# Arbitrary-Word Mod-Count Invariants and Exact Prefix-Trie DFA Constructions

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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-20261014-advance-20261102-r07-regular-language-foundations · 2026-10-14 · 100 questions

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



## 1. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=2. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1], [[0, 1], [1, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1], [[0, 1], [1, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0001/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 2. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=3. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2], [[0, 1], [1, 2], [2, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2], [[0, 1], [1, 2], [2, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0002/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 3. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=4. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3], [[0, 1], [1, 2], [2, 3], [3, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3], [[0, 1], [1, 2], [2, 3], [3, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0003/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 4. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=5. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0004/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 5. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=6. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0005/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 6. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=7. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0006/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 7. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=8. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0007/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 8. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=9. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7, 8], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0008/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 9. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=10. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0009/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 10. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=11. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0010/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 11. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=12. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0011/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 12. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=13. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0012/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 13. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=14. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0013/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 14. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=15. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0014/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 15. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=16. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0015/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 16. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=17. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0016/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 17. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=18. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0017/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 18. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=19. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 18], [18, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 18], [18, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0018/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 19. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=20. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 18], [18, 19], [19, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19], [[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9], [9, 10], [10, 11], [11, 12], [12, 13], [13, 14], [14, 15], [15, 16], [16, 17], [17, 18], [18, 19], [19, 0]], 0, [0], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0019/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 20. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=2. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1], [[0, 1, 0], [1, 0, 1]], 0, [1], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1], [[0, 1, 0], [1, 0, 1]], 0, [1], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0020/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 21. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=3. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2], [[0, 1, 0], [1, 2, 1], [2, 0, 2]], 0, [1], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2], [[0, 1, 0], [1, 2, 1], [2, 0, 2]], 0, [1], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0021/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 22. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=4. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 0, 3]], 0, [1], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 0, 3]], 0, [1], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0022/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 23. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=5. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 4, 3], [4, 0, 4]], 0, [1], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 4, 3], [4, 0, 4]], 0, [1], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0023/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 24. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=6. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 4, 3], [4, 5, 4], [5, 0, 5]], 0, [1], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 4, 3], [4, 5, 4], [5, 0, 5]], 0, [1], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0024/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 25. A State Invariant Covers Every Word, Not Just Test Cases

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=7. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Give (states,transition,start,accepting,regular?,state_equals_target_symbol_count_mod_modulus_for_all_prefixes). Rows follow state order and columns symbol order.

View answer and explanation([0, 1, 2, 3, 4, 5, 6], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 4, 3], [4, 5, 4], [5, 6, 5], [6, 0, 6]], 0, [1], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes)

The invariant holds initially with count zero and is preserved by each symbol transition. Induction over arbitrary word length makes the accepting residue characterize the whole declared language. This explicit DFA/invariant is the certificate, not a finite suite or a new Lean proof. The exact result, retaining the requested order, is ([0, 1, 2, 3, 4, 5, 6], [[0, 1, 0], [1, 2, 1], [2, 3, 2], [3, 4, 3], [4, 5, 4], [5, 6, 5], [6, 0, 6]], 0, [1], Yes, state_equals_target_symbol_count_mod_modulus_for_all_prefixes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0025/)

Regular language · qf-mod33-regular-mod-count-certificate · candidate





## 26. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=2. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[0, 0, 0, 0, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], 0, Yes)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], 0, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0026/)

Regular language · qf-mod33-regular-word-membership · candidate





## 27. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=3. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[1, 0, 0, 0, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([1, 0, 0, 0, 0], [0, 1, 1, 1, 1, 1], 1, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([1, 0, 0, 0, 0], [0, 1, 1, 1, 1, 1], 1, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0027/)

Regular language · qf-mod33-regular-word-membership · candidate





## 28. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=4. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[0, 1, 0, 0, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([0, 1, 0, 0, 0], [0, 0, 1, 1, 1, 1], 1, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([0, 1, 0, 0, 0], [0, 0, 1, 1, 1, 1], 1, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0028/)

Regular language · qf-mod33-regular-word-membership · candidate





## 29. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=5. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[1, 1, 0, 0, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([1, 1, 0, 0, 0], [0, 1, 2, 2, 2, 2], 2, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([1, 1, 0, 0, 0], [0, 1, 2, 2, 2, 2], 2, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0029/)

Regular language · qf-mod33-regular-word-membership · candidate





## 30. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=6. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[0, 0, 1, 0, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([0, 0, 1, 0, 0], [0, 0, 0, 1, 1, 1], 1, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([0, 0, 1, 0, 0], [0, 0, 0, 1, 1, 1], 1, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0030/)

