^23b^24, No], [14, 2, 8, a^22b^24, No], [14, 3, 7, a^21b^24, No], [14, 4, 6, a^20b^24, No], [14, 5, 5, a^19b^24, No], [14, 6, 4, a^18b^24, No], [14, 7, 3, a^17b^24, No], [14, 8, 2, a^16b^24, No], [14, 9, 1, a^15b^24, No], [14, 10, 0, a^14b^24, No], [15, 1, 8, a^23b^24, No], [15, 2, 7, a^22b^24, No], [15, 3, 6, a^21b^24, No], [15, 4, 5, a^20b^24, No], [15, 5, 4, a^19b^24, No], [15, 6, 3, a^18b^24, No], [15, 7, 2, a^17b^24, No], [15, 8, 1, a^16b^24, No], [15, 9, 0, a^15b^24, No], [16, 1, 7, a^23b^24, No], [16, 2, 6, a^22b^24, No], [16, 3, 5, a^21b^24, No], [16, 4, 4, a^20b^24, No], [16, 5, 3, a^19b^24, No], [16, 6, 2, a^18b^24, No], [16, 7, 1, a^17b^24, No], [16, 8, 0, a^16b^24, No], [17, 1, 6, a^23b^24, No], [17, 2, 5, a^22b^24, No], [17, 3, 4, a^21b^24, No], [17, 4, 3, a^20b^24, No], [17, 5, 2, a^19b^24, No], [17, 6, 1, a^18b^24, No], [17, 7, 0, a^17b^24, No], [18, 1, 5, a^23b^24, No], [18, 2, 4, a^22b^24, No], [18, 3, 3, a^21b^24, No], [18, 4, 2, a^20b^24, No], [18, 5, 1, a^19b^24, No], [18, 6, 0, a^18b^24, No], [19, 1, 4, a^23b^24, No], [19, 2, 3, a^22b^24, No], [19, 3, 2, a^21b^24, No], [19, 4, 1, a^20b^24, No], [19, 5, 0, a^19b^24, No], [20, 1, 3, a^23b^24, No], [20, 2, 2, a^22b^24, No], [20, 3, 1, a^21b^24, No], [20, 4, 0, a^20b^24, No], [21, 1, 2, a^23b^24, No], [21, 2, 1, a^22b^24, No], [21, 3, 0, a^21b^24, No], [22, 1, 1, a^23b^24, No], [22, 2, 0, a^22b^24, No], [23, 1, 0, a^23b^24, No]], 300, 300, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (24, a^24b^24, [[0, 1, 23, a^23b^24, No], [0, 2, 22, a^22b^24, No], [0, 3, 21, a^21b^24, No], [0, 4, 20, a^20b^24, No], [0, 5, 19, a^19b^24, No], [0, 6, 18, a^18b^24, No], [0, 7, 17, a^17b^24, No], [0, 8, 16, a^16b^24, No], [0, 9, 15, a^15b^24, No], [0, 10, 14, a^14b^24, No], [0, 11, 13, a^13b^24, No], [0, 12, 12, a^12b^24, No], [0, 13, 11, a^11b^24, No], [0, 14, 10, a^10b^24, No], [0, 15, 9, a^9b^24, No], [0, 16, 8, a^8b^24, No], [0, 17, 7, a^7b^24, No], [0, 18, 6, a^6b^24, No], [0, 19, 5, a^5b^24, No], [0, 20, 4, a^4b^24, No], [0, 21, 3, a^3b^24, No], [0, 22, 2, a^2b^24, No], [0, 23, 1, a^1b^24, No], [0, 24, 0, a^0b^24, No], [1, 1, 22, a^23b^24, No], [1, 2, 21, a^22b^24, No], [1, 3, 20, a^21b^24, No], [1, 4, 19, a^20b^24, No], [1, 5, 18, a^19b^24, No], [1, 6, 17, a^18b^24, No], [1, 7, 16, a^17b^24, No], [1, 8, 15, a^16b^24, No], [1, 9, 14, a^15b^24, No], [1, 10, 13, a^14b^24, No], [1, 11, 12, a^13b^24, No], [1, 12, 11, a^12b^24, No], [1, 13, 10, a^11b^24, No], [1, 14, 9, a^10b^24, No], [1, 15, 8, a^9b^24, No], [1, 16, 7, a^8b^24, No], [1, 17, 6, a^7b^24, No], [1, 18, 5, a^6b^24, No], [1, 19, 4, a^5b^24, No], [1, 20, 3, a^4b^24, No], [1, 21, 2, a^3b^24, No], [1, 22, 1, a^2b^24, No], [1, 23, 0, a^1b^24, No], [2, 1, 21, a^23b^24, No], [2, 2, 20, a^22b^24, No], [2, 3, 19, a^21b^24, No], [2, 4, 18, a^20b^24, No], [2, 5, 17, a^19b^24, No], [2, 6, 16, a^18b^24, No], [2, 7, 15, a^17b^24, No], [2, 8, 14, a^16b^24, No], [2, 9, 13, a^15b^24, No], [2, 10, 12, a^14b^24, No], [2, 11, 11, a^13b^24, No], [2, 12, 10, a^12b^24, No], [2, 13, 9, a^11b^24, No], [2, 14, 8, a^10b^24, No], [2, 15, 7, a^9b^24, No], [2, 16, 6, a^8b^24, No], [2, 17, 5, a^7b^24, No], [2, 18, 4, a^6b^24, No], [2, 19, 3, a^5b^24, No], [2, 20, 2, a^4b^24, No], [2, 21, 1, a^3b^24, No], [2, 22, 0, a^2b^24, No], [3, 1, 20, a^23b^24, No], [3, 2, 19, a^22b^24, No], [3, 3, 18, a^21b^24, No], [3, 4, 17, a^20b^24, No], [3, 5, 16, a^19b^24, No], [3, 6, 15, a^18b^24, No], [3, 7, 14, a^17b^24, No], [3, 8, 13, a^16b^24, No], [3, 9, 12, a^15b^24, No], [3, 10, 11, a^14b^24, No], [3, 11, 10, a^13b^24, No], [3, 12, 9, a^12b^24, No], [3, 13, 8, a^11b^24, No], [3, 14, 7, a^10b^24, No], [3, 15, 6, a^9b^24, No], [3, 16, 5, a^8b^24, No], [3, 17, 4, a^7b^24, No], [3, 18, 3, a^6b^24, No], [3, 19, 2, a^5b^24, No], [3, 20, 1, a^4b^24, No], [3, 21, 0, a^3b^24, No], [4, 1, 19, a^23b^24, No], [4, 2, 18, a^22b^24, No], [4, 3, 17, a^21b^24, No], [4, 4, 16, a^20b^24, No], [4, 5, 15, a^19b^24, No], [4, 6, 14, a^18b^24, No], [4, 7, 13, a^17b^24, No], [4, 8, 12, a^16b^24, No], [4, 9, 11, a^15b^24, No], [4, 10, 10, a^14b^24, No], [4, 11, 9, a^13b^24, No], [4, 12, 8, a^12b^24, No], [4, 13, 7, a^11b^24, No], [4, 14, 6, a^10b^24, No], [4, 15, 5, a^9b^24, No], [4, 16, 4, a^8b^24, No], [4, 17, 3, a^7b^24, No], [4, 18, 2, a^6b^24, No], [4, 19, 1, a^5b^24, No], [4, 20, 0, a^4b^24, No], [5, 1, 18, a^23b^24, No], [5, 2, 17, a^22b^24, No], [5, 3, 16, a^21b^24, No], [5, 4, 15, a^20b^24, No], [5, 5, 14, a^19b^24, No], [5, 6, 13, a^18b^24, No], [5, 7, 12, a^17b^24, No], [5, 8, 11, a^16b^24, No], [5, 9, 10, a^15b^24, No], [5, 10, 9, a^14b^24, No], [5, 11, 8, a^13b^24, No], [5, 12, 7, a^12b^24, No], [5, 13, 6, a^11b^24, No], [5, 