# Complete Grammars, Explicit Derivations and Bounded-Search Nonconclusions

This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.

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

batch-20261014-advance-20261109-r07-context-free-grammar-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.

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

[Field-order and source-binding data](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S0, A0], terminals=[a0, b0], start=S0, productions=[[S0, [a0, A0]], [A0, [b0, A0]], [A0, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S0, A0], [a0, b0], S0, [[S0, [a0, A0]], [A0, [b0, A0]], [A0, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S0, A0], [a0, b0], S0, [[S0, [a0, A0]], [A0, [b0, A0]], [A0, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0001/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 2. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S1, A1], terminals=[a1, b1], start=S1, productions=[[S1, [a1, A1]], [A1, [b1, A1]], [A1, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S1, A1], [a1, b1], S1, [[S1, [a1, A1]], [A1, [b1, A1]], [A1, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S1, A1], [a1, b1], S1, [[S1, [a1, A1]], [A1, [b1, A1]], [A1, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0002/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 3. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S2, A2], terminals=[a2, b2], start=S2, productions=[[S2, [a2, A2]], [A2, [b2, A2]], [A2, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S2, A2], [a2, b2], S2, [[S2, [a2, A2]], [A2, [b2, A2]], [A2, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S2, A2], [a2, b2], S2, [[S2, [a2, A2]], [A2, [b2, A2]], [A2, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0003/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 4. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S3, A3], terminals=[a3, b3], start=S3, productions=[[S3, [a3, A3]], [A3, [b3, A3]], [A3, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S3, A3], [a3, b3], S3, [[S3, [a3, A3]], [A3, [b3, A3]], [A3, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S3, A3], [a3, b3], S3, [[S3, [a3, A3]], [A3, [b3, A3]], [A3, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0004/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 5. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S4, A4], terminals=[a4, b4], start=S4, productions=[[S4, [a4, A4]], [A4, [b4, A4]], [A4, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S4, A4], [a4, b4], S4, [[S4, [a4, A4]], [A4, [b4, A4]], [A4, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S4, A4], [a4, b4], S4, [[S4, [a4, A4]], [A4, [b4, A4]], [A4, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0005/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 6. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S5, A5], terminals=[a5, b5], start=S5, productions=[[S5, [a5, A5]], [A5, [b5, A5]], [A5, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S5, A5], [a5, b5], S5, [[S5, [a5, A5]], [A5, [b5, A5]], [A5, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S5, A5], [a5, b5], S5, [[S5, [a5, A5]], [A5, [b5, A5]], [A5, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0006/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 7. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S6, A6], terminals=[a6, b6], start=S6, productions=[[S6, [a6, A6]], [A6, [b6, A6]], [A6, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S6, A6], [a6, b6], S6, [[S6, [a6, A6]], [A6, [b6, A6]], [A6, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S6, A6], [a6, b6], S6, [[S6, [a6, A6]], [A6, [b6, A6]], [A6, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0007/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 8. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S7, A7], terminals=[a7, b7], start=S7, productions=[[S7, [a7, A7]], [A7, [b7, A7]], [A7, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S7, A7], [a7, b7], S7, [[S7, [a7, A7]], [A7, [b7, A7]], [A7, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S7, A7], [a7, b7], S7, [[S7, [a7, A7]], [A7, [b7, A7]], [A7, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0008/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 9. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S8, A8], terminals=[a8, b8], start=S8, productions=[[S8, [a8, A8]], [A8, [b8, A8]], [A8, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S8, A8], [a8, b8], S8, [[S8, [a8, A8]], [A8, [b8, A8]], [A8, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S8, A8], [a8, b8], S8, [[S8, [a8, A8]], [A8, [b8, A8]], [A8, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0009/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 10. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S9, A9], terminals=[a9, b9], start=S9, productions=[[S9, [a9, A9]], [A9, [b9, A9]], [A9, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S9, A9], [a9, b9], S9, [[S9, [a9, A9]], [A9, [b9, A9]], [A9, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S9, A9], [a9, b9], S9, [[S9, [a9, A9]], [A9, [b9, A9]], [A9, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0010/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 11. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S10, A10], terminals=[a10, b10], start=S10, productions=[[S10, [a10, A10]], [A10, [b10, A10]], [A10, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S10, A10], [a10, b10], S10, [[S10, [a10, A10]], [A10, [b10, A10]], [A10, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S10, A10], [a10, b10], S10, [[S10, [a10, A10]], [A10, [b10, A10]], [A10, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0011/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 12. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S11, A11], terminals=[a11, b11], start=S11, productions=[[S11, [a11, A11]], [A11, [b11, A11]], [A11, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S11, A11], [a11, b11], S11, [[S11, [a11, A11]], [A11, [b11, A11]], [A11, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S11, A11], [a11, b11], S11, [[S11, [a11, A11]], [A11, [b11, A11]], [A11, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0012/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 13. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S12, A12], terminals=[a12, b12], start=S12, productions=[[S12, [a12, A12]], [A12, [b12, A12]], [A12, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S12, A12], [a12, b12], S12, [[S12, [a12, A12]], [A12, [b12, A12]], [A12, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S12, A12], [a12, b12], S12, [[S12, [a12, A12]], [A12, [b12, A12]], [A12, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0013/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 14. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S13, A13], terminals=[a13, b13], start=S13, productions=[[S13, [a13, A13]], [A13, [b13, A13]], [A13, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S13, A13], [a13, b13], S13, [[S13, [a13, A13]], [A13, [b13, A13]], [A13, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S13, A13], [a13, b13], S13, [[S13, [a13, A13]], [A13, [b13, A13]], [A13, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0014/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 15. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S14, A14], terminals=[a14, b14], start=S14, productions=[[S14, [a14, A14]], [A14, [b14, A14]], [A14, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S14, A14], [a14, b14], S14, [[S14, [a14, A14]], [A14, [b14, A14]], [A14, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S14, A14], [a14, b14], S14, [[S14, [a14, A14]], [A14, [b14, A14]], [A14, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0015/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 16. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S15, A15], terminals=[a15, b15], start=S15, productions=[[S15, [a15, A15]], [A15, [b15, A15]], [A15, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S15, A15], [a15, b15], S15, [[S15, [a15, A15]], [A15, [b15, A15]], [A15, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S15, A15], [a15, b15], S15, [[S15, [a15, A15]], [A15, [b15, A15]], [A15, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0016/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 17. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S16, A16], terminals=[a16, b16], start=S16, productions=[[S16, [a16, A16]], [A16, [b16, A16]], [A16, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S16, A16], [a16, b16], S16, [[S16, [a16, A16]], [A16, [b16, A16]], [A16, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S16, A16], [a16, b16], S16, [[S16, [a16, A16]], [A16, [b16, A16]], [A16, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0017/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 18. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S17, A17], terminals=[a17, b17], start=S17, productions=[[S17, [a17, A17]], [A17, [b17, A17]], [A17, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S17, A17], [a17, b17], S17, [[S17, [a17, A17]], [A17, [b17, A17]], [A17, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S17, A17], [a17, b17], S17, [[S17, [a17, A17]], [A17, [b17, A17]], [A17, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0018/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 19. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S18, A18], terminals=[a18, b18], start=S18, productions=[[S18, [a18, A18]], [A18, [b18, A18]], [A18, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S18, A18], [a18, b18], S18, [[S18, [a18, A18]], [A18, [b18, A18]], [A18, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S18, A18], [a18, b18], S18, [[S18, [a18, A18]], [A18, [b18, A18]], [A18, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0019/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 20. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S19, A19], terminals=[a19, b19], start=S19, productions=[[S19, [a19, A19]], [A19, [b19, A19]], [A19, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S19, A19], [a19, b19], S19, [[S19, [a19, A19]], [A19, [b19, A19]], [A19, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S19, A19], [a19, b19], S19, [[S19, [a19, A19]], [A19, [b19, A19]], [A19, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0020/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 21. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S20, A20], terminals=[a20, b20], start=S20, productions=[[S20, [a20, A20]], [A20, [b20, A20]], [A20, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S20, A20], [a20, b20], S20, [[S20, [a20, A20]], [A20, [b20, A20]], [A20, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S20, A20], [a20, b20], S20, [[S20, [a20, A20]], [A20, [b20, A20]], [A20, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0021/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 22. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S21, A21], terminals=[a21, b21], start=S21, productions=[[S21, [a21, A21]], [A21, [b21, A21]], [A21, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S21, A21], [a21, b21], S21, [[S21, [a21, A21]], [A21, [b21, A21]], [A21, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S21, A21], [a21, b21], S21, [[S21, [a21, A21]], [A21, [b21, A21]], [A21, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0022/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 23. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S22, A22], terminals=[a22, b22], start=S22, productions=[[S22, [a22, A22]], [A22, [b22, A22]], [A22, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S22, A22], [a22, b22], S22, [[S22, [a22, A22]], [A22, [b22, A22]], [A22, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S22, A22], [a22, b22], S22, [[S22, [a22, A22]], [A22, [b22, A22]], [A22, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0023/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 24. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S23, A23], terminals=[a23, b23], start=S23, productions=[[S23, [a23, A23]], [A23, [b23, A23]], [A23, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S23, A23], [a23, b23], S23, [[S23, [a23, A23]], [A23, [b23, A23]], [A23, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S23, A23], [a23, b23], S23, [[S23, [a23, A23]], [A23, [b23, A23]], [A23, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0024/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 25. Grammar Qualification Requires Complete Alphabets, Start and Productions