Regular language · qf-mod33-regular-word-membership · candidate





## 31. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=7. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[1, 0, 1, 0, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([1, 0, 1, 0, 0], [0, 1, 1, 2, 2, 2], 2, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([1, 0, 1, 0, 0], [0, 1, 1, 2, 2, 2], 2, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0031/)

Regular language · qf-mod33-regular-word-membership · candidate





## 32. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=8. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[0, 1, 1, 0, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([0, 1, 1, 0, 0], [0, 0, 1, 2, 2, 2], 2, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([0, 1, 1, 0, 0], [0, 0, 1, 2, 2, 2], 2, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0032/)

Regular language · qf-mod33-regular-word-membership · candidate





## 33. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=9. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[1, 1, 1, 0, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([1, 1, 1, 0, 0], [0, 1, 2, 3, 3, 3], 3, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([1, 1, 1, 0, 0], [0, 1, 2, 3, 3, 3], 3, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0033/)

Regular language · qf-mod33-regular-word-membership · candidate





## 34. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=10. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[0, 0, 0, 1, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 1], 1, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 1], 1, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0034/)

Regular language · qf-mod33-regular-word-membership · candidate





## 35. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=11. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[1, 0, 0, 1, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([1, 0, 0, 1, 0], [0, 1, 1, 1, 2, 2], 2, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([1, 0, 0, 1, 0], [0, 1, 1, 1, 2, 2], 2, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0035/)

Regular language · qf-mod33-regular-word-membership · candidate





## 36. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=12. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[0, 1, 0, 1, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([0, 1, 0, 1, 0], [0, 0, 1, 1, 2, 2], 2, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([0, 1, 0, 1, 0], [0, 0, 1, 1, 2, 2], 2, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0036/)

Regular language · qf-mod33-regular-word-membership · candidate





## 37. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=13. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[1, 1, 0, 1, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([1, 1, 0, 1, 0], [0, 1, 2, 2, 3, 3], 3, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([1, 1, 0, 1, 0], [0, 1, 2, 2, 3, 3], 3, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0037/)

Regular language · qf-mod33-regular-word-membership · candidate





## 38. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=14. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[0, 0, 1, 1, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([0, 0, 1, 1, 0], [0, 0, 0, 1, 2, 2], 2, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([0, 0, 1, 1, 0], [0, 0, 0, 1, 2, 2], 2, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0038/)

Regular language · qf-mod33-regular-word-membership · candidate





## 39. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=15. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[1, 0, 1, 1, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([1, 0, 1, 1, 0], [0, 1, 1, 2, 3, 3], 3, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([1, 0, 1, 1, 0], [0, 1, 1, 2, 3, 3], 3, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0039/)

Regular language · qf-mod33-regular-word-membership · candidate





## 40. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=16. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[0, 1, 1, 1, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([0, 1, 1, 1, 0], [0, 0, 1, 2, 3, 3], 3, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([0, 1, 1, 1, 0], [0, 0, 1, 2, 3, 3], 3, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0040/)

Regular language · qf-mod33-regular-word-membership · candidate





## 41. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=17. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[1, 1, 1, 1, 0]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([1, 1, 1, 1, 0], [0, 1, 2, 3, 4, 4], 4, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([1, 1, 1, 1, 0], [0, 1, 2, 3, 4, 4], 4, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0041/)

Regular language · qf-mod33-regular-word-membership · candidate





## 42. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=18. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[0, 0, 0, 0, 1]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], 1, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 1], 1, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0042/)

Regular language · qf-mod33-regular-word-membership · candidate





## 43. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=19. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[1, 0, 0, 0, 1]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([1, 0, 0, 0, 1], [0, 1, 1, 1, 1, 2], 2, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([1, 0, 0, 0, 1], [0, 1, 1, 1, 1, 2], 2, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0043/)

Regular language · qf-mod33-regular-word-membership · candidate





## 44. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=20. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[0, 1, 0, 0, 1]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([0, 1, 0, 0, 1], [0, 0, 1, 1, 1, 2], 2, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([0, 1, 0, 0, 1], [0, 0, 1, 1, 1, 2], 2, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0044/)

Regular language · qf-mod33-regular-word-membership · candidate





## 45. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=2. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[1, 1, 0, 0, 1]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([1, 1, 0, 0, 1], [0, 1, 0, 0, 0, 1], 1, Yes)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([1, 1, 0, 0, 1], [0, 1, 0, 0, 0, 1], 1, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0045/)