14, 5, a^10b^24, No], [5, 15, 4, a^9b^24, No], [5, 16, 3, a^8b^24, No], [5, 17, 2, a^7b^24, No], [5, 18, 1, a^6b^24, No], [5, 19, 0, a^5b^24, No], [6, 1, 17, a^23b^24, No], [6, 2, 16, a^22b^24, No], [6, 3, 15, a^21b^24, No], [6, 4, 14, a^20b^24, No], [6, 5, 13, a^19b^24, No], [6, 6, 12, a^18b^24, No], [6, 7, 11, a^17b^24, No], [6, 8, 10, a^16b^24, No], [6, 9, 9, a^15b^24, No], [6, 10, 8, a^14b^24, No], [6, 11, 7, a^13b^24, No], [6, 12, 6, a^12b^24, No], [6, 13, 5, a^11b^24, No], [6, 14, 4, a^10b^24, No], [6, 15, 3, a^9b^24, No], [6, 16, 2, a^8b^24, No], [6, 17, 1, a^7b^24, No], [6, 18, 0, a^6b^24, No], [7, 1, 16, a^23b^24, No], [7, 2, 15, a^22b^24, No], [7, 3, 14, a^21b^24, No], [7, 4, 13, a^20b^24, No], [7, 5, 12, a^19b^24, No], [7, 6, 11, a^18b^24, No], [7, 7, 10, a^17b^24, No], [7, 8, 9, a^16b^24, No], [7, 9, 8, a^15b^24, No], [7, 10, 7, a^14b^24, No], [7, 11, 6, a^13b^24, No], [7, 12, 5, a^12b^24, No], [7, 13, 4, a^11b^24, No], [7, 14, 3, a^10b^24, No], [7, 15, 2, a^9b^24, No], [7, 16, 1, a^8b^24, No], [7, 17, 0, a^7b^24, No], [8, 1, 15, a^23b^24, No], [8, 2, 14, a^22b^24, No], [8, 3, 13, a^21b^24, No], [8, 4, 12, a^20b^24, No], [8, 5, 11, a^19b^24, No], [8, 6, 10, a^18b^24, No], [8, 7, 9, a^17b^24, No], [8, 8, 8, a^16b^24, No], [8, 9, 7, a^15b^24, No], [8, 10, 6, a^14b^24, No], [8, 11, 5, a^13b^24, No], [8, 12, 4, a^12b^24, No], [8, 13, 3, a^11b^24, No], [8, 14, 2, a^10b^24, No], [8, 15, 1, a^9b^24, No], [8, 16, 0, a^8b^24, No], [9, 1, 14, a^23b^24, No], [9, 2, 13, a^22b^24, No], [9, 3, 12, a^21b^24, No], [9, 4, 11, a^20b^24, No], [9, 5, 10, a^19b^24, No], [9, 6, 9, a^18b^24, No], [9, 7, 8, a^17b^24, No], [9, 8, 7, a^16b^24, No], [9, 9, 6, a^15b^24, No], [9, 10, 5, a^14b^24, No], [9, 11, 4, a^13b^24, No], [9, 12, 3, a^12b^24, No], [9, 13, 2, a^11b^24, No], [9, 14, 1, a^10b^24, No], [9, 15, 0, a^9b^24, No], [10, 1, 13, a^23b^24, No], [10, 2, 12, a^22b^24, No], [10, 3, 11, a^21b^24, No], [10, 4, 10, a^20b^24, No], [10, 5, 9, a^19b^24, No], [10, 6, 8, a^18b^24, No], [10, 7, 7, a^17b^24, No], [10, 8, 6, a^16b^24, No], [10, 9, 5, a^15b^24, No], [10, 10, 4, a^14b^24, No], [10, 11, 3, a^13b^24, No], [10, 12, 2, a^12b^24, No], [10, 13, 1, a^11b^24, No], [10, 14, 0, a^10b^24, No], [11, 1, 12, a^23b^24, No], [11, 2, 11, a^22b^24, No], [11, 3, 10, a^21b^24, No], [11, 4, 9, a^20b^24, No], [11, 5, 8, a^19b^24, No], [11, 6, 7, a^18b^24, No], [11, 7, 6, a^17b^24, No], [11, 8, 5, a^16b^24, No], [11, 9, 4, a^15b^24, No], [11, 10, 3, a^14b^24, No], [11, 11, 2, a^13b^24, No], [11, 12, 1, a^12b^24, No], [11, 13, 0, a^11b^24, No], [12, 1, 11, a^23b^24, No], [12, 2, 10, a^22b^24, No], [12, 3, 9, a^21b^24, No], [12, 4, 8, a^20b^24, No], [12, 5, 7, a^19b^24, No], [12, 6, 6, a^18b^24, No], [12, 7, 5, a^17b^24, No], [12, 8, 4, a^16b^24, No], [12, 9, 3, a^15b^24, No], [12, 10, 2, a^14b^24, No], [12, 11, 1, a^13b^24, No], [12, 12, 0, a^12b^24, No], [13, 1, 10, a^23b^24, No], [13, 2, 9, a^22b^24, No], [13, 3, 8, a^21b^24, No], [13, 4, 7, a^20b^24, No], [13, 5, 6, a^19b^24, No], [13, 6, 5, a^18b^24, No], [13, 7, 4, a^17b^24, No], [13, 8, 3, a^16b^24, No], [13, 9, 2, a^15b^24, No], [13, 10, 1, a^14b^24, No], [13, 11, 0, a^13b^24, No], [14, 1, 9, a^23b^24, No], [14, 2, 8, a^22b^24, No], [14, 3, 7, a^21b^24, No], [14, 4, 6, a^20b^24, No], [14, 5, 5, a^19b^24, No], [14, 6, 4, a^18b^24, No], [14, 7, 3, a^17b^24, No], [14, 8, 2, a^16b^24, No], [14, 9, 1, a^15b^24, No], [14, 10, 0, a^14b^24, No], [15, 1, 8, a^23b^24, No], [15, 2, 7, a^22b^24, No], [15, 3, 6, a^21b^24, No], [15, 4, 5, a^20b^24, No], [15, 5, 4, a^19b^24, No], [15, 6, 3, a^18b^24, No], [15, 7, 2, a^17b^24, No], [15, 8, 1, a^16b^24, No], [15, 9, 0, a^15b^24, No], [16, 1, 7, a^23b^24, No], [16, 2, 6, a^22b^24, No], [16, 3, 5, a^21b^24, No], [16, 4, 4, a^20b^24, No], [16, 5, 3, a^19b^24, No], [16, 6, 2, a^18b^24, No], [16, 7, 1, a^17b^24, No], [16, 8, 0, a^16b^24, No], [17, 1, 6, a^23b^24, No], [17, 2, 5, a^22b^24, No], [17, 3, 4, a^21b^24, No], [17, 4, 3, a^20b^24, No], [17, 5, 2, a^19b^24, No], [17, 6, 1, a^18b^24, No], [17, 7, 0, a^17b^24, No], [18, 1, 5, a^23b^24, No], [18, 2, 4, a^22b^24, No], [18, 3, 3, a^21b^24, No], [18, 4, 2, a^20b^24, No], [18, 5, 1, a^19b^24, No], [18, 6, 0, a^18b^24, No], [19, 1, 4, a^23b^24, No], [19, 2, 3, a^22b^24, No], [19, 3, 2, a^21b^24, No], [19, 4, 1, a^20b^24, No], [19, 5, 0, a^19b^24, No], [20, 1, 3, a^23b^24, No], [20, 2, 2, a^22b^24, No], [20, 3, 1, a^21b^24, No], [20, 4, 0, a^20b^24, No], [21, 1, 2, a^23b^24, No], [21, 2, 1, a^22b^24, No], [21, 3, 0, a^21b^24, No], [22, 1, 1, a^23b^24, No], [22, 2, 0, a^22b^24, No], [23, 1, 0, a^23b^24, No]], 300, 300, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