Use the complete finite grammar: nonterminals=[S24, A24], terminals=[a24, b24], start=S24, productions=[[S24, [a24, A24]], [A24, [b24, A24]], [A24, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Give (nonterminals, terminals, start, productions, allow_epsilon, grammar_qualified, certificate).

View answer and explanation([S24, A24], [a24, b24], S24, [[S24, [a24, A24]], [A24, [b24, A24]], [A24, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs)

Report the original alphabets, start, production list and explicit epsilon flag, followed by the qualification certificate. A legal grammar definition alone is not a derivation or a word-membership witness. The exact result, retaining the requested order, is ([S24, A24], [a24, b24], S24, [[S24, [a24, A24]], [A24, [b24, A24]], [A24, []]], Yes, Yes, finite_disjoint_alphabets_single_nonterminal_lhs).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0025/)

Context-free grammar · qf-mod40-cfg-qualification · candidate





## 26. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S0, A0], terminals=[a0, b0], start=S0, productions=[[S0, [a0, A0]], [A0, [b0, A0]], [A0, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S0], [a0, A0], [a0]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S0], [a0, A0], [a0]], [[0, 0, S0, [a0, A0]], [1, 2, A0, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S0], [a0, A0], [a0]], [[0, 0, S0, [a0, A0]], [1, 2, A0, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0026/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 27. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S1, A1], terminals=[a1, b1], start=S1, productions=[[S1, [a1, A1]], [A1, [b1, A1]], [A1, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S1], [a1, A1], [a1, b1, A1], [a1, b1]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S1], [a1, A1], [a1, b1, A1], [a1, b1]], [[0, 0, S1, [a1, A1]], [1, 1, A1, [b1, A1]], [2, 2, A1, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S1], [a1, A1], [a1, b1, A1], [a1, b1]], [[0, 0, S1, [a1, A1]], [1, 1, A1, [b1, A1]], [2, 2, A1, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0027/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 28. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S2, A2], terminals=[a2, b2], start=S2, productions=[[S2, [a2, A2]], [A2, [b2, A2]], [A2, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S2], [a2, A2], [a2, b2, A2], [a2, b2, b2, A2], [a2, b2, b2]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S2], [a2, A2], [a2, b2, A2], [a2, b2, b2, A2], [a2, b2, b2]], [[0, 0, S2, [a2, A2]], [1, 1, A2, [b2, A2]], [2, 1, A2, [b2, A2]], [3, 2, A2, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S2], [a2, A2], [a2, b2, A2], [a2, b2, b2, A2], [a2, b2, b2]], [[0, 0, S2, [a2, A2]], [1, 1, A2, [b2, A2]], [2, 1, A2, [b2, A2]], [3, 2, A2, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0028/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 29. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S3, A3], terminals=[a3, b3], start=S3, productions=[[S3, [a3, A3]], [A3, [b3, A3]], [A3, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S3], [a3, A3], [a3, b3, A3], [a3, b3, b3, A3], [a3, b3, b3, b3, A3], [a3, b3, b3, b3]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S3], [a3, A3], [a3, b3, A3], [a3, b3, b3, A3], [a3, b3, b3, b3, A3], [a3, b3, b3, b3]], [[0, 0, S3, [a3, A3]], [1, 1, A3, [b3, A3]], [2, 1, A3, [b3, A3]], [3, 1, A3, [b3, A3]], [4, 2, A3, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S3], [a3, A3], [a3, b3, A3], [a3, b3, b3, A3], [a3, b3, b3, b3, A3], [a3, b3, b3, b3]], [[0, 0, S3, [a3, A3]], [1, 1, A3, [b3, A3]], [2, 1, A3, [b3, A3]], [3, 1, A3, [b3, A3]], [4, 2, A3, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0029/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 30. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S4, A4], terminals=[a4, b4], start=S4, productions=[[S4, [a4, A4]], [A4, [b4, A4]], [A4, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S4], [a4, A4], [a4, b4, A4], [a4, b4, b4, A4], [a4, b4, b4, b4, A4], [a4, b4, b4, b4, b4, A4], [a4, b4, b4, b4, b4]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S4], [a4, A4], [a4, b4, A4], [a4, b4, b4, A4], [a4, b4, b4, b4, A4], [a4, b4, b4, b4, b4, A4], [a4, b4, b4, b4, b4]], [[0, 0, S4, [a4, A4]], [1, 1, A4, [b4, A4]], [2, 1, A4, [b4, A4]], [3, 1, A4, [b4, A4]], [4, 1, A4, [b4, A4]], [5, 2, A4, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S4], [a4, A4], [a4, b4, A4], [a4, b4, b4, A4], [a4, b4, b4, b4, A4], [a4, b4, b4, b4, b4, A4], [a4, b4, b4, b4, b4]], [[0, 0, S4, [a4, A4]], [1, 1, A4, [b4, A4]], [2, 1, A4, [b4, A4]], [3, 1, A4, [b4, A4]], [4, 1, A4, [b4, A4]], [5, 2, A4, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0030/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 31. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S5, A5], terminals=[a5, b5], start=S5, productions=[[S5, [a5, A5]], [A5, [b5, A5]], [A5, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S5], [a5, A5], [a5]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S5], [a5, A5], [a5]], [[0, 0, S5, [a5, A5]], [1, 2, A5, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S5], [a5, A5], [a5]], [[0, 0, S5, [a5, A5]], [1, 2, A5, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0031/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 32. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S6, A6], terminals=[a6, b6], start=S6, productions=[[S6, [a6, A6]], [A6, [b6, A6]], [A6, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S6], [a6, A6], [a6, b6, A6], [a6, b6]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S6], [a6, A6], [a6, b6, A6], [a6, b6]], [[0, 0, S6, [a6, A6]], [1, 1, A6, [b6, A6]], [2, 2, A6, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S6], [a6, A6], [a6, b6, A6], [a6, b6]], [[0, 0, S6, [a6, A6]], [1, 1, A6, [b6, A6]], [2, 2, A6, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0032/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 33. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S7, A7], terminals=[a7, b7], start=S7, productions=[[S7, [a7, A7]], [A7, [b7, A7]], [A7, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S7], [a7, A7], [a7, b7, A7], [a7, b7, b7, A7], [a7, b7, b7]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S7], [a7, A7], [a7, b7, A7], [a7, b7, b7, A7], [a7, b7, b7]], [[0, 0, S7, [a7, A7]], [1, 1, A7, [b7, A7]], [2, 1, A7, [b7, A7]], [3, 2, A7, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S7], [a7, A7], [a7, b7, A7], [a7, b7, b7, A7], [a7, b7, b7]], [[0, 0, S7, [a7, A7]], [1, 1, A7, [b7, A7]], [2, 1, A7, [b7, A7]], [3, 2, A7, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0033/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 34. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S8, A8], terminals=[a8, b8], start=S8, productions=[[S8, [a8, A8]], [A8, [b8, A8]], [A8, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S8], [a8, A8], [a8, b8, A8], [a8, b8, b8, A8], [a8, b8, b8, b8, A8], [a8, b8, b8, b8]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S8], [a8, A8], [a8, b8, A8], [a8, b8, b8, A8], [a8, b8, b8, b8, A8], [a8, b8, b8, b8]], [[0, 0, S8, [a8, A8]], [1, 1, A8, [b8, A8]], [2, 1, A8, [b8, A8]], [3, 1, A8, [b8, A8]], [4, 2, A8, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S8], [a8, A8], [a8, b8, A8], [a8, b8, b8, A8], [a8, b8, b8, b8, A8], [a8, b8, b8, b8]], [[0, 0, S8, [a8, A8]], [1, 1, A8, [b8, A8]], [2, 1, A8, [b8, A8]], [3, 1, A8, [b8, A8]], [4, 2, A8, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0034/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 35. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S9, A9], terminals=[a9, b9], start=S9, productions=[[S9, [a9, A9]], [A9, [b9, A9]], [A9, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S9], [a9, A9], [a9, b9, A9], [a9, b9, b9, A9], [a9, b9, b9, b9, A9], [a9, b9, b9, b9, b9, A9], [a9, b9, b9, b9, b9]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S9], [a9, A9], [a9, b9, A9], [a9, b9, b9, A9], [a9, b9, b9, b9, A9], [a9, b9, b9, b9, b9, A9], [a9, b9, b9, b9, b9]], [[0, 0, S9, [a9, A9]], [1, 1, A9, [b9, A9]], [2, 1, A9, [b9, A9]], [3, 1, A9, [b9, A9]], [4, 1, A9, [b9, A9]], [5, 2, A9, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S9], [a9, A9], [a9, b9, A9], [a9, b9, b9, A9], [a9, b9, b9, b9, A9], [a9, b9, b9, b9, b9, A9], [a9, b9, b9, b9, b9]], [[0, 0, S9, [a9, A9]], [1, 1, A9, [b9, A9]], [2, 1, A9, [b9, A9]], [3, 1, A9, [b9, A9]], [4, 1, A9, [b9, A9]], [5, 2, A9, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0035/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 36. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S10, A10], terminals=[a10, b10], start=S10, productions=[[S10, [a10, A10]], [A10, [b10, A10]], [A10, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S10], [a10, A10], [a10]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S10], [a10, A10], [a10]], [[0, 0, S10, [a10, A10]], [1, 2, A10, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S10], [a10, A10], [a10]], [[0, 0, S10, [a10, A10]], [1, 2, A10, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0036/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 37. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S11, A11], terminals=[a11, b11], start=S11, productions=[[S11, [a11, A11]], [A11, [b11, A11]], [A11, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S11], [a11, A11], [a11, b11, A11], [a11, b11]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S11], [a11, A11], [a11, b11, A11], [a11, b11]], [[0, 0, S11, [a11, A11]], [1, 1, A11, [b11, A11]], [2, 2, A11, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S11], [a11, A11], [a11, b11, A11], [a11, b11]], [[0, 0, S11, [a11, A11]], [1, 1, A11, [b11, A11]], [2, 2, A11, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0037/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 38. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S12, A12], terminals=[a12, b12], start=S12, productions=[[S12, [a12, A12]], [A12, [b12, A12]], [A12, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S12], [a12, A12], [a12, b12, A12], [a12, b12, b12, A12], [a12, b12, b12]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S12], [a12, A12], [a12, b12, A12], [a12, b12, b12, A12], [a12, b12, b12]], [[0, 0, S12, [a12, A12]], [1, 1, A12, [b12, A12]], [2, 1, A12, [b12, A12]], [3, 2, A12, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S12], [a12, A12], [a12, b12, A12], [a12, b12, b12, A12], [a12, b12, b12]], [[0, 0, S12, [a12, A12]], [1, 1, A12, [b12, A12]], [2, 1, A12, [b12, A12]], [3, 2, A12, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0038/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 39. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S13, A13], terminals=[a13, b13], start=S13, productions=[[S13, [a13, A13]], [A13, [b13, A13]], [A13, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S13], [a13, A13], [a13, b13, A13], [a13, b13, b13, A13], [a13, b13, b13, b13, A13], [a13, b13, b13, b13]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S13], [a13, A13], [a13, b13, A13], [a13, b13, b13, A13], [a13, b13, b13, b13, A13], [a13, b13, b13, b13]], [[0, 0, S13, [a13, A13]], [1, 1, A13, [b13, A13]], [2, 1, A13, [b13, A13]], [3, 1, A13, [b13, A13]], [4, 2, A13, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S13], [a13, A13], [a13, b13, A13], [a13, b13, b13, A13], [a13, b13, b13, b13, A13], [a13, b13, b13, b13]], [[0, 0, S13, [a13, A13]], [1, 1, A13, [b13, A13]], [2, 1, A13, [b13, A13]], [3, 1, A13, [b13, A13]], [4, 2, A13, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0039/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 40. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S14, A14], terminals=[a14, b14], start=S14, productions=[[S14, [a14, A14]], [A14, [b14, A14]], [A14, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S14], [a14, A14], [a14, b14, A14], [a14, b14, b14, A14], [a14, b14, b14, b14, A14], [a14, b14, b14, b14, b14, A14], [a14, b14, b14, b14, b14]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S14], [a14, A14], [a14, b14, A14], [a14, b14, b14, A14], [a14, b14, b14, b14, A14], [a14, b14, b14, b14, b14, A14], [a14, b14, b14, b14, b14]], [[0, 0, S14, [a14, A14]], [1, 1, A14, [b14, A14]], [2, 1, A14, [b14, A14]], [3, 1, A14, [b14, A14]], [4, 1, A14, [b14, A14]], [5, 2, A14, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S14], [a14, A14], [a14, b14, A14], [a14, b14, b14, A14], [a14, b14, b14, b14, A14], [a14, b14, b14, b14, b14, A14], [a14, b14, b14, b14, b14]], [[0, 0, S14, [a14, A14]], [1, 1, A14, [b14, A14]], [2, 1, A14, [b14, A14]], [3, 1, A14, [b14, A14]], [4, 1, A14, [b14, A14]], [5, 2, A14, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0040/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 41. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S15, A15], terminals=[a15, b15], start=S15, productions=[[S15, [a15, A15]], [A15, [b15, A15]], [A15, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S15], [a15, A15], [a15]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S15], [a15, A15], [a15]], [[0, 0, S15, [a15, A15]], [1, 2, A15, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S15], [a15, A15], [a15]], [[0, 0, S15, [a15, A15]], [1, 2, A15, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0041/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 42. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S16, A16], terminals=[a16, b16], start=S16, productions=[[S16, [a16, A16]], [A16, [b16, A16]], [A16, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S16], [a16, A16], [a16, b16, A16], [a16, b16]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S16], [a16, A16], [a16, b16, A16], [a16, b16]], [[0, 0, S16, [a16, A16]], [1, 1, A16, [b16, A16]], [2, 2, A16, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S16], [a16, A16], [a16, b16, A16], [a16, b16]], [[0, 0, S16, [a16, A16]], [1, 1, A16, [b16, A16]], [2, 2, A16, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0042/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 43. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S17, A17], terminals=[a17, b17], start=S17, productions=[[S17, [a17, A17]], [A17, [b17, A17]], [A17, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S17], [a17, A17], [a17, b17, A17], [a17, b17, b17, A17], [a17, b17, b17]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S17], [a17, A17], [a17, b17, A17], [a17, b17, b17, A17], [a17, b17, b17]], [[0, 0, S17, [a17, A17]], [1, 1, A17, [b17, A17]], [2, 1, A17, [b17, A17]], [3, 2, A17, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S17], [a17, A17], [a17, b17, A17], [a17, b17, b17, A17], [a17, b17, b17]], [[0, 0, S17, [a17, A17]], [1, 1, A17, [b17, A17]], [2, 1, A17, [b17, A17]], [3, 2, A17, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0043/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 44. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S18, A18], terminals=[a18, b18], start=S18, productions=[[S18, [a18, A18]], [A18, [b18, A18]], [A18, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S18], [a18, A18], [a18, b18, A18], [a18, b18, b18, A18], [a18, b18, b18, b18, A18], [a18, b18, b18, b18]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S18], [a18, A18], [a18, b18, A18], [a18, b18, b18, A18], [a18, b18, b18, b18, A18], [a18, b18, b18, b18]], [[0, 0, S18, [a18, A18]], [1, 1, A18, [b18, A18]], [2, 1, A18, [b18, A18]], [3, 1, A18, [b18, A18]], [4, 2, A18, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S18], [a18, A18], [a18, b18, A18], [a18, b18, b18, A18], [a18, b18, b18, b18, A18], [a18, b18, b18, b18]], [[0, 0, S18, [a18, A18]], [1, 1, A18, [b18, A18]], [2, 1, A18, [b18, A18]], [3, 1, A18, [b18, A18]], [4, 2, A18, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0044/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 45. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S19, A19], terminals=[a19, b19], start=S19, productions=[[S19, [a19, A19]], [A19, [b19, A19]], [A19, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S19], [a19, A19], [a19, b19, A19], [a19, b19, b19, A19], [a19, b19, b19, b19, A19], [a19, b19, b19, b19, b19, A19], [a19, b19, b19, b19, b19]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S19], [a19, A19], [a19, b19, A19], [a19, b19, b19, A19], [a19, b19, b19, b19, A19], [a19, b19, b19, b19, b19, A19], [a19, b19, b19, b19, b19]], [[0, 0, S19, [a19, A19]], [1, 1, A19, [b19, A19]], [2, 1, A19, [b19, A19]], [3, 1, A19, [b19, A19]], [4, 1, A19, [b19, A19]], [5, 2, A19, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S19], [a19, A19], [a19, b19, A19], [a19, b19, b19, A19], [a19, b19, b19, b19, A19], [a19, b19, b19, b19, b19, A19], [a19, b19, b19, b19, b19]], [[0, 0, S19, [a19, A19]], [1, 1, A19, [b19, A19]], [2, 1, A19, [b19, A19]], [3, 1, A19, [b19, A19]], [4, 1, A19, [b19, A19]], [5, 2, A19, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0045/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 46. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S20, A20], terminals=[a20, b20], start=S20, productions=[[S20, [a20, A20]], [A20, [b20, A20]], [A20, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S20], [a20, A20], [a20]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S20], [a20, A20], [a20]], [[0, 0, S20, [a20, A20]], [1, 2, A20, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S20], [a20, A20], [a20]], [[0, 0, S20, [a20, A20]], [1, 2, A20, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0046/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 47. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S21, A21], terminals=[a21, b21], start=S21, productions=[[S21, [a21, A21]], [A21, [b21, A21]], [A21, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S21], [a21, A21], [a21, b21, A21], [a21, b21]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S21], [a21, A21], [a21, b21, A21], [a21, b21]], [[0, 0, S21, [a21, A21]], [1, 1, A21, [b21, A21]], [2, 2, A21, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S21], [a21, A21], [a21, b21, A21], [a21, b21]], [[0, 0, S21, [a21, A21]], [1, 1, A21, [b21, A21]], [2, 2, A21, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0047/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 48. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S22, A22], terminals=[a22, b22], start=S22, productions=[[S22, [a22, A22]], [A22, [b22, A22]], [A22, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S22], [a22, A22], [a22, b22, A22], [a22, b22, b22, A22], [a22, b22, b22]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S22], [a22, A22], [a22, b22, A22], [a22, b22, b22, A22], [a22, b22, b22]], [[0, 0, S22, [a22, A22]], [1, 1, A22, [b22, A22]], [2, 1, A22, [b22, A22]], [3, 2, A22, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S22], [a22, A22], [a22, b22, A22], [a22, b22, b22, A22], [a22, b22, b22]], [[0, 0, S22, [a22, A22]], [1, 1, A22, [b22, A22]], [2, 1, A22, [b22, A22]], [3, 2, A22, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0048/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 49. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S23, A23], terminals=[a23, b23], start=S23, productions=[[S23, [a23, A23]], [A23, [b23, A23]], [A23, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S23], [a23, A23], [a23, b23, A23], [a23, b23, b23, A23], [a23, b23, b23, b23, A23], [a23, b23, b23, b23]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S23], [a23, A23], [a23, b23, A23], [a23, b23, b23, A23], [a23, b23, b23, b23, A23], [a23, b23, b23, b23]], [[0, 0, S23, [a23, A23]], [1, 1, A23, [b23, A23]], [2, 1, A23, [b23, A23]], [3, 1, A23, [b23, A23]], [4, 2, A23, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S23], [a23, A23], [a23, b23, A23], [a23, b23, b23, A23], [a23, b23, b23, b23, A23], [a23, b23, b23, b23]], [[0, 0, S23, [a23, A23]], [1, 1, A23, [b23, A23]], [2, 1, A23, [b23, A23]], [3, 1, A23, [b23, A23]], [4, 2, A23, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0049/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 50. Each Derivation Step Replaces One Nonterminal by One Declared Right Side