Regular language · qf-mod33-regular-word-membership · candidate





## 46. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=3. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[0, 0, 1, 0, 1]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([0, 0, 1, 0, 1], [0, 0, 0, 1, 1, 2], 2, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([0, 0, 1, 0, 1], [0, 0, 0, 1, 1, 2], 2, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0046/)

Regular language · qf-mod33-regular-word-membership · candidate





## 47. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=4. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[1, 0, 1, 0, 1]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([1, 0, 1, 0, 1], [0, 1, 1, 2, 2, 3], 3, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([1, 0, 1, 0, 1], [0, 1, 1, 2, 2, 3], 3, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0047/)

Regular language · qf-mod33-regular-word-membership · candidate





## 48. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=5. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[0, 1, 1, 0, 1]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([0, 1, 1, 0, 1], [0, 0, 1, 2, 2, 3], 3, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([0, 1, 1, 0, 1], [0, 0, 1, 2, 2, 3], 3, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0048/)

Regular language · qf-mod33-regular-word-membership · candidate





## 49. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=6. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[1, 1, 1, 0, 1]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([1, 1, 1, 0, 1], [0, 1, 2, 3, 3, 4], 4, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([1, 1, 1, 0, 1], [0, 1, 2, 3, 3, 4], 4, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0049/)

Regular language · qf-mod33-regular-word-membership · candidate





## 50. Trace One Word under the Already Declared Mod-Count Language

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=7. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Word=[0, 0, 0, 1, 1]. Give (word,state_trace,final_state,accepted?), retaining the initial state and every transition.

View answer and explanation([0, 0, 0, 1, 1], [0, 0, 0, 0, 1, 2], 2, No)

Read symbols in their given order and accept exactly at the target residue. For the empty word, no transition occurs and the final state remains zero. This one membership result is not itself the full-language certificate. The exact result, retaining the requested order, is ([0, 0, 0, 1, 1], [0, 0, 0, 0, 1, 2], 2, No).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0050/)