75. Every Legal Split of a^p b^p Is Defeated by Pumping Down

Use positive pumping length p=25 and adversary word a^p b^p. Give (p, adversary_word, all_legal_split_rows, split_count, triangular_count, every_split_defeated, pump_down_scope).

Each split row gives|x|,|y|,remaining a-count p−|x|−|y|,the i=0 word,and membership in a^n b^n. The third entry is not the full|z|: z also contains all p b-symbols.

View answer and explanation

(25, a^25b^25, [[0, 1, 24, a^24b^25, No], [0, 2, 23, a^23b^25, No], [0, 3, 22, a^22b^25, No], [0, 4, 21, a^21b^25, No], [0, 5, 20, a^20b^25, No], [0, 6, 19, a^19b^25, No], [0, 7, 18, a^18b^25, No], [0, 8, 17, a^17b^25, No], [0, 9, 16, a^16b^25, No], [0, 10, 15, a^15b^25, No], [0, 11, 14, a^14b^25, No], [0, 12, 13, a^13b^25, No], [0, 13, 12, a^12b^25, No], [0, 14, 11, a^11b^25, No], [0, 15, 10, a^10b^25, No], [0, 16, 9, a^9b^25, No], [0, 17, 8, a^8b^25, No], [0, 18, 7, a^7b^25, No], [0, 19, 6, a^6b^25, No], [0, 20, 5, a^5b^25, No], [0, 21, 4, a^4b^25, No], [0, 22, 3, a^3b^25, No], [0, 23, 2, a^2b^25, No], [0, 24, 1, a^1b^25, No], [0, 25, 0, a^0b^25, No], [1, 1, 23, a^24b^25, No], [1, 2, 22, a^23b^25, No], [1, 3, 21, a^22b^25, No], [1, 4, 20, a^21b^25, No], [1, 5, 19, a^20b^25, No], [1, 6, 18, a^19b^25, No], [1, 7, 17, a^18b^25, No], [1, 8, 16, a^17b^25, No], [1, 9, 15, a^16b^25, No], [1, 10, 14, a^15b^25, No], [1, 11, 13, a^14b^25, No], [1, 12, 12, a^13b^25, No], [1, 13, 11, a^12b^25, No], [1, 14, 10, a^11b^25, No], [1, 15, 9, a^10b^25, No], [1, 16, 8, a^9b^25, No], [1, 17, 7, a^8b^25, No], [1, 18, 6, a^7b^25, No], [1, 19, 5, a^6b^25, No], [1, 20, 4, a^5b^25, No], [1, 21, 3, a^4b^25, No], [1, 22, 2, a^3b^25, No], [1, 23, 1, a^2b^25, No], [1, 24, 0, a^1b^25, No], [2, 1, 22, a^24b^25, No], [2, 2, 21, a^23b^25, No], [2, 3, 20, a^22b^25, No], [2, 4, 19, a^21b^25, No], [2, 5, 18, a^20b^25, No], [2, 6, 17, a^19b^25, No], [2, 7, 16, a^18b^25, No], [2, 8, 15, a^17b^25, No], [2, 9, 14, a^16b^25, No], [2, 10, 13, a^15b^25, No], [2, 11, 12, a^14b^25, No], [2, 12, 11, a^13b^25, No], [2, 13, 10, a^12b^25, No], [2, 14, 9, a^11b^25, No], [2, 15, 8, a^10b^25, No], [2, 16, 7, a^9b^25, No], [2, 17, 6, a^8b^25, No], [2, 18, 5, a^7b^25, No], [2, 19, 4, a^6b^25, No], [2, 20, 3, a^5b^25, No], [2, 21, 2, a^4b^25, No], [2, 22, 1, a^3b^25, No], [2, 23, 0, a^2b^25, No], [3, 1, 21, a^24b^25, No], [3, 2, 20, a^23b^25, No], [3, 3, 19, a^22b^25, No], [3, 4, 18, a^21b^25, No], [3, 5, 17, a^20b^25, No], [3, 6, 16, a^19b^25, No], [3, 7, 15, a^18b^25, No], [3, 8, 14, a^17b^25, No], [3, 9, 13, a^16b^25, No], [3, 10, 12, a^15b^25, No], [3, 11, 11, a^14b^25, No], [3, 12, 10, a^13b^25, No], [3, 13, 9, a^12b^25, No], [3, 14, 8, a^11b^25, No], [3, 15, 7, a^10b^25, No], [3, 16, 6, a^9b^25, No], [3, 17, 5, a^8b^25, No], [3, 18, 4, a^7b^25, No], [3, 19, 3, a^6b^25, No], [3, 20, 2, a^5b^25, No], [3, 21, 1, a^4b^25, No], [3, 22, 0, a^3b^25, No], [4, 1, 20, a^24b^25, No], [4, 2, 19, a^23b^25, No], [4, 3, 18, a^22b^25, No], [4, 4, 17, a^21b^25, No], [4, 5, 16, a^20b^25, No], [4, 6, 15, a^19b^25, No], [4, 7, 14, a^18b^25, No], [4, 8, 13, a^17b^25, No], [4, 9, 12, a^16b^25, No], [4, 10, 11, a^15b^25, No], [4, 11, 10, a^14b^25, No], [4, 12, 9, a^13b^25, No], [4, 13, 8, a^12b^25, No], [4, 14, 7, a^11b^25, No], [4, 15, 6, a^10b^25, No], [4, 16, 5, a^9b^25, No], [4, 17, 4, a^8b^25, No], [4, 18, 3, a^7b^25, No], [4, 19, 2, a^6b^25, No], [4, 20, 1, a^5b^25, No], [4, 21, 0, a^4b^25, No], [5, 1, 19, a^24b^25, No], [5, 2, 18, a^23b^25, No], [5, 3, 17, a^22b^25, No], [5, 4, 16, a^21b^25, No], [5, 5, 15, a^20b^25, No], [5, 6, 14, a^19b^25, No], [5, 7, 13, a^18b^25, No], [5, 8, 12, a^17b^25, No], [5, 9, 11, a^16b^25, No], [5, 10, 10, a^15b^25, No], [5, 11, 9, a^14b^25, No], [5, 12, 8, a^13b^25, No], [5, 13, 7, a^12b^25, No], [5, 14, 6, a^11b^25, No], [5, 15, 5, a^10b^25, No], [5, 16, 4, a^9b^25, No], [5, 17, 3, a^8b^25, No], [5, 18, 2, a^7b^25, No], [5, 19, 1, a^6b^25, No], [5, 20, 0, a^5b^25, No], [6, 1, 18, a^24b^25, No], [6, 2, 17, a^23b^25, No], [6, 3, 16, a^22b^25, No], [6, 4, 15, a^21b^25, No], [6, 5, 14, a^20b^25, No], [6, 6, 13, a^19b^25, No], [6, 7, 12, a^18b^25, No], [6, 8, 11, a^17b^25, No], [6, 9, 10, a^16b^25, No], [6, 10, 9, a^15b^25, No], [6, 11, 8, a^14b^25, No], [6, 12, 7, a^13b^25, No], [6, 13, 6, a^12b^25, No], [6, 14, 5, a^11b^25, No], [6, 15, 4, a^10b^25, No], [6, 16, 3, a^9b^25, No], [6, 17, 2, a^8b^25, No], [6, 18, 1, a^7b^25, No], [6, 19, 0, a^6b^25, No], [7, 1, 17, a^24b^25, No], [7, 2, 16, a^23b^25, No], [7, 