Use the complete finite grammar: nonterminals=[S24, A24], terminals=[a24, b24], start=S24, productions=[[S24, [a24, A24]], [A24, [b24, A24]], [A24, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Supplied forms=[[S24], [a24, A24], [a24, b24, A24], [a24, b24, b24, A24], [a24, b24, b24, b24, A24], [a24, b24, b24, b24, b24, A24], [a24, b24, b24, b24, b24]]. Give (forms, step_witnesses, valid_derivation, final_form_terminal_only, certificate).

final_form_terminal_only checks the last form separately. Even Yes needs valid_derivation to certify a derivation from the start symbol.

View answer and explanation([[S24], [a24, A24], [a24, b24, A24], [a24, b24, b24, A24], [a24, b24, b24, b24, A24], [a24, b24, b24, b24, b24, A24], [a24, b24, b24, b24, b24]], [[0, 0, S24, [a24, A24]], [1, 1, A24, [b24, A24]], [2, 1, A24, [b24, A24]], [3, 1, A24, [b24, A24]], [4, 1, A24, [b24, A24]], [5, 2, A24, []]], Yes, Yes, finite_production_sequence)

A derivation must begin with [start]; each successive form replaces exactly one matching lhs occurrence. Each step witness is [position,production index,lhs,rhs], using the first matching witness in source order. Keep validity separate from whether the last form is terminal-only. The exact result, retaining the requested order, is ([[S24], [a24, A24], [a24, b24, A24], [a24, b24, b24, A24], [a24, b24, b24, b24, A24], [a24, b24, b24, b24, b24, A24], [a24, b24, b24, b24, b24]], [[0, 0, S24, [a24, A24]], [1, 1, A24, [b24, A24]], [2, 1, A24, [b24, A24]], [3, 1, A24, [b24, A24]], [4, 1, A24, [b24, A24]], [5, 2, A24, []]], Yes, Yes, finite_production_sequence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0050/)