Regular language · qf-mod33-regular-word-membership · candidate





## 51. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 0, 0, 0, 0, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[1, 0, 0, 0, 0, 0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 0, 0, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 0, 0, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0051/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 52. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 1, 0, 0, 0, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 0], [1, 1, 0, 0, 0], [1, 1, 0, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 0], [1, 1, 0, 0, 0], [1, 1, 0, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0052/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 53. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 0, 1, 0, 0, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[1, 0, 1, 0, 0, 0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 0], [1, 0, 1, 0, 0], [1, 0, 1, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 0, 0, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 0], [1, 0, 1, 0, 0], [1, 0, 1, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 0, 0, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0053/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 54. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 1, 1, 0, 0, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [5, 8], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0054/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 55. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 0, 0, 1, 0, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[1, 0, 0, 1, 0, 0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 1], [1, 0, 0, 1, 0], [1, 0, 0, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 1, 0, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 1], [1, 0, 0, 1, 0], [1, 0, 0, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 1, 0, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0055/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 56. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 1, 0, 1, 0, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 1], [1, 1, 0, 1, 0], [1, 1, 0, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 1], [1, 1, 0, 1, 0], [1, 1, 0, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0056/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 57. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 0, 1, 1, 0, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[1, 0, 1, 1, 0, 0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 1], [1, 0, 1, 1, 0], [1, 0, 1, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 1, 0, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 1], [1, 0, 1, 1, 0], [1, 0, 1, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 1, 0, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0057/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 58. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 1, 1, 1, 0, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1, 0], [1, 1, 1, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1, 0], [1, 1, 1, 1, 0, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [8, 5], [6, 8], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0058/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 59. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 0, 0, 0, 1, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[1, 0, 0, 0, 1, 0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 1], [1, 0, 0, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 0, 1, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 1], [1, 0, 0, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 0, 1, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0059/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 60. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 1, 0, 0, 1, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 0], [1, 1, 0, 0, 1], [1, 1, 0, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 0], [1, 1, 0, 0, 1], [1, 1, 0, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0060/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 61. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 0, 1, 0, 1, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[1, 0, 1, 0, 1, 0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 0], [1, 0, 1, 0, 1], [1, 0, 1, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 0, 1, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 0], [1, 0, 1, 0, 1], [1, 0, 1, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 0, 1, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0061/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 62. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 1, 1, 0, 1, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 0], [1, 1, 1, 0, 1], [1, 1, 1, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 0], [1, 1, 1, 0, 1], [1, 1, 1, 0, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [5, 8], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0062/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 63. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 0, 0, 1, 1, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[1, 0, 0, 1, 1, 0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 1], [1, 0, 0, 1, 1], [1, 0, 0, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 1, 1, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 1], [1, 0, 0, 1, 1], [1, 0, 0, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 0, 1, 1, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0063/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 64. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 1, 0, 1, 1, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 1], [1, 1, 0, 1, 1], [1, 1, 0, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 1], [1, 1, 0, 1, 1], [1, 1, 0, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0064/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 65. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 0, 1, 1, 1, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[1, 0, 1, 1, 1, 0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 1], [1, 0, 1, 1, 1], [1, 0, 1, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 1, 1, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 1], [1, 0, 1, 1, 1], [1, 0, 1, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [1, 0, 1, 1, 1, 0], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0065/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 66. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 1, 1, 1, 1, 0]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 0]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [8, 5], [8, 6], [7, 8], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0066/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 67. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 0, 0, 0, 0, 1]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[1, 0, 0, 0, 0, 1]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 0, 0, 0, 1], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 0, 0, 0, 1], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0067/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 68. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 1, 0, 0, 0, 1]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 0], [1, 1, 0, 0, 0], [1, 1, 0, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 0], [1, 1, 0, 0, 0], [1, 1, 0, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0068/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 69. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 0, 1, 0, 0, 1]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[1, 0, 1, 0, 0, 1]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 0], [1, 0, 1, 0, 0], [1, 0, 1, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 1, 0, 0, 1], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 0], [1, 0, 1, 0, 0], [1, 0, 1, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 1, 0, 0, 1], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0069/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 70. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 1, 1, 0, 0, 1]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 0], [1, 1, 1, 0, 0], [1, 1, 1, 0, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [5, 8], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0070/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 71. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 0, 0, 1, 0, 1]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[1, 0, 0, 1, 0, 1]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 1], [1, 0, 0, 1, 0], [1, 0, 0, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 0, 1, 0, 1], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 1], [1, 0, 0, 1, 0], [1, 0, 0, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 0, 1, 0, 1], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0071/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 72. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 1, 0, 1, 0, 1]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 1], [1, 1, 0, 1, 0], [1, 1, 0, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 1], [1, 1, 0], [1, 1, 0, 1], [1, 1, 0, 1, 0], [1, 1, 0, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [4, 8], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0072/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 73. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 0, 1, 1, 0, 1]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[1, 0, 1, 1, 0, 1]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 1], [1, 0, 1, 1, 0], [1, 0, 1, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 1, 1, 0, 1], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 0], [1, 0, 1], [1, 0, 1, 1], [1, 0, 1, 1, 0], [1, 0, 1, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [8, 4], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [1, 0, 1, 1, 0, 1], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0073/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 74. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 1, 1, 1, 0, 1]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[0]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1, 0], [1, 1, 1, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 1], [1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1, 0], [1, 1, 1, 1, 0, 1]], 8, [1, 7], [[1, 2], [8, 8], [8, 3], [8, 4], [8, 5], [6, 8], [8, 7], [8, 8], [8, 8]], [0], [0, 1], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0074/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 75. Construct a Complete Prefix-Trie DFA for an Exact Finite Language

Use alphabet=Fin(2)={0,…,1} and the complete nonempty duplicate-free word family=[[0], [1, 0, 0, 0, 1, 1]]. The family may include the empty word; this constructor does not handle an empty family. Probe=[1, 0, 0, 0, 1, 1]. Give (prefix_states,dead_state,accepting,transition,probe,trace,accepted?,explicit_DFA_recognizes_exact_finite_language).

View answer and explanation([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 1], [1, 0, 0, 0, 1, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [8, 6], [8, 7], [8, 8], [8, 8]], [1, 0, 0, 0, 1, 1], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language)

Collect all prefixes including the empty prefix, sort first by length then by their JSON lexicographic order, and use those positions as states. Append a dead state, send absent extensions to it and loop there. Accept exactly the states equal to full supplied words. Retain all table rows and the complete probe trace. This construction covers arbitrary words of the declared alphabet without claiming minimization. The exact result, retaining the requested order, is ([[], [0], [1], [1, 0], [1, 0, 0], [1, 0, 0, 0], [1, 0, 0, 0, 1], [1, 0, 0, 0, 1, 1]], 8, [1, 7], [[1, 2], [8, 8], [3, 8], [4, 8], [5, 8], [8, 6], [8, 7], [8, 8], [8, 8]], [1, 0, 0, 0, 1, 1], [0, 2, 3, 4, 5, 6, 7], Yes, explicit_DFA_recognizes_exact_finite_language).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0075/)

