Arbitrary-Word Mod-Count Invariants and Exact Prefix-Trie DFA Constructions
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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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Regular language · qf-mod33-regular-finite-test-boundary · candidate
Canonical question JSONL · English JSONL · Validation report