Context-free grammar · qf-mod40-cfg-derivation-audit · candidate





## 51. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S0, A0], terminals=[a0, b0], start=S0, productions=[[S0, [a0, A0]], [A0, [b0, A0]], [A0, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a0], supplied forms=[[S0], [a0, A0], [a0]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a0], [[0, 0, S0, [a0, A0]], [1, 2, A0, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a0], [[0, 0, S0, [a0, A0]], [1, 2, A0, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0051/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 52. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S1, A1], terminals=[a1, b1], start=S1, productions=[[S1, [a1, A1]], [A1, [b1, A1]], [A1, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a1, b1], supplied forms=[[S1], [a1, A1], [a1, b1, A1], [a1, b1]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a1, b1], [[0, 0, S1, [a1, A1]], [1, 1, A1, [b1, A1]], [2, 2, A1, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a1, b1], [[0, 0, S1, [a1, A1]], [1, 1, A1, [b1, A1]], [2, 2, A1, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0052/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 53. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S2, A2], terminals=[a2, b2], start=S2, productions=[[S2, [a2, A2]], [A2, [b2, A2]], [A2, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a2, b2, b2], supplied forms=[[S2], [a2, A2], [a2, b2, A2], [a2, b2, b2, A2], [a2, b2, b2]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a2, b2, b2], [[0, 0, S2, [a2, A2]], [1, 1, A2, [b2, A2]], [2, 1, A2, [b2, A2]], [3, 2, A2, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a2, b2, b2], [[0, 0, S2, [a2, A2]], [1, 1, A2, [b2, A2]], [2, 1, A2, [b2, A2]], [3, 2, A2, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0053/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 54. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S3, A3], terminals=[a3, b3], start=S3, productions=[[S3, [a3, A3]], [A3, [b3, A3]], [A3, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a3, b3, b3, b3], supplied forms=[[S3], [a3, A3], [a3, b3, A3], [a3, b3, b3, A3], [a3, b3, b3, b3, A3], [a3, b3, b3, b3]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a3, b3, b3, b3], [[0, 0, S3, [a3, A3]], [1, 1, A3, [b3, A3]], [2, 1, A3, [b3, A3]], [3, 1, A3, [b3, A3]], [4, 2, A3, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a3, b3, b3, b3], [[0, 0, S3, [a3, A3]], [1, 1, A3, [b3, A3]], [2, 1, A3, [b3, A3]], [3, 1, A3, [b3, A3]], [4, 2, A3, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0054/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 55. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S4, A4], terminals=[a4, b4], start=S4, productions=[[S4, [a4, A4]], [A4, [b4, A4]], [A4, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a4, b4, b4, b4, b4], supplied forms=[[S4], [a4, A4], [a4, b4, A4], [a4, b4, b4, A4], [a4, b4, b4, b4, A4], [a4, b4, b4, b4, b4, A4], [a4, b4, b4, b4, b4]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a4, b4, b4, b4, b4], [[0, 0, S4, [a4, A4]], [1, 1, A4, [b4, A4]], [2, 1, A4, [b4, A4]], [3, 1, A4, [b4, A4]], [4, 1, A4, [b4, A4]], [5, 2, A4, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a4, b4, b4, b4, b4], [[0, 0, S4, [a4, A4]], [1, 1, A4, [b4, A4]], [2, 1, A4, [b4, A4]], [3, 1, A4, [b4, A4]], [4, 1, A4, [b4, A4]], [5, 2, A4, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0055/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 56. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S5, A5], terminals=[a5, b5], start=S5, productions=[[S5, [a5, A5]], [A5, [b5, A5]], [A5, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a5], supplied forms=[[S5], [a5, A5], [a5]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a5], [[0, 0, S5, [a5, A5]], [1, 2, A5, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a5], [[0, 0, S5, [a5, A5]], [1, 2, A5, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0056/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 57. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S6, A6], terminals=[a6, b6], start=S6, productions=[[S6, [a6, A6]], [A6, [b6, A6]], [A6, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a6, b6], supplied forms=[[S6], [a6, A6], [a6, b6, A6], [a6, b6]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a6, b6], [[0, 0, S6, [a6, A6]], [1, 1, A6, [b6, A6]], [2, 2, A6, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a6, b6], [[0, 0, S6, [a6, A6]], [1, 1, A6, [b6, A6]], [2, 2, A6, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0057/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 58. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S7, A7], terminals=[a7, b7], start=S7, productions=[[S7, [a7, A7]], [A7, [b7, A7]], [A7, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a7, b7, b7], supplied forms=[[S7], [a7, A7], [a7, b7, A7], [a7, b7, b7, A7], [a7, b7, b7]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a7, b7, b7], [[0, 0, S7, [a7, A7]], [1, 1, A7, [b7, A7]], [2, 1, A7, [b7, A7]], [3, 2, A7, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a7, b7, b7], [[0, 0, S7, [a7, A7]], [1, 1, A7, [b7, A7]], [2, 1, A7, [b7, A7]], [3, 2, A7, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0058/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 59. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S8, A8], terminals=[a8, b8], start=S8, productions=[[S8, [a8, A8]], [A8, [b8, A8]], [A8, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a8, b8, b8, b8], supplied forms=[[S8], [a8, A8], [a8, b8, A8], [a8, b8, b8, A8], [a8, b8, b8, b8, A8], [a8, b8, b8, b8]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a8, b8, b8, b8], [[0, 0, S8, [a8, A8]], [1, 1, A8, [b8, A8]], [2, 1, A8, [b8, A8]], [3, 1, A8, [b8, A8]], [4, 2, A8, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a8, b8, b8, b8], [[0, 0, S8, [a8, A8]], [1, 1, A8, [b8, A8]], [2, 1, A8, [b8, A8]], [3, 1, A8, [b8, A8]], [4, 2, A8, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0059/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 60. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S9, A9], terminals=[a9, b9], start=S9, productions=[[S9, [a9, A9]], [A9, [b9, A9]], [A9, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a9, b9, b9, b9, b9], supplied forms=[[S9], [a9, A9], [a9, b9, A9], [a9, b9, b9, A9], [a9, b9, b9, b9, A9], [a9, b9, b9, b9, b9, A9], [a9, b9, b9, b9, b9]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a9, b9, b9, b9, b9], [[0, 0, S9, [a9, A9]], [1, 1, A9, [b9, A9]], [2, 1, A9, [b9, A9]], [3, 1, A9, [b9, A9]], [4, 1, A9, [b9, A9]], [5, 2, A9, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a9, b9, b9, b9, b9], [[0, 0, S9, [a9, A9]], [1, 1, A9, [b9, A9]], [2, 1, A9, [b9, A9]], [3, 1, A9, [b9, A9]], [4, 1, A9, [b9, A9]], [5, 2, A9, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0060/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 61. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S10, A10], terminals=[a10, b10], start=S10, productions=[[S10, [a10, A10]], [A10, [b10, A10]], [A10, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a10], supplied forms=[[S10], [a10, A10], [a10]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a10], [[0, 0, S10, [a10, A10]], [1, 2, A10, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a10], [[0, 0, S10, [a10, A10]], [1, 2, A10, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0061/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 62. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S11, A11], terminals=[a11, b11], start=S11, productions=[[S11, [a11, A11]], [A11, [b11, A11]], [A11, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a11, b11], supplied forms=[[S11], [a11, A11], [a11, b11, A11], [a11, b11]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a11, b11], [[0, 0, S11, [a11, A11]], [1, 1, A11, [b11, A11]], [2, 2, A11, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a11, b11], [[0, 0, S11, [a11, A11]], [1, 1, A11, [b11, A11]], [2, 2, A11, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0062/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 63. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S12, A12], terminals=[a12, b12], start=S12, productions=[[S12, [a12, A12]], [A12, [b12, A12]], [A12, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a12, b12, b12], supplied forms=[[S12], [a12, A12], [a12, b12, A12], [a12, b12, b12, A12], [a12, b12, b12]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a12, b12, b12], [[0, 0, S12, [a12, A12]], [1, 1, A12, [b12, A12]], [2, 1, A12, [b12, A12]], [3, 2, A12, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a12, b12, b12], [[0, 0, S12, [a12, A12]], [1, 1, A12, [b12, A12]], [2, 1, A12, [b12, A12]], [3, 2, A12, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0063/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 64. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S13, A13], terminals=[a13, b13], start=S13, productions=[[S13, [a13, A13]], [A13, [b13, A13]], [A13, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a13, b13, b13, b13], supplied forms=[[S13], [a13, A13], [a13, b13, A13], [a13, b13, b13, A13], [a13, b13, b13, b13, A13], [a13, b13, b13, b13]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a13, b13, b13, b13], [[0, 0, S13, [a13, A13]], [1, 1, A13, [b13, A13]], [2, 1, A13, [b13, A13]], [3, 1, A13, [b13, A13]], [4, 2, A13, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a13, b13, b13, b13], [[0, 0, S13, [a13, A13]], [1, 1, A13, [b13, A13]], [2, 1, A13, [b13, A13]], [3, 1, A13, [b13, A13]], [4, 2, A13, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0064/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 65. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S14, A14], terminals=[a14, b14], start=S14, productions=[[S14, [a14, A14]], [A14, [b14, A14]], [A14, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a14, b14, b14, b14, b14], supplied forms=[[S14], [a14, A14], [a14, b14, A14], [a14, b14, b14, A14], [a14, b14, b14, b14, A14], [a14, b14, b14, b14, b14, A14], [a14, b14, b14, b14, b14]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a14, b14, b14, b14, b14], [[0, 0, S14, [a14, A14]], [1, 1, A14, [b14, A14]], [2, 1, A14, [b14, A14]], [3, 1, A14, [b14, A14]], [4, 1, A14, [b14, A14]], [5, 2, A14, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a14, b14, b14, b14, b14], [[0, 0, S14, [a14, A14]], [1, 1, A14, [b14, A14]], [2, 1, A14, [b14, A14]], [3, 1, A14, [b14, A14]], [4, 1, A14, [b14, A14]], [5, 2, A14, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0065/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 66. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S15, A15], terminals=[a15, b15], start=S15, productions=[[S15, [a15, A15]], [A15, [b15, A15]], [A15, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a15], supplied forms=[[S15], [a15, A15], [a15]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a15], [[0, 0, S15, [a15, A15]], [1, 2, A15, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a15], [[0, 0, S15, [a15, A15]], [1, 2, A15, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0066/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 67. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S16, A16], terminals=[a16, b16], start=S16, productions=[[S16, [a16, A16]], [A16, [b16, A16]], [A16, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a16, b16], supplied forms=[[S16], [a16, A16], [a16, b16, A16], [a16, b16]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a16, b16], [[0, 0, S16, [a16, A16]], [1, 1, A16, [b16, A16]], [2, 2, A16, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a16, b16], [[0, 0, S16, [a16, A16]], [1, 1, A16, [b16, A16]], [2, 2, A16, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0067/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 68. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S17, A17], terminals=[a17, b17], start=S17, productions=[[S17, [a17, A17]], [A17, [b17, A17]], [A17, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a17, b17, b17], supplied forms=[[S17], [a17, A17], [a17, b17, A17], [a17, b17, b17, A17], [a17, b17, b17]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a17, b17, b17], [[0, 0, S17, [a17, A17]], [1, 1, A17, [b17, A17]], [2, 1, A17, [b17, A17]], [3, 2, A17, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a17, b17, b17], [[0, 0, S17, [a17, A17]], [1, 1, A17, [b17, A17]], [2, 1, A17, [b17, A17]], [3, 2, A17, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0068/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 69. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S18, A18], terminals=[a18, b18], start=S18, productions=[[S18, [a18, A18]], [A18, [b18, A18]], [A18, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a18, b18, b18, b18], supplied forms=[[S18], [a18, A18], [a18, b18, A18], [a18, b18, b18, A18], [a18, b18, b18, b18, A18], [a18, b18, b18, b18]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a18, b18, b18, b18], [[0, 0, S18, [a18, A18]], [1, 1, A18, [b18, A18]], [2, 1, A18, [b18, A18]], [3, 1, A18, [b18, A18]], [4, 2, A18, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a18, b18, b18, b18], [[0, 0, S18, [a18, A18]], [1, 1, A18, [b18, A18]], [2, 1, A18, [b18, A18]], [3, 1, A18, [b18, A18]], [4, 2, A18, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0069/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 70. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S19, A19], terminals=[a19, b19], start=S19, productions=[[S19, [a19, A19]], [A19, [b19, A19]], [A19, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a19, b19, b19, b19, b19], supplied forms=[[S19], [a19, A19], [a19, b19, A19], [a19, b19, b19, A19], [a19, b19, b19, b19, A19], [a19, b19, b19, b19, b19, A19], [a19, b19, b19, b19, b19]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a19, b19, b19, b19, b19], [[0, 0, S19, [a19, A19]], [1, 1, A19, [b19, A19]], [2, 1, A19, [b19, A19]], [3, 1, A19, [b19, A19]], [4, 1, A19, [b19, A19]], [5, 2, A19, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a19, b19, b19, b19, b19], [[0, 0, S19, [a19, A19]], [1, 1, A19, [b19, A19]], [2, 1, A19, [b19, A19]], [3, 1, A19, [b19, A19]], [4, 1, A19, [b19, A19]], [5, 2, A19, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0070/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 71. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S20, A20], terminals=[a20, b20], start=S20, productions=[[S20, [a20, A20]], [A20, [b20, A20]], [A20, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a20], supplied forms=[[S20], [a20, A20], [a20]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a20], [[0, 0, S20, [a20, A20]], [1, 2, A20, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a20], [[0, 0, S20, [a20, A20]], [1, 2, A20, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0071/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 72. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S21, A21], terminals=[a21, b21], start=S21, productions=[[S21, [a21, A21]], [A21, [b21, A21]], [A21, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a21, b21], supplied forms=[[S21], [a21, A21], [a21, b21, A21], [a21, b21]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a21, b21], [[0, 0, S21, [a21, A21]], [1, 1, A21, [b21, A21]], [2, 2, A21, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a21, b21], [[0, 0, S21, [a21, A21]], [1, 1, A21, [b21, A21]], [2, 2, A21, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0072/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 73. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S22, A22], terminals=[a22, b22], start=S22, productions=[[S22, [a22, A22]], [A22, [b22, A22]], [A22, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a22, b22, b22], supplied forms=[[S22], [a22, A22], [a22, b22, A22], [a22, b22, b22, A22], [a22, b22, b22]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a22, b22, b22], [[0, 0, S22, [a22, A22]], [1, 1, A22, [b22, A22]], [2, 1, A22, [b22, A22]], [3, 2, A22, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a22, b22, b22], [[0, 0, S22, [a22, A22]], [1, 1, A22, [b22, A22]], [2, 1, A22, [b22, A22]], [3, 2, A22, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0073/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 74. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S23, A23], terminals=[a23, b23], start=S23, productions=[[S23, [a23, A23]], [A23, [b23, A23]], [A23, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a23, b23, b23, b23], supplied forms=[[S23], [a23, A23], [a23, b23, A23], [a23, b23, b23, A23], [a23, b23, b23, b23, A23], [a23, b23, b23, b23]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a23, b23, b23, b23], [[0, 0, S23, [a23, A23]], [1, 1, A23, [b23, A23]], [2, 1, A23, [b23, A23]], [3, 1, A23, [b23, A23]], [4, 2, A23, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a23, b23, b23, b23], [[0, 0, S23, [a23, A23]], [1, 1, A23, [b23, A23]], [2, 1, A23, [b23, A23]], [3, 1, A23, [b23, A23]], [4, 2, A23, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0074/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 75. A Valid Start-to-Terminal Derivation Proves Word Membership