Regular language · qf-mod33-regular-finite-language-construction · candidate





## 76. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=2. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [0, 0, 0, 0, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[0, 0, 0, 0, 0], Yes]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[0, 0, 0, 0, 0], Yes]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0076/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 77. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=3. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [1, 0, 0, 0, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[1, 0, 0, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[1, 0, 0, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0077/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 78. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=4. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [0, 1, 0, 0, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[0, 1, 0, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[0, 1, 0, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0078/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 79. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=5. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [1, 1, 0, 0, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[1, 1, 0, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[1, 1, 0, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0079/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 80. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=6. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [0, 0, 1, 0, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[0, 0, 1, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[0, 0, 1, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0080/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 81. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=7. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [1, 0, 1, 0, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[1, 0, 1, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[1, 0, 1, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0081/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 82. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=8. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [0, 1, 1, 0, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[0, 1, 1, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[0, 1, 1, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0082/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 83. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=9. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [1, 1, 1, 0, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[1, 1, 1, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[1, 1, 1, 0, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0083/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 84. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=10. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [0, 0, 0, 1, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[0, 0, 0, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[0, 0, 0, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0084/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 85. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=11. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [1, 0, 0, 1, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[1, 0, 0, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[1, 0, 0, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0085/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 86. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=12. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [0, 1, 0, 1, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[0, 1, 0, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[0, 1, 0, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0086/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 87. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=13. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [1, 1, 0, 1, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[1, 1, 0, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[1, 1, 0, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0087/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 88. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=14. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [0, 0, 1, 1, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[0, 0, 1, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[0, 0, 1, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0088/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 89. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=15. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [1, 0, 1, 1, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[1, 0, 1, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[1, 0, 1, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0089/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 90. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=16. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [0, 1, 1, 1, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[0, 1, 1, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[0, 1, 1, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0090/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 91. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=17. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [1, 1, 1, 1, 0]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[1, 1, 1, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[1, 1, 1, 1, 0], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0091/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 92. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=18. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [0, 0, 0, 0, 1]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[0, 0, 0, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[0, 0, 0, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0092/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 93. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=19. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [1, 0, 0, 0, 1]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[1, 0, 0, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[1, 0, 0, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0093/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 94. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(2)={0,…,1} and the language whose number of occurrences of target_symbol=1 is congruent to residue=0 modulo modulus=20. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [0, 1, 0, 0, 1]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], Yes], [[0, 1, 0, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], Yes], [[0, 1, 0, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0094/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 95. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=2. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [1, 1, 0, 0, 1]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], No], [[1, 1, 0, 0, 1], Yes]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], No], [[1, 1, 0, 0, 1], Yes]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0095/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 96. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=3. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [0, 0, 1, 0, 1]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], No], [[0, 0, 1, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], No], [[0, 0, 1, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0096/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 97. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=4. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [1, 0, 1, 0, 1]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], No], [[1, 0, 1, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], No], [[1, 0, 1, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0097/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 98. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=5. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [0, 1, 1, 0, 1]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], No], [[0, 1, 1, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], No], [[0, 1, 1, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0098/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 99. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=6. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [1, 1, 1, 0, 1]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], No], [[1, 1, 1, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], No], [[1, 1, 1, 0, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0099/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate





## 100. A Finite Test Suite Alone Is Not a Language or Equivalence Certificate

Use alphabet=Fin(3)={0,…,2} and the language whose number of occurrences of target_symbol=1 is congruent to residue=1 modulo modulus=7. Construct the complete DFA on states 0 through modulus−1, start 0 and accepting subset [residue]. A target symbol increments the state modulo the modulus; every other symbol leaves it unchanged. The invariant is state equals target-symbol count modulo the modulus for every prefix of every finite word, not just selected tests. No regex, NFA, epsilon move, minimization or pumping-lemma result is added. Ordered test_words=[[], [0, 0, 0, 1, 1]]. Give (tested_words,sufficient?,finite_test_suite_not_regular_language_or_equivalence_certificate), with each tested row [word,accepted?].

View answer and explanation([[[], No], [[0, 0, 0, 1, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate)

Report membership of each test under the supplied DFA. The No concerns inferring full-language regularity or equivalence from tests alone, not denial of the separate explicit DFA/invariant certificate already available for this particular model. The exact result, retaining the requested order, is ([[[], No], [[0, 0, 0, 1, 1], No]], No, finite_test_suite_not_regular_language_or_equivalence_certificate).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r07-regular-language-0100/)

Regular language · qf-mod33-regular-finite-test-boundary · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261102-r07-regular-language-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261014-advance-20261102-r07-regular-language-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261102-r07-regular-language-foundations/validation-report.json)