3, 15, a^22b^25, No], [7, 4, 14, a^21b^25, No], [7, 5, 13, a^20b^25, No], [7, 6, 12, a^19b^25, No], [7, 7, 11, a^18b^25, No], [7, 8, 10, a^17b^25, No], [7, 9, 9, a^16b^25, No], [7, 10, 8, a^15b^25, No], [7, 11, 7, a^14b^25, No], [7, 12, 6, a^13b^25, No], [7, 13, 5, a^12b^25, No], [7, 14, 4, a^11b^25, No], [7, 15, 3, a^10b^25, No], [7, 16, 2, a^9b^25, No], [7, 17, 1, a^8b^25, No], [7, 18, 0, a^7b^25, No], [8, 1, 16, a^24b^25, No], [8, 2, 15, a^23b^25, No], [8, 3, 14, a^22b^25, No], [8, 4, 13, a^21b^25, No], [8, 5, 12, a^20b^25, No], [8, 6, 11, a^19b^25, No], [8, 7, 10, a^18b^25, No], [8, 8, 9, a^17b^25, No], [8, 9, 8, a^16b^25, No], [8, 10, 7, a^15b^25, No], [8, 11, 6, a^14b^25, No], [8, 12, 5, a^13b^25, No], [8, 13, 4, a^12b^25, No], [8, 14, 3, a^11b^25, No], [8, 15, 2, a^10b^25, No], [8, 16, 1, a^9b^25, No], [8, 17, 0, a^8b^25, No], [9, 1, 15, a^24b^25, No], [9, 2, 14, a^23b^25, No], [9, 3, 13, a^22b^25, No], [9, 4, 12, a^21b^25, No], [9, 5, 11, a^20b^25, No], [9, 6, 10, a^19b^25, No], [9, 7, 9, a^18b^25, No], [9, 8, 8, a^17b^25, No], [9, 9, 7, a^16b^25, No], [9, 10, 6, a^15b^25, No], [9, 11, 5, a^14b^25, No], [9, 12, 4, a^13b^25, No], [9, 13, 3, a^12b^25, No], [9, 14, 2, a^11b^25, No], [9, 15, 1, a^10b^25, No], [9, 16, 0, a^9b^25, No], [10, 1, 14, a^24b^25, No], [10, 2, 13, a^23b^25, No], [10, 3, 12, a^22b^25, No], [10, 4, 11, a^21b^25, No], [10, 5, 10, a^20b^25, No], [10, 6, 9, a^19b^25, No], [10, 7, 8, a^18b^25, No], [10, 8, 7, a^17b^25, No], [10, 9, 6, a^16b^25, No], [10, 10, 5, a^15b^25, No], [10, 11, 4, a^14b^25, No], [10, 12, 3, a^13b^25, No], [10, 13, 2, a^12b^25, No], [10, 14, 1, a^11b^25, No], [10, 15, 0, a^10b^25, No], [11, 1, 13, a^24b^25, No], [11, 2, 12, a^23b^25, No], [11, 3, 11, a^22b^25, No], [11, 4, 10, a^21b^25, No], [11, 5, 9, a^20b^25, No], [11, 6, 8, a^19b^25, No], [11, 7, 7, a^18b^25, No], [11, 8, 6, a^17b^25, No], [11, 9, 5, a^16b^25, No], [11, 10, 4, a^15b^25, No], [11, 11, 3, a^14b^25, No], [11, 12, 2, a^13b^25, No], [11, 13, 1, a^12b^25, No], [11, 14, 0, a^11b^25, No], [12, 1, 12, a^24b^25, No], [12, 2, 11, a^23b^25, No], [12, 3, 10, a^22b^25, No], [12, 4, 9, a^21b^25, No], [12, 5, 8, a^20b^25, No], [12, 6, 7, a^19b^25, No], [12, 7, 6, a^18b^25, No], [12, 8, 5, a^17b^25, No], [12, 9, 4, a^16b^25, No], [12, 10, 3, a^15b^25, No], [12, 11, 2, a^14b^25, No], [12, 12, 1, a^13b^25, No], [12, 13, 0, a^12b^25, No], [13, 1, 11, a^24b^25, No], [13, 2, 10, a^23b^25, No], [13, 3, 9, a^22b^25, No], [13, 4, 8, a^21b^25, No], [13, 5, 7, a^20b^25, No], [13, 6, 6, a^19b^25, No], [13, 7, 5, a^18b^25, No], [13, 8, 4, a^17b^25, No], [13, 9, 3, a^16b^25, No], [13, 10, 2, a^15b^25, No], [13, 11, 1, a^14b^25, No], [13, 12, 0, a^13b^25, No], [14, 1, 10, a^24b^25, No], [14, 2, 9, a^23b^25, No], [14, 3, 8, a^22b^25, No], [14, 4, 7, a^21b^25, No], [14, 5, 6, a^20b^25, No], [14, 6, 5, a^19b^25, No], [14, 7, 4, a^18b^25, No], [14, 8, 3, a^17b^25, No], [14, 9, 2, a^16b^25, No], [14, 10, 1, a^15b^25, No], [14, 11, 0, a^14b^25, No], [15, 1, 9, a^24b^25, No], [15, 2, 8, a^23b^25, No], [15, 3, 7, a^22b^25, No], [15, 4, 6, a^21b^25, No], [15, 5, 5, a^20b^25, No], [15, 6, 4, a^19b^25, No], [15, 7, 3, a^18b^25, No], [15, 8, 2, a^17b^25, No], [15, 9, 1, a^16b^25, No], [15, 10, 0, a^15b^25, No], [16, 1, 8, a^24b^25, No], [16, 2, 7, a^23b^25, No], [16, 3, 6, a^22b^25, No], [16, 4, 5, a^21b^25, No], [16, 5, 4, a^20b^25, No], [16, 6, 3, a^19b^25, No], [16, 7, 2, a^18b^25, No], [16, 8, 1, a^17b^25, No], [16, 9, 0, a^16b^25, No], [17, 1, 7, a^24b^25, No], [17, 2, 6, a^23b^25, No], [17, 3, 5, a^22b^25, No], [17, 4, 4, a^21b^25, No], [17, 5, 3, a^20b^25, No], [17, 6, 2, a^19b^25, No], [17, 7, 1, a^18b^25, No], [17, 8, 0, a^17b^25, No], [18, 1, 6, a^24b^25, No], [18, 2, 5, a^23b^25, No], [18, 3, 4, a^22b^25, No], [18, 4, 3, a^21b^25, No], [18, 5, 2, a^20b^25, No], [18, 6, 1, a^19b^25, No], [18, 7, 0, a^18b^25, No], [19, 1, 5, a^24b^25, No], [19, 2, 4, a^23b^25, No], [19, 3, 3, a^22b^25, No], [19, 4, 2, a^21b^25, No], [19, 5, 1, a^20b^25, No], [19, 6, 0, a^19b^25, No], [20, 1, 4, a^24b^25, No], [20, 2, 3, a^23b^25, No], [20, 3, 2, a^22b^25, No], [20, 4, 1, a^21b^25, No], [20, 5, 0, a^20b^25, No], [21, 1, 3, a^24b^25, No], [21, 2, 2, a^23b^25, No], [21, 3, 1, a^22b^25, No], [21, 4, 0, a^21b^25, No], [22, 1, 2, a^24b^25, No], [22, 2, 1, a^23b^25, No], [22, 3, 0, a^22b^25, No], [23, 1, 1, a^24b^25, No], [23, 2, 0, a^23b^25, No], [24, 1, 0, a^24b^25, No]], 325, 325, Yes, every_legal_split_is_defeated_by_i=0)