Use the complete finite grammar: nonterminals=[S24, A24], terminals=[a24, b24], start=S24, productions=[[S24, [a24, A24]], [A24, [b24, A24]], [A24, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Candidate word=[a24, b24, b24, b24, b24], supplied forms=[[S24], [a24, A24], [a24, b24, A24], [a24, b24, b24, A24], [a24, b24, b24, b24, A24], [a24, b24, b24, b24, b24, A24], [a24, b24, b24, b24, b24]]. Give (word, step_witnesses, valid_derivation, terminal_word, membership_certificate, certificate).

View answer and explanation([a24, b24, b24, b24, b24], [[0, 0, S24, [a24, A24]], [1, 1, A24, [b24, A24]], [2, 1, A24, [b24, A24]], [3, 1, A24, [b24, A24]], [4, 1, A24, [b24, A24]], [5, 2, A24, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership)

Require a valid derivation from [start], a terminal-only word and equality of the final form with that word. Empty word membership needs an actual allowed epsilon derivation; the empty language is different from a language containing the empty word. No ambiguity, regularity or CFG-PDA equivalence follows. The exact result, retaining the requested order, is ([a24, b24, b24, b24, b24], [[0, 0, S24, [a24, A24]], [1, 1, A24, [b24, A24]], [2, 1, A24, [b24, A24]], [3, 1, A24, [b24, A24]], [4, 1, A24, [b24, A24]], [5, 2, A24, []]], Yes, Yes, Yes, derivation_from_start_to_terminal_only_word_proves_membership).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0075/)