Cover every x≥0,y≥1 with x+y≤p. Each y contains only a symbols; pumping with i=0 leaves fewer a symbols than b symbols. The split-row third entry is the remaining a count p−x−y, not the full z length, because z also contains all p b symbols. A fixed-p table supports this adversary pattern; the general nonregularity argument quantifies over every proposed p. The exact result, retaining the requested order, is (25, a^25b^25, [[0, 1, 24, a^24b^25, No], [0, 2, 23, a^23b^25, No], [0, 3, 22, a^22b^25, No], [0, 4, 21, a^21b^25, No], [0, 5, 20, a^20b^25, No], [0, 6, 19, a^19b^25, No], [0, 7, 18, a^18b^25, No], [0, 8, 17, a^17b^25, No], [0, 9, 16, a^16b^25, No], [0, 10, 15, a^15b^25, No], [0, 11, 14, a^14b^25, No], [0, 12, 13, a^13b^25, No], [0, 13, 12, a^12b^25, No], [0, 14, 11, a^11b^25, No], [0, 15, 10, a^10b^25, No], [0, 16, 9, a^9b^25, No], [0, 17, 8, a^8b^25, No], [0, 18, 7, a^7b^25, No], [0, 19, 6, a^6b^25, No], [0, 20, 5, a^5b^25, No], [0, 21, 4, a^4b^25, No], [0, 22, 3, a^3b^25, No], [0, 23, 2, a^2b^25, No], [0, 24, 1, a^1b^25, No], [0, 25, 0, a^0b^25, No], [1, 1, 23, a^24b^25, No], [1, 2, 22, a^23b^25, No], [1, 3, 21, a^22b^25, No], [1, 4, 20, a^21b^25, No], [1, 5, 19, a^20b^25, No], [1, 6, 18, a^19b^25, No], [1, 7, 17, a^18b^25, No], [1, 8, 16, a^17b^25, No], [1, 9, 15, a^16b^25, No], [1, 10, 14, a^15b^25, No], [1, 11, 13, a^14b^25, No], [1, 12, 12, a^13b^25, No], [1, 13, 11, a^12b^25, No], [1, 14, 10, a^11b^25, No], [1, 15, 9, a^10b^25, No], [1, 16, 8, a^9b^25, No], [1, 17, 7, a^8b^25, No], [1, 18, 6, a^7b^25, No], [1, 19, 5, a^6b^25, No], [1, 20, 4, a^5b^25, No], [1, 21, 3, a^4b^25, No], [1, 22, 2, a^3b^25, No], [1, 23, 1, a^2b^25, No], [1, 24, 0, a^1b^25, No], [2, 1, 22, a^24b^25, No], [2, 2, 21, a^23b^25, No], [2, 3, 20, a^22b^25, No], [2, 4, 19, a^21b^25, No], [2, 5, 18, a^20b^25, No], [2, 6, 17, a^19b^25, No], [2, 7, 16, a^18b^25, No], [2, 8, 15, a^17b^25, No], [2, 9, 14, a^16b^25, No], [2, 10, 13, a^15b^25, No], [2, 11, 12, a^14b^25, No], [2, 12, 11, a^13b^25, No], [2, 13, 10, a^12b^25, No], [2, 14, 9, a^11b^25, No], [2, 15, 8, a^10b^25, No], [2, 16, 7, a^9b^25, No], [2, 17, 6, a^8b^25, No], [2, 18, 5, a^7b^25, No], [2, 19, 4, a^6b^25, No], [2, 20, 3, a^5b^25, No], [2, 21, 2, a^4b^25, No], [2, 22, 1, a^3b^25, No], [2, 23, 0, a^2b^25, No], [3, 1, 21, a^24b^25, No], [3, 2, 20, a^23b^25, No], [3, 3, 19, a^22b^25, No], [3, 4, 18, a^21b^25, No], [3, 5, 17, a^20b^25, No], [3, 6, 16, a^19b^25, No], [3, 7, 15, a^18b^25, No], [3, 8, 14, a^17b^25, No], [3, 9, 13, a^16b^25, No], [3, 10, 12, a^15b^25, No], [3, 11, 11, a^14b^25, No], [3, 12, 10, a^13b^25, No], [3, 13, 9, a^12b^25, No], [3, 14, 8, a^11b^25, No], [3, 15, 7, a^10b^25, No], [3, 16, 6, a^9b^25, No], [3, 17, 5, a^8b^25, No], [3, 18, 4, a^7b^25, No], [3, 19, 3, a^6b^25, No], [3, 20, 2, a^5b^25, No], [3, 21, 1, a^4b^25, No], [3, 22, 0, a^3b^25, No], [4, 1, 20, a^24b^25, No], [4, 2, 19, a^23b^25, No], [4, 3, 18, a^22b^25, No], [4, 4, 17, a^21b^25, No], [4, 5, 16, a^20b^25, No], [4, 6, 15, a^19b^25, No], [4, 7, 14, a^18b^25, No], [4, 8, 13, a^17b^25, No], [4, 9, 12, a^16b^25, No], [4, 10, 11, a^15b^25, No], [4, 11, 10, a^14b^25, No], [4, 12, 9, a^13b^25, No], [4, 13, 8, a^12b^25, No], [4, 14, 7, a^11b^25, No], [4, 15, 6, a^10b^25, No], [4, 16, 5, a^9b^25, No], [4, 17, 4, a^8b^25, No], [4, 18, 3, a^7b^25, No], [4, 19, 2, a^6b^25, No], [4, 20, 1, a^5b^25, No], [4, 21, 0, a^4b^25, No], [5, 1, 19, a^24b^25, No], [5, 2, 18, a^23b^25, No], [5, 3, 17, a^22b^25, No], [5, 4, 16, a^21b^25, No], [5, 5, 15, a^20b^25, No], [5, 6, 14, a^19b^25, No], [5, 7, 13, a^18b^25, No], [5, 8, 12, a^17b^25, No], [5, 9, 11, a^16b^25, No], [5, 10, 10, a^15b^25, No], [5, 11, 9, a^14b^25, No], [5, 12, 8, a^13b^25, No], [5, 13, 7, a^12b^25, No], [5, 14, 6, a^11b^25, No], [5, 15, 5, a^10b^25, No], [5, 16, 4, a^9b^25, No], [5, 17, 3, a^8b^25, No], [5, 18, 2, a^7b^25, No], [5, 19, 1, a^6b^25, No], [5, 20, 0, a^5b^25, No], [6, 1, 18, a^24b^25, No], [6, 2, 17, a^23b^25, No], [6, 3, 16, a^22b^25, No], [6, 4, 15, a^21b^25, No], [6, 5, 14, a^20b^25, No], [6, 6, 13, a^19b^25, No], [6, 7, 12, a^18b^25, No], [6, 8, 11, a^17b^25, No], [6, 9, 10, a^16b^25, No], [6, 10, 9, a^15b^25, No], [6, 11, 8, a^14b^25, No], [6, 12, 7, a^13b^25, No], [6, 13, 6, a^12b^25, No], [6, 14, 5, a^11b^25, No], [6, 15, 4, a^10b^25, No], [6, 16, 3, a^9b^25, No], [6, 17, 2, a^8b^25, No], [6, 18, 1, a^7b^25, No], [6, 19, 0, a^6b^25, No], [7, 1, 17, a^24b^25, No], [7, 2, 16, a^23b^25, No], [7, 3, 15, a^22b^25, No], [7, 4, 14, a^21b^25, No], [7, 