Context-free grammar · qf-mod40-cfg-membership-witness · candidate





## 76. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S0, A0], terminals=[a0, b0], start=S0, productions=[[S0, [a0, A0]], [A0, [b0, A0]], [A0, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a0, b0] with max_steps=1. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a0, b0], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a0, b0], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0076/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 77. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S1, A1], terminals=[a1, b1], start=S1, productions=[[S1, [a1, A1]], [A1, [b1, A1]], [A1, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a1, b1, b1] with max_steps=2. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a1, b1, b1], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a1, b1, b1], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0077/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 78. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S2, A2], terminals=[a2, b2], start=S2, productions=[[S2, [a2, A2]], [A2, [b2, A2]], [A2, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a2, b2, b2, b2] with max_steps=1. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a2, b2, b2, b2], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a2, b2, b2, b2], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0078/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 79. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S3, A3], terminals=[a3, b3], start=S3, productions=[[S3, [a3, A3]], [A3, [b3, A3]], [A3, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a3, b3, b3, b3, b3] with max_steps=2. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a3, b3, b3, b3, b3], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a3, b3, b3, b3, b3], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0079/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 80. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S4, A4], terminals=[a4, b4], start=S4, productions=[[S4, [a4, A4]], [A4, [b4, A4]], [A4, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a4, b4, b4, b4, b4, b4] with max_steps=1. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a4, b4, b4, b4, b4, b4], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a4, b4, b4, b4, b4, b4], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0080/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 81. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S5, A5], terminals=[a5, b5], start=S5, productions=[[S5, [a5, A5]], [A5, [b5, A5]], [A5, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a5, b5] with max_steps=2. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a5, b5], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a5, b5], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0081/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 82. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S6, A6], terminals=[a6, b6], start=S6, productions=[[S6, [a6, A6]], [A6, [b6, A6]], [A6, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a6, b6, b6] with max_steps=1. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a6, b6, b6], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a6, b6, b6], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0082/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 83. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S7, A7], terminals=[a7, b7], start=S7, productions=[[S7, [a7, A7]], [A7, [b7, A7]], [A7, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a7, b7, b7, b7] with max_steps=2. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a7, b7, b7, b7], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a7, b7, b7, b7], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0083/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 84. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S8, A8], terminals=[a8, b8], start=S8, productions=[[S8, [a8, A8]], [A8, [b8, A8]], [A8, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a8, b8, b8, b8, b8] with max_steps=1. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a8, b8, b8, b8, b8], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a8, b8, b8, b8, b8], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0084/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 85. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S9, A9], terminals=[a9, b9], start=S9, productions=[[S9, [a9, A9]], [A9, [b9, A9]], [A9, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a9, b9, b9, b9, b9, b9] with max_steps=2. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a9, b9, b9, b9, b9, b9], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a9, b9, b9, b9, b9, b9], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0085/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 86. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S10, A10], terminals=[a10, b10], start=S10, productions=[[S10, [a10, A10]], [A10, [b10, A10]], [A10, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a10, b10] with max_steps=1. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a10, b10], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a10, b10], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0086/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 87. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S11, A11], terminals=[a11, b11], start=S11, productions=[[S11, [a11, A11]], [A11, [b11, A11]], [A11, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a11, b11, b11] with max_steps=2. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a11, b11, b11], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a11, b11, b11], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0087/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 88. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S12, A12], terminals=[a12, b12], start=S12, productions=[[S12, [a12, A12]], [A12, [b12, A12]], [A12, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a12, b12, b12, b12] with max_steps=1. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a12, b12, b12, b12], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a12, b12, b12, b12], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0088/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 89. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S13, A13], terminals=[a13, b13], start=S13, productions=[[S13, [a13, A13]], [A13, [b13, A13]], [A13, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a13, b13, b13, b13, b13] with max_steps=2. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a13, b13, b13, b13, b13], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a13, b13, b13, b13, b13], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0089/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 90. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S14, A14], terminals=[a14, b14], start=S14, productions=[[S14, [a14, A14]], [A14, [b14, A14]], [A14, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a14, b14, b14, b14, b14, b14] with max_steps=1. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a14, b14, b14, b14, b14, b14], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a14, b14, b14, b14, b14, b14], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0090/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 91. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S15, A15], terminals=[a15, b15], start=S15, productions=[[S15, [a15, A15]], [A15, [b15, A15]], [A15, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a15, b15] with max_steps=2. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a15, b15], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a15, b15], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0091/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 92. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S16, A16], terminals=[a16, b16], start=S16, productions=[[S16, [a16, A16]], [A16, [b16, A16]], [A16, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a16, b16, b16] with max_steps=1. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a16, b16, b16], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a16, b16, b16], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0092/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 93. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S17, A17], terminals=[a17, b17], start=S17, productions=[[S17, [a17, A17]], [A17, [b17, A17]], [A17, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a17, b17, b17, b17] with max_steps=2. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a17, b17, b17, b17], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a17, b17, b17, b17], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0093/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 94. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S18, A18], terminals=[a18, b18], start=S18, productions=[[S18, [a18, A18]], [A18, [b18, A18]], [A18, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a18, b18, b18, b18, b18] with max_steps=1. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a18, b18, b18, b18, b18], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a18, b18, b18, b18, b18], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0094/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 95. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S19, A19], terminals=[a19, b19], start=S19, productions=[[S19, [a19, A19]], [A19, [b19, A19]], [A19, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a19, b19, b19, b19, b19, b19] with max_steps=2. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a19, b19, b19, b19, b19, b19], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a19, b19, b19, b19, b19, b19], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0095/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 96. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S20, A20], terminals=[a20, b20], start=S20, productions=[[S20, [a20, A20]], [A20, [b20, A20]], [A20, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a20, b20] with max_steps=1. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a20, b20], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a20, b20], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0096/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 97. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S21, A21], terminals=[a21, b21], start=S21, productions=[[S21, [a21, A21]], [A21, [b21, A21]], [A21, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a21, b21, b21] with max_steps=2. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a21, b21, b21], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a21, b21, b21], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0097/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 98. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S22, A22], terminals=[a22, b22], start=S22, productions=[[S22, [a22, A22]], [A22, [b22, A22]], [A22, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a22, b22, b22, b22] with max_steps=1. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a22, b22, b22, b22], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a22, b22, b22, b22], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0098/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 99. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S23, A23], terminals=[a23, b23], start=S23, productions=[[S23, [a23, A23]], [A23, [b23, A23]], [A23, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a23, b23, b23, b23, b23] with max_steps=2. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a23, b23, b23, b23, b23], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a23, b23, b23, b23, b23], 2, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0099/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate





## 100. A Bounded Search No-Hit Is Not a Nonmembership Proof

Use the complete finite grammar: nonterminals=[S24, A24], terminals=[a24, b24], start=S24, productions=[[S24, [a24, A24]], [A24, [b24, A24]], [A24, []]] and allow_epsilon=true. Alphabets must be unique and disjoint, the start belongs to nonterminals, and each production [lhs,rhs] has exactly one nonterminal lhs and a finite legal-symbol rhs. Empty rhs is permitted only under the explicit epsilon convention. Keep original symbol and production order. Search word=[a24, b24, b24, b24, b24, b24] with max_steps=1. Give (word, max_steps, found_in_bounded_search, nonmembership_claim, boundary).

found reports this depth- and length-bounded search only. The fixed No does not assert nonmembership; a failed search is not proof that the word is outside the language.

View answer and explanation([a24, b24, b24, b24, b24, b24], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound)

Run the declared finite production search from [start], retaining only intermediate forms of length at most word.length+max_steps+1. Report whether this bounded search found the exact form. The final fixed No makes no nonmembership claim, even on a no-hit; without a completeness bound, a longer derivation may prove membership. A found nonterminal form is not by itself a terminal-word membership certificate. The exact result, retaining the requested order, is ([a24, b24, b24, b24, b24, b24], 1, No, No, bounded_search_failure_cannot_prove_nonmembership_without_complete_bound).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261109-r07-context-free-grammar-0100/)

Context-free grammar · qf-mod40-cfg-bounded-search-boundary · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261109-r07-context-free-grammar-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261014-advance-20261109-r07-context-free-grammar-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261109-r07-context-free-grammar-foundations/validation-report.json)