5, 13, a^20b^25, No], [7, 6, 12, a^19b^25, No], [7, 7, 11, a^18b^25, No], [7, 8, 10, a^17b^25, No], [7, 9, 9, a^16b^25, No], [7, 10, 8, a^15b^25, No], [7, 11, 7, a^14b^25, No], [7, 12, 6, a^13b^25, No], [7, 13, 5, a^12b^25, No], [7, 14, 4, a^11b^25, No], [7, 15, 3, a^10b^25, No], [7, 16, 2, a^9b^25, No], [7, 17, 1, a^8b^25, No], [7, 18, 0, a^7b^25, No], [8, 1, 16, a^24b^25, No], [8, 2, 15, a^23b^25, No], [8, 3, 14, a^22b^25, No], [8, 4, 13, a^21b^25, No], [8, 5, 12, a^20b^25, No], [8, 6, 11, a^19b^25, No], [8, 7, 10, a^18b^25, No], [8, 8, 9, a^17b^25, No], [8, 9, 8, a^16b^25, No], [8, 10, 7, a^15b^25, No], [8, 11, 6, a^14b^25, No], [8, 12, 5, a^13b^25, No], [8, 13, 4, a^12b^25, No], [8, 14, 3, a^11b^25, No], [8, 15, 2, a^10b^25, No], [8, 16, 1, a^9b^25, No], [8, 17, 0, a^8b^25, No], [9, 1, 15, a^24b^25, No], [9, 2, 14, a^23b^25, No], [9, 3, 13, a^22b^25, No], [9, 4, 12, a^21b^25, No], [9, 5, 11, a^20b^25, No], [9, 6, 10, a^19b^25, No], [9, 7, 9, a^18b^25, No], [9, 8, 8, a^17b^25, No], [9, 9, 7, a^16b^25, No], [9, 10, 6, a^15b^25, No], [9, 11, 5, a^14b^25, No], [9, 12, 4, a^13b^25, No], [9, 13, 3, a^12b^25, No], [9, 14, 2, a^11b^25, No], [9, 15, 1, a^10b^25, No], [9, 16, 0, a^9b^25, No], [10, 1, 14, a^24b^25, No], [10, 2, 13, a^23b^25, No], [10, 3, 12, a^22b^25, No], [10, 4, 11, a^21b^25, No], [10, 5, 10, a^20b^25, No], [10, 6, 9, a^19b^25, No], [10, 7, 8, a^18b^25, No], [10, 8, 7, a^17b^25, No], [10, 9, 6, a^16b^25, No], [10, 10, 5, a^15b^25, No], [10, 11, 4, a^14b^25, No], [10, 12, 3, a^13b^25, No], [10, 13, 2, a^12b^25, No], [10, 14, 1, a^11b^25, No], [10, 15, 0, a^10b^25, No], [11, 1, 13, a^24b^25, No], [11, 2, 12, a^23b^25, No], [11, 3, 11, a^22b^25, No], [11, 4, 10, a^21b^25, No], [11, 5, 9, a^20b^25, No], [11, 6, 8, a^19b^25, No], [11, 7, 7, a^18b^25, No], [11, 8, 6, a^17b^25, No], [11, 9, 5, a^16b^25, No], [11, 10, 4, a^15b^25, No], [11, 11, 3, a^14b^25, No], [11, 12, 2, a^13b^25, No], [11, 13, 1, a^12b^25, No], [11, 14, 0, a^11b^25, No], [12, 1, 12, a^24b^25, No], [12, 2, 11, a^23b^25, No], [12, 3, 10, a^22b^25, No], [12, 4, 9, a^21b^25, No], [12, 5, 8, a^20b^25, No], [12, 6, 7, a^19b^25, No], [12, 7, 6, a^18b^25, No], [12, 8, 5, a^17b^25, No], [12, 9, 4, a^16b^25, No], [12, 10, 3, a^15b^25, No], [12, 11, 2, a^14b^25, No], [12, 12, 1, a^13b^25, No], [12, 13, 0, a^12b^25, No], [13, 1, 11, a^24b^25, No], [13, 2, 10, a^23b^25, No], [13, 3, 9, a^22b^25, No], [13, 4, 8, a^21b^25, No], [13, 5, 7, a^20b^25, No], [13, 6, 6, a^19b^25, No], [13, 7, 5, a^18b^25, No], [13, 8, 4, a^17b^25, No], [13, 9, 3, a^16b^25, No], [13, 10, 2, a^15b^25, No], [13, 11, 1, a^14b^25, No], [13, 12, 0, a^13b^25, No], [14, 1, 10, a^24b^25, No], [14, 2, 9, a^23b^25, No], [14, 3, 8, a^22b^25, No], [14, 4, 7, a^21b^25, No], [14, 5, 6, a^20b^25, No], [14, 6, 5, a^19b^25, No], [14, 7, 4, a^18b^25, No], [14, 8, 3, a^17b^25, No], [14, 9, 2, a^16b^25, No], [14, 10, 1, a^15b^25, No], [14, 11, 0, a^14b^25, No], [15, 1, 9, a^24b^25, No], [15, 2, 8, a^23b^25, No], [15, 3, 7, a^22b^25, No], [15, 4, 6, a^21b^25, No], [15, 5, 5, a^20b^25, No], [15, 6, 4, a^19b^25, No], [15, 7, 3, a^18b^25, No], [15, 8, 2, a^17b^25, No], [15, 9, 1, a^16b^25, No], [15, 10, 0, a^15b^25, No], [16, 1, 8, a^24b^25, No], [16, 2, 7, a^23b^25, No], [16, 3, 6, a^22b^25, No], [16, 4, 5, a^21b^25, No], [16, 5, 4, a^20b^25, No], [16, 6, 3, a^19b^25, No], [16, 7, 2, a^18b^25, No], [16, 8, 1, a^17b^25, No], [16, 9, 0, a^16b^25, No], [17, 1, 7, a^24b^25, No], [17, 2, 6, a^23b^25, No], [17, 3, 5, a^22b^25, No], [17, 4, 4, a^21b^25, No], [17, 5, 3, a^20b^25, No], [17, 6, 2, a^19b^25, No], [17, 7, 1, a^18b^25, No], [17, 8, 0, a^17b^25, No], [18, 1, 6, a^24b^25, No], [18, 2, 5, a^23b^25, No], [18, 3, 4, a^22b^25, No], [18, 4, 3, a^21b^25, No], [18, 5, 2, a^20b^25, No], [18, 6, 1, a^19b^25, No], [18, 7, 0, a^18b^25, No], [19, 1, 5, a^24b^25, No], [19, 2, 4, a^23b^25, No], [19, 3, 3, a^22b^25, No], [19, 4, 2, a^21b^25, No], [19, 5, 1, a^20b^25, No], [19, 6, 0, a^19b^25, No], [20, 1, 4, a^24b^25, No], [20, 2, 3, a^23b^25, No], [20, 3, 2, a^22b^25, No], [20, 4, 1, a^21b^25, No], [20, 5, 0, a^20b^25, No], [21, 1, 3, a^24b^25, No], [21, 2, 2, a^23b^25, No], [21, 3, 1, a^22b^25, No], [21, 4, 0, a^21b^25, No], [22, 1, 2, a^24b^25, No], [22, 2, 1, a^23b^25, No], [22, 3, 0, a^22b^25, No], [23, 1, 1, a^24b^25, No], [23, 2, 0, a^23b^25, No], [24, 1, 0, a^24b^25, No]], 325, 325, Yes, every_legal_split_is_defeated_by_i=0).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-anbn-all-splits · candidate

76. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=single_split, complete model={"p":1,"word":["a","b"],"x_length":0,"y_length":1}. Give (mode, p, word, x_length, y_length, z_length, split_legal, pumping_quantifiers_resolved_claim, boundary).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(single_split, 1, [a, b], 0, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (single_split, 1, [a, b], 0, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

77. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=wrong_quantifiers, complete model={"p":2,"word":["a","a","b"]}. Give (mode, p, word, legal_split_count, adversarial_choice_order, reversed_quantifier_claim).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(wrong_quantifiers, 2, [a, a, b], 3, adversary_word_then_prover_split_then_adversary_i, No)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (wrong_quantifiers, 2, [a, a, b], 3, adversary_word_then_prover_split_then_adversary_i, No).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

78. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=bounded_words, complete model={"alphabet":["a","b","c"],"max_checked_length":3}. Give (mode, alphabet, max_checked_length, checked_word_count, regularity_proof_claim, boundary).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(bounded_words, [a, b, c], 3, 40, No, bounded_words_do_not_prove_regular)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (bounded_words, [a, b, c], 3, 40, No, bounded_words_do_not_prove_regular).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

79. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=decomposition_order, complete model={"p":4,"word":["a","a","a","a","b"]}. Give (mode, p, word, legal_split_count, all_splits_obligation, preselected_split_suffices_claim).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(decomposition_order, 4, [a, a, a, a, b], 10, all_legal_splits_must_be_addressed_not_a_preselected_decomposition, No)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (decomposition_order, 4, [a, a, a, a, b], 10, all_legal_splits_must_be_addressed_not_a_preselected_decomposition, No).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

80. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=single_split, complete model={"p":5,"word":["a","a","a","a","a","b"],"x_length":4,"y_length":1}. Give (mode, p, word, x_length, y_length, z_length, split_legal, pumping_quantifiers_resolved_claim, boundary).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(single_split, 5, [a, a, a, a, a, b], 4, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (single_split, 5, [a, a, a, a, a, b], 4, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

81. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=wrong_quantifiers, complete model={"p":6,"word":["a","a","a","a","a","a","b"]}. Give (mode, p, word, legal_split_count, adversarial_choice_order, reversed_quantifier_claim).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(wrong_quantifiers, 6, [a, a, a, a, a, a, b], 21, adversary_word_then_prover_split_then_adversary_i, No)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (wrong_quantifiers, 6, [a, a, a, a, a, a, b], 21, adversary_word_then_prover_split_then_adversary_i, No).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

82. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=bounded_words, complete model={"alphabet":["a"],"max_checked_length":2}. Give (mode, alphabet, max_checked_length, checked_word_count, regularity_proof_claim, boundary).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(bounded_words, [a], 2, 3, No, bounded_words_do_not_prove_regular)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (bounded_words, [a], 2, 3, No, bounded_words_do_not_prove_regular).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

83. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=decomposition_order, complete model={"p":8,"word":["a","a","a","a","a","a","a","a","b"]}. Give (mode, p, word, legal_split_count, all_splits_obligation, preselected_split_suffices_claim).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(decomposition_order, 8, [a, a, a, a, a, a, a, a, b], 36, all_legal_splits_must_be_addressed_not_a_preselected_decomposition, No)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (decomposition_order, 8, [a, a, a, a, a, a, a, a, b], 36, all_legal_splits_must_be_addressed_not_a_preselected_decomposition, No).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

84. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=single_split, complete model={"p":9,"word":["a","a","a","a","a","a","a","a","a","b"],"x_length":8,"y_length":1}. Give (mode, p, word, x_length, y_length, z_length, split_legal, pumping_quantifiers_resolved_claim, boundary).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(single_split, 9, [a, a, a, a, a, a, a, a, a, b], 8, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (single_split, 9, [a, a, a, a, a, a, a, a, a, b], 8, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

85. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=wrong_quantifiers, complete model={"p":10,"word":["a","a","a","a","a","a","a","a","a","a","b"]}. Give (mode, p, word, legal_split_count, adversarial_choice_order, reversed_quantifier_claim).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(wrong_quantifiers, 10, [a, a, a, a, a, a, a, a, a, a, b], 55, adversary_word_then_prover_split_then_adversary_i, No)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (wrong_quantifiers, 10, [a, a, a, a, a, a, a, a, a, a, b], 55, adversary_word_then_prover_split_then_adversary_i, No).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

86. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=bounded_words, complete model={"alphabet":["a","b"],"max_checked_length":1}. Give (mode, alphabet, max_checked_length, checked_word_count, regularity_proof_claim, boundary).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(bounded_words, [a, b], 1, 3, No, bounded_words_do_not_prove_regular)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (bounded_words, [a, b], 1, 3, No, bounded_words_do_not_prove_regular).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

87. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=decomposition_order, complete model={"p":12,"word":["a","a","a","a","a","a","a","a","a","a","a","a","b"]}. Give (mode, p, word, legal_split_count, all_splits_obligation, preselected_split_suffices_claim).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(decomposition_order, 12, [a, a, a, a, a, a, a, a, a, a, a, a, b], 78, all_legal_splits_must_be_addressed_not_a_preselected_decomposition, No)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (decomposition_order, 12, [a, a, a, a, a, a, a, a, a, a, a, a, b], 78, all_legal_splits_must_be_addressed_not_a_preselected_decomposition, No).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

88. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=single_split, complete model={"p":13,"word":["a","a","a","a","a","a","a","a","a","a","a","a","a","b"],"x_length":12,"y_length":1}. Give (mode, p, word, x_length, y_length, z_length, split_legal, pumping_quantifiers_resolved_claim, boundary).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(single_split, 13, [a, a, a, a, a, a, a, a, a, a, a, a, a, b], 12, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (single_split, 13, [a, a, a, a, a, a, a, a, a, a, a, a, a, b], 12, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

89. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=wrong_quantifiers, complete model={"p":14,"word":["a","a","a","a","a","a","a","a","a","a","a","a","a","a","b"]}. Give (mode, p, word, legal_split_count, adversarial_choice_order, reversed_quantifier_claim).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(wrong_quantifiers, 14, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 105, adversary_word_then_prover_split_then_adversary_i, No)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (wrong_quantifiers, 14, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 105, adversary_word_then_prover_split_then_adversary_i, No).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

90. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=bounded_words, complete model={"alphabet":["a","b","c"],"max_checked_length":5}. Give (mode, alphabet, max_checked_length, checked_word_count, regularity_proof_claim, boundary).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(bounded_words, [a, b, c], 5, 364, No, bounded_words_do_not_prove_regular)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (bounded_words, [a, b, c], 5, 364, No, bounded_words_do_not_prove_regular).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

91. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=decomposition_order, complete model={"p":16,"word":["a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","b"]}. Give (mode, p, word, legal_split_count, all_splits_obligation, preselected_split_suffices_claim).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(decomposition_order, 16, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 136, all_legal_splits_must_be_addressed_not_a_preselected_decomposition, No)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (decomposition_order, 16, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 136, all_legal_splits_must_be_addressed_not_a_preselected_decomposition, No).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

92. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=single_split, complete model={"p":17,"word":["a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","b"],"x_length":16,"y_length":1}. Give (mode, p, word, x_length, y_length, z_length, split_legal, pumping_quantifiers_resolved_claim, boundary).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(single_split, 17, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 16, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (single_split, 17, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 16, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

93. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=wrong_quantifiers, complete model={"p":18,"word":["a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","b"]}. Give (mode, p, word, legal_split_count, adversarial_choice_order, reversed_quantifier_claim).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(wrong_quantifiers, 18, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 171, adversary_word_then_prover_split_then_adversary_i, No)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (wrong_quantifiers, 18, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 171, adversary_word_then_prover_split_then_adversary_i, No).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

94. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=bounded_words, complete model={"alphabet":["a"],"max_checked_length":4}. Give (mode, alphabet, max_checked_length, checked_word_count, regularity_proof_claim, boundary).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(bounded_words, [a], 4, 5, No, bounded_words_do_not_prove_regular)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (bounded_words, [a], 4, 5, No, bounded_words_do_not_prove_regular).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

95. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=decomposition_order, complete model={"p":20,"word":["a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","b"]}. Give (mode, p, word, legal_split_count, all_splits_obligation, preselected_split_suffices_claim).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(decomposition_order, 20, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 210, all_legal_splits_must_be_addressed_not_a_preselected_decomposition, No)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (decomposition_order, 20, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 210, all_legal_splits_must_be_addressed_not_a_preselected_decomposition, No).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

96. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=single_split, complete model={"p":21,"word":["a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","b"],"x_length":20,"y_length":1}. Give (mode, p, word, x_length, y_length, z_length, split_legal, pumping_quantifiers_resolved_claim, boundary).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(single_split, 21, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 20, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (single_split, 21, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 20, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

97. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=wrong_quantifiers, complete model={"p":22,"word":["a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","b"]}. Give (mode, p, word, legal_split_count, adversarial_choice_order, reversed_quantifier_claim).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(wrong_quantifiers, 22, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 253, adversary_word_then_prover_split_then_adversary_i, No)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (wrong_quantifiers, 22, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 253, adversary_word_then_prover_split_then_adversary_i, No).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

98. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=bounded_words, complete model={"alphabet":["a","b"],"max_checked_length":3}. Give (mode, alphabet, max_checked_length, checked_word_count, regularity_proof_claim, boundary).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(bounded_words, [a, b], 3, 15, No, bounded_words_do_not_prove_regular)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (bounded_words, [a, b], 3, 15, No, bounded_words_do_not_prove_regular).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

99. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=decomposition_order, complete model={"p":24,"word":["a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","b"]}. Give (mode, p, word, legal_split_count, all_splits_obligation, preselected_split_suffices_claim).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(decomposition_order, 24, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 300, all_legal_splits_must_be_addressed_not_a_preselected_decomposition, No)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (decomposition_order, 24, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 300, all_legal_splits_must_be_addressed_not_a_preselected_decomposition, No).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

100. One Split or Bounded Words Cannot Resolve the Pumping Quantifiers or Prove Regularity

Use mode=single_split, complete model={"p":25,"word":["a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","a","b"],"x_length":24,"y_length":1}. Give (mode, p, word, x_length, y_length, z_length, split_legal, pumping_quantifiers_resolved_claim, boundary).

The four modes have9/6/6/6 fields. The pumping property is necessary, not sufficient, for regularity. No rejects the quantifier or method inference, not the possible legality of a single split.

View answer and explanation

(single_split, 25, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 24, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i)

A single split may be legal without resolving exists split forall i. The adversary chooses the word, the prover then chooses a split, and the adversary then chooses i. A nonregularity argument must address all legal splits, not a preselected one. Bounded word checks do not prove regularity, and the pumping property is necessary but not sufficient. The exact result, retaining the requested order, is (single_split, 25, [a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, b], 24, 1, 1, Yes, No, one_decomposition_does_not_resolve_exists_split_forall_i).

Reviewed step-by-step derivation

Pumping lemma for regular languages · qf-mod49-pump-boundary · candidate

Canonical question JSONL · English JSONL · Validation report