# Typed Regular Expressions, Complete Bounded Languages and Membership 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-20261015-advance-20261118-r06-regular-expression-foundations · 2026-10-15 · 100 questions This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt. Field order uses a source-bound presentation supplement; canonical JSONL and mathematical answers are unchanged. [Field-order and source-binding data](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json) ## 1. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"symbol","value":"a"}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {type:"symbol",value:"a"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {type:"symbol",value:"a"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0001/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 2. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0002/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 3. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0003/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 4. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0004/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 5. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0005/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 6. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"symbol","value":"c"},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{type:"symbol",value:"c"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{type:"symbol",value:"c"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0006/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 7. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0007/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 8. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0008/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 9. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0009/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 10. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0010/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 11. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"symbol","value":"b"},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{type:"symbol",value:"b"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{type:"symbol",value:"b"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0011/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 12. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0012/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 13. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0013/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 14. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0014/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 15. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0015/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 16. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"symbol","value":"a"},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{type:"symbol",value:"a"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{type:"symbol",value:"a"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0016/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 17. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0017/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 18. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0018/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 19. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0019/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 20. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0020/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 21. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"symbol","value":"c"},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{type:"symbol",value:"c"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{type:"symbol",value:"c"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0021/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 22. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0022/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 23. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0023/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 24. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0024/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 25. Regular Expressions Have Typed Nodes and an Explicit Alphabet Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Give (alphabet, canonical_typed_AST, AST_qualified, qualification_statement). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet) Validate each node type, exact key set, arity and symbol membership. Canonicalize unions extensionally while retaining concatenation order. A shell wildcard is not one of these core regular-expression AST nodes. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, Yes, typed_AST_symbols_in_alphabet). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0025/) Regular expression · qf-mod49-regex-ast-qualification · candidate ## 26. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"symbol","value":"a"}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=1. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {type:"symbol",value:"a"}, 1, [[a]], 1, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {type:"symbol",value:"a"}, 1, [[a]], 1, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0026/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 27. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=2. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, 2, [[b, c]], 1, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, 2, [[b, c]], 1, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0027/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 28. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=3. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, 3, [[c, a, b]], 1, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, 3, [[c, a, b]], 1, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0028/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 29. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=4. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}, 4, [[a, b, c, a]], 1, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}, 4, [[a, b, c, a]], 1, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0029/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 30. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=5. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, 5, [[b, c, a, b, c]], 1, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, 5, [[b, c, a, b, c]], 1, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0030/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 31. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"symbol","value":"c"},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=1. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{type:"symbol",value:"c"}],type:"union"}, 1, [[], [a], [c]], 3, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{type:"symbol",value:"c"}],type:"union"}, 1, [[], [a], [c]], 3, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0031/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 32. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=2. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}, 2, [[], [b], [a, b], [b, b]], 4, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}, 2, [[], [b], [a, b], [b, b]], 4, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0032/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 33. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=3. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}, 3, [[], [c], [c, c], [b, c, a], [c, c, c]], 5, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}, 3, [[], [c], [c, c], [b, c, a], [c, c, c]], 5, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0033/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 34. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=4. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}, 4, [[], [a], [a, a], [a, a, a], [a, a, a, a], [c, a, b, c]], 6, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}, 4, [[], [a], [a, a], [a, a, a], [a, a, a, a], [c, a, b, c]], 6, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0034/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 35. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=5. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}, 5, [[], [b], [b, b], [b, b, b], [b, b, b, b], [a, b, c, a, b], [b, b, b, b, b]], 7, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}, 5, [[], [b], [b, b], [b, b, b], [b, b, b, b], [a, b, c, a, b], [b, b, b, b, b]], 7, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0035/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 36. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"symbol","value":"b"},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=1. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{type:"symbol",value:"b"}],type:"union"}],type:"union"}, 1, [[], [a], [b], [c]], 4, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{type:"symbol",value:"b"}],type:"union"}],type:"union"}, 1, [[], [a], [b], [c]], 4, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0036/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 37. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=2. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}, 2, [[], [a], [b], [a, a], [b, b], [c, a]], 6, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}, 2, [[], [a], [b], [a, a], [b, b], [c, a]], 6, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0037/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 38. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=3. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}, 3, [[], [b], [c], [b, b], [c, c], [a, b, c], [b, b, b], [c, c, c]], 8, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}, 3, [[], [b], [c], [b, b], [c, c], [a, b, c], [b, b, b], [c, c, c]], 8, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0038/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 39. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=4. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}, 4, [[], [a], [c], [a, a], [c, c], [a, a, a], [c, c, c], [a, a, a, a], [b, c, a, b], [c, c, c, c]], 10, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}, 4, [[], [a], [c], [a, a], [c, c], [a, a, a], [c, c, c], [a, a, a, a], [b, c, a, b], [c, c, c, c]], 10, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0039/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 40. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=5. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}, 5, [[], [a], [b], [a, a], [b, b], [a, a, a], [b, b, b], [a, a, a, a], [b, b, b, b], [a, a, a, a, a], [b, b, b, b, b], [c, a, b, c, a]], 12, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}, 5, [[], [a], [b], [a, a], [b, b], [a, a, a], [b, b, b], [a, a, a, a], [b, b, b, b], [a, a, a, a, a], [b, b, b, b, b], [c, a, b, c, a]], 12, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0040/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 41. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"symbol","value":"a"},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=1. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{type:"symbol",value:"a"}],type:"union"}],type:"union"}],type:"union"}, 1, [[], [a], [b], [c]], 4, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{type:"symbol",value:"a"}],type:"union"}],type:"union"}],type:"union"}, 1, [[], [a], [b], [c]], 4, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0041/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 42. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=2. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, 2, [[], [a], [b], [c], [a, a], [b, b], [b, c], [c, c]], 8, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, 2, [[], [a], [b], [c], [a, a], [b, b], [b, c], [c, c]], 8, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0042/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 43. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=3. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, 3, [[], [a], [b], [c], [a, a], [b, b], [c, c], [a, a, a], [b, b, b], [c, a, b], [c, c, c]], 11, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, 3, [[], [a], [b], [c], [a, a], [b, b], [c, c], [a, a, a], [b, b, b], [c, a, b], [c, c, c]], 11, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0043/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 44. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=4. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, 4, [[], [a], [b], [c], [a, a], [b, b], [c, c], [a, a, a], [b, b, b], [c, c, c], [a, a, a, a], [a, b, c, a], [b, b, b, b], [c, c, c, c]], 14, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, 4, [[], [a], [b], [c], [a, a], [b, b], [c, c], [a, a, a], [b, b, b], [c, c, c], [a, a, a, a], [a, b, c, a], [b, b, b, b], [c, c, c, c]], 14, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0044/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 45. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=5. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, 5, [[], [a], [b], [c], [a, a], [b, b], [c, c], [a, a, a], [b, b, b], [c, c, c], [a, a, a, a], [b, b, b, b], [c, c, c, c], [a, a, a, a, a], [b, b, b, b, b], [b, c, a, b, c], [c, c, c, c, c]], 17, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, 5, [[], [a], [b], [c], [a, a], [b, b], [c, c], [a, a, a], [b, b, b], [c, c, c], [a, a, a, a], [b, b, b, b], [c, c, c, c], [a, a, a, a, a], [b, b, b, b, b], [b, c, a, b, c], [c, c, c, c, c]], 17, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0045/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 46. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"symbol","value":"c"},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=1. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{type:"symbol",value:"c"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, 1, [[], [a], [b], [c]], 4, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{type:"symbol",value:"c"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, 1, [[], [a], [b], [c]], 4, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0046/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 47. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=2. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, 2, [[], [a], [b], [c], [a, a], [a, b], [b, b], [c, c]], 8, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, 2, [[], [a], [b], [c], [a, a], [a, b], [b, b], [c, c]], 8, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0047/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 48. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=3. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, 3, [[], [a], [b], [c], [a, a], [b, b], [c, c], [a, a, a], [b, b, b], [b, c, a], [c, c, c]], 11, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, 3, [[], [a], [b], [c], [a, a], [b, b], [c, c], [a, a, a], [b, b, b], [b, c, a], [c, c, c]], 11, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0048/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 49. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=4. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, 4, [[], [a], [b], [c], [a, a], [b, b], [c, c], [a, a, a], [b, b, b], [c, c, c], [a, a, a, a], [b, b, b, b], [c, a, b, c], [c, c, c, c]], 14, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, 4, [[], [a], [b], [c], [a, a], [b, b], [c, c], [a, a, a], [b, b, b], [c, c, c], [a, a, a, a], [b, b, b, b], [c, a, b, c], [c, c, c, c]], 14, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0049/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 50. The Complete Bounded Language Includes Epsilon and Terminates Epsilon Cycles Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Enumerate every accepted word of length at most=5. Give (alphabet, canonical_typed_AST, max_length, complete_bounded_language, word_count, bounded_enumeration_complete, epsilon_cycle_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, 5, [[], [a], [b], [c], [a, a], [b, b], [c, c], [a, a, a], [b, b, b], [c, c, c], [a, a, a, a], [b, b, b, b], [c, c, c, c], [a, a, a, a, a], [a, b, c, a, b], [b, b, b, b, b], [c, c, c, c, c]], 17, Yes, epsilon_cycles_terminate_by_extensional_dedup) Retain all words in length/canonical order and the exact count. Star closure includes the empty word and uses extensional deduplication to terminate epsilon cycles. Completeness is only through the declared bound; it does not establish equivalence of the full unbounded languages. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, 5, [[], [a], [b], [c], [a, a], [b, b], [c, c], [a, a, a], [b, b, b], [c, c, c], [a, a, a, a], [b, b, b, b], [c, c, c, c], [a, a, a, a, a], [a, b, c, a, b], [b, b, b, b, b], [c, c, c, c, c]], 17, Yes, epsilon_cycles_terminate_by_extensional_dedup). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0050/) Regular expression · qf-mod49-regex-bounded-language · candidate ## 51. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"symbol","value":"a"}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {type:"symbol",value:"a"}, [], No, [], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {type:"symbol",value:"a"}, [], No, [], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0051/) Regular expression · qf-mod49-regex-word-membership · candidate ## 52. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[b]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, [b], No, [], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, [b], No, [], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0052/) Regular expression · qf-mod49-regex-word-membership · candidate ## 53. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[c, a]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, [c, a], No, [], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, [c, a], No, [], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0053/) Regular expression · qf-mod49-regex-word-membership · candidate ## 54. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[a, b, c]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}, [a, b, c], No, [], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}, [a, b, c], No, [], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0054/) Regular expression · qf-mod49-regex-word-membership · candidate ## 55. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[b, c, a, b]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, [b, c, a, b], No, [], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, [b, c, a, b], No, [], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0055/) Regular expression · qf-mod49-regex-word-membership · candidate ## 56. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"symbol","value":"c"},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{type:"symbol",value:"c"}],type:"union"}, [], Yes, [[]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{type:"symbol",value:"c"}],type:"union"}, [], Yes, [[]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0056/) Regular expression · qf-mod49-regex-word-membership · candidate ## 57. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[a]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}, [a], No, [[b]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}, [a], No, [[b]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0057/) Regular expression · qf-mod49-regex-word-membership · candidate ## 58. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[b, c]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}, [b, c], No, [[c, c]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}, [b, c], No, [[c, c]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0058/) Regular expression · qf-mod49-regex-word-membership · candidate ## 59. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[c, a, b]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}, [c, a, b], No, [[a, a, a]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}, [c, a, b], No, [[a, a, a]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0059/) Regular expression · qf-mod49-regex-word-membership · candidate ## 60. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[a, b, c, a]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}, [a, b, c, a], No, [[b, b, b, b]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}, [a, b, c, a], No, [[b, b, b, b]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0060/) Regular expression · qf-mod49-regex-word-membership · candidate ## 61. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"symbol","value":"b"},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{type:"symbol",value:"b"}],type:"union"}],type:"union"}, [], Yes, [[]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{type:"symbol",value:"b"}],type:"union"}],type:"union"}, [], Yes, [[]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0061/) Regular expression · qf-mod49-regex-word-membership · candidate ## 62. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[c]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}, [c], No, [[a], [b]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}, [c], No, [[a], [b]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0062/) Regular expression · qf-mod49-regex-word-membership · candidate ## 63. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[a, b]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}, [a, b], No, [[b, b], [c, c]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}, [a, b], No, [[b, b], [c, c]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0063/) Regular expression · qf-mod49-regex-word-membership · candidate ## 64. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[b, c, a]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}, [b, c, a], No, [[a, a, a], [c, c, c]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}, [b, c, a], No, [[a, a, a], [c, c, c]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0064/) Regular expression · qf-mod49-regex-word-membership · candidate ## 65. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[c, a, b, c]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}, [c, a, b, c], No, [[a, a, a, a], [b, b, b, b]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}, [c, a, b, c], No, [[a, a, a, a], [b, b, b, b]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0065/) Regular expression · qf-mod49-regex-word-membership · candidate ## 66. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"symbol","value":"a"},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{type:"symbol",value:"a"}],type:"union"}],type:"union"}],type:"union"}, [], Yes, [[]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{type:"symbol",value:"a"}],type:"union"}],type:"union"}],type:"union"}, [], Yes, [[]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0066/) Regular expression · qf-mod49-regex-word-membership · candidate ## 67. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[b]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, [b], Yes, [[a], [b], [c]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, [b], Yes, [[a], [b], [c]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0067/) Regular expression · qf-mod49-regex-word-membership · candidate ## 68. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[c, a]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, [c, a], No, [[a, a], [b, b], [c, c]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, [c, a], No, [[a, a], [b, b], [c, c]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0068/) Regular expression · qf-mod49-regex-word-membership · candidate ## 69. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[a, b, c]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, [a, b, c], No, [[a, a, a], [b, b, b], [c, c, c]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, [a, b, c], No, [[a, a, a], [b, b, b], [c, c, c]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0069/) Regular expression · qf-mod49-regex-word-membership · candidate ## 70. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[b, c, a, b]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, [b, c, a, b], No, [[a, a, a, a], [b, b, b, b], [c, c, c, c]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}, [b, c, a, b], No, [[a, a, a, a], [b, b, b, b], [c, c, c, c]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0070/) Regular expression · qf-mod49-regex-word-membership · candidate ## 71. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"symbol","value":"c"},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{type:"symbol",value:"c"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, [], Yes, [[]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{type:"symbol",value:"c"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, [], Yes, [[]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0071/) Regular expression · qf-mod49-regex-word-membership · candidate ## 72. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[a]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, [a], Yes, [[a], [b], [c]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, [a], Yes, [[a], [b], [c]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0072/) Regular expression · qf-mod49-regex-word-membership · candidate ## 73. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[b, c]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, [b, c], No, [[a, a], [b, b], [c, c]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, [b, c], No, [[a, a], [b, b], [c, c]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0073/) Regular expression · qf-mod49-regex-word-membership · candidate ## 74. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[c, a, b]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, [c, a, b], No, [[a, a, a], [b, b, b], [c, c, c]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, [c, a, b], No, [[a, a, a], [b, b, b], [c, c, c]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0074/) Regular expression · qf-mod49-regex-word-membership · candidate ## 75. Whole-Word Membership Uses a Sufficient Complete Length Slice Use alphabet=[a, b, c] and typed regular-expression AST={"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"union","children":[{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},{"type":"star","child":{"type":"symbol","value":"b"}}]},{"type":"star","child":{"type":"symbol","value":"c"}}]},{"type":"star","child":{"type":"symbol","value":"a"}}]},{"type":"star","child":{"type":"symbol","value":"b"}}]}. Words are arrays of alphabet symbols; distinguish empty language from the language containing the empty word. Decide word=[a, b, c, a]. Give (alphabet, canonical_typed_AST, word, member, accepted_words_at_same_length, membership_scope). View answer and explanation([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, [a, b, c, a], No, [[a, a, a, a], [b, b, b, b], [c, c, c, c]], complete_membership_not_absence_from_an_insufficient_bound) Compute membership at exactly this word length, with a complete accepted-word slice for that length. Do not infer nonmembership merely because an insufficient shorter bound omitted the word. [] is an empty word when used as a word, while [] as a language contains no words. The exact result, retaining the requested order, is ([a, b, c], {children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{child:{type:"symbol",value:"a"},type:"star"},{children:[{child:{type:"symbol",value:"c"},type:"star"},{children:[{child:{type:"symbol",value:"b"},type:"star"},{children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"},{type:"symbol",value:"c"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}],type:"union"}],type:"union"}],type:"union"}],type:"union"}, [a, b, c, a], No, [[a, a, a, a], [b, b, b, b], [c, c, c, c]], complete_membership_not_absence_from_an_insufficient_bound). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0075/) Regular expression · qf-mod49-regex-word-membership · candidate ## 76. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=empty_vs_epsilon, complete model={"alphabet":[]}. Give (mode, alphabet, empty_language, epsilon_language, language_identity_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(empty_vs_epsilon, [], [], [[]], No, empty_language_not_language_containing_empty_word) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (empty_vs_epsilon, [], [], [[]], No, empty_language_not_language_containing_empty_word). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0076/) Regular expression · qf-mod49-regex-boundary · candidate ## 77. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=wildcard_confusion, complete model={"alphabet":["a","b"],"token":"*"}. Give (mode, alphabet, wildcard_token, core_AST_node_qualification, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(wildcard_confusion, [a, b], *, No, shell_wildcard_is_not_a_core_regex_AST_node) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (wildcard_confusion, [a, b], *, No, shell_wildcard_is_not_a_core_regex_AST_node). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0077/) Regular expression · qf-mod49-regex-boundary · candidate ## 78. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=wildcard_confusion, complete model={"alphabet":["a","b"],"token":"?"}. Give (mode, alphabet, wildcard_token, core_AST_node_qualification, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(wildcard_confusion, [a, b], ?, No, shell_wildcard_is_not_a_core_regex_AST_node) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (wildcard_confusion, [a, b], ?, No, shell_wildcard_is_not_a_core_regex_AST_node). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0078/) Regular expression · qf-mod49-regex-boundary · candidate ## 79. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=bounded_equality, complete model={"alphabet":["a"],"left":{"type":"epsilon"},"right":{"type":"union","children":[{"type":"epsilon"},{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"a"}]}]},"max_length":1}. Give (mode, bound, left_bounded_language, right_bounded_language, equal_through_bound, next_bound, left_next_language, right_next_language, differs_at_next_bound, unbounded_equivalence_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(bounded_equality, 1, [[]], [[]], Yes, 2, [[]], [[], [a, a]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (bounded_equality, 1, [[]], [[]], Yes, 2, [[]], [[], [a, a]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0079/) Regular expression · qf-mod49-regex-boundary · candidate ## 80. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=concat_order, complete model={"alphabet":["a","b","c"],"left":{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},"right":{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"a"}]},"max_length":2}. Give (mode, left_AST, right_AST, left_bounded_language, right_bounded_language, bounded_languages_equal, order_irrelevance_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(concat_order, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"a"}],type:"concat"}, [[a, b]], [[b, a]], No, No, concatenation_order_is_semantic) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (concat_order, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"a"}],type:"concat"}, [[a, b]], [[b, a]], No, No, concatenation_order_is_semantic). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0080/) Regular expression · qf-mod49-regex-boundary · candidate ## 81. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=bounded_equality, complete model={"alphabet":["b"],"left":{"type":"epsilon"},"right":{"type":"union","children":[{"type":"epsilon"},{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"}]}]},"max_length":2}. Give (mode, bound, left_bounded_language, right_bounded_language, equal_through_bound, next_bound, left_next_language, right_next_language, differs_at_next_bound, unbounded_equivalence_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(bounded_equality, 2, [[]], [[]], Yes, 3, [[]], [[], [b, b, b]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (bounded_equality, 2, [[]], [[]], Yes, 3, [[]], [[], [b, b, b]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0081/) Regular expression · qf-mod49-regex-boundary · candidate ## 82. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=concat_order, complete model={"alphabet":["a","b","c"],"left":{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},"right":{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"}]},"max_length":3}. Give (mode, left_AST, right_AST, left_bounded_language, right_bounded_language, bounded_languages_equal, order_irrelevance_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(concat_order, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"b"},{type:"symbol",value:"b"}],type:"concat"}, [[b, b, c]], [[c, b, b]], No, No, concatenation_order_is_semantic) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (concat_order, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"b"},{type:"symbol",value:"b"}],type:"concat"}, [[b, b, c]], [[c, b, b]], No, No, concatenation_order_is_semantic). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0082/) Regular expression · qf-mod49-regex-boundary · candidate ## 83. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=bounded_equality, complete model={"alphabet":["c"],"left":{"type":"epsilon"},"right":{"type":"union","children":[{"type":"epsilon"},{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"c"},{"type":"symbol","value":"c"},{"type":"symbol","value":"c"}]}]},"max_length":3}. Give (mode, bound, left_bounded_language, right_bounded_language, equal_through_bound, next_bound, left_next_language, right_next_language, differs_at_next_bound, unbounded_equivalence_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(bounded_equality, 3, [[]], [[]], Yes, 4, [[]], [[], [c, c, c, c]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (bounded_equality, 3, [[]], [[]], Yes, 4, [[]], [[], [c, c, c, c]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0083/) Regular expression · qf-mod49-regex-boundary · candidate ## 84. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=concat_order, complete model={"alphabet":["a","b","c"],"left":{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"c"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},"right":{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"c"},{"type":"symbol","value":"c"},{"type":"symbol","value":"c"}]},"max_length":4}. Give (mode, left_AST, right_AST, left_bounded_language, right_bounded_language, bounded_languages_equal, order_irrelevance_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(concat_order, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"c"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"c"},{type:"symbol",value:"c"},{type:"symbol",value:"c"}],type:"concat"}, [[c, c, c, a]], [[a, c, c, c]], No, No, concatenation_order_is_semantic) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (concat_order, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"c"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"c"},{type:"symbol",value:"c"},{type:"symbol",value:"c"}],type:"concat"}, [[c, c, c, a]], [[a, c, c, c]], No, No, concatenation_order_is_semantic). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0084/) Regular expression · qf-mod49-regex-boundary · candidate ## 85. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=bounded_equality, complete model={"alphabet":["a"],"left":{"type":"epsilon"},"right":{"type":"union","children":[{"type":"epsilon"},{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"}]}]},"max_length":4}. Give (mode, bound, left_bounded_language, right_bounded_language, equal_through_bound, next_bound, left_next_language, right_next_language, differs_at_next_bound, unbounded_equivalence_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(bounded_equality, 4, [[]], [[]], Yes, 5, [[]], [[], [a, a, a, a, a]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (bounded_equality, 4, [[]], [[]], Yes, 5, [[]], [[], [a, a, a, a, a]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0085/) Regular expression · qf-mod49-regex-boundary · candidate ## 86. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=concat_order, complete model={"alphabet":["a","b","c"],"left":{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},"right":{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"}]},"max_length":5}. Give (mode, left_AST, right_AST, left_bounded_language, right_bounded_language, bounded_languages_equal, order_irrelevance_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(concat_order, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"a"}],type:"concat"}, [[a, a, a, a, b]], [[b, a, a, a, a]], No, No, concatenation_order_is_semantic) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (concat_order, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"a"}],type:"concat"}, [[a, a, a, a, b]], [[b, a, a, a, a]], No, No, concatenation_order_is_semantic). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0086/) Regular expression · qf-mod49-regex-boundary · candidate ## 87. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=bounded_equality, complete model={"alphabet":["b"],"left":{"type":"epsilon"},"right":{"type":"union","children":[{"type":"epsilon"},{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"}]}]},"max_length":5}. Give (mode, bound, left_bounded_language, right_bounded_language, equal_through_bound, next_bound, left_next_language, right_next_language, differs_at_next_bound, unbounded_equivalence_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(bounded_equality, 5, [[]], [[]], Yes, 6, [[]], [[], [b, b, b, b, b, b]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (bounded_equality, 5, [[]], [[]], Yes, 6, [[]], [[], [b, b, b, b, b, b]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0087/) Regular expression · qf-mod49-regex-boundary · candidate ## 88. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=concat_order, complete model={"alphabet":["a","b","c"],"left":{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},"right":{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"b"}]},"max_length":2}. Give (mode, left_AST, right_AST, left_bounded_language, right_bounded_language, bounded_languages_equal, order_irrelevance_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(concat_order, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"b"}],type:"concat"}, [[b, c]], [[c, b]], No, No, concatenation_order_is_semantic) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (concat_order, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"b"}],type:"concat"}, [[b, c]], [[c, b]], No, No, concatenation_order_is_semantic). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0088/) Regular expression · qf-mod49-regex-boundary · candidate ## 89. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=bounded_equality, complete model={"alphabet":["c"],"left":{"type":"epsilon"},"right":{"type":"union","children":[{"type":"epsilon"},{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"c"},{"type":"symbol","value":"c"},{"type":"symbol","value":"c"},{"type":"symbol","value":"c"},{"type":"symbol","value":"c"},{"type":"symbol","value":"c"}]}]},"max_length":6}. Give (mode, bound, left_bounded_language, right_bounded_language, equal_through_bound, next_bound, left_next_language, right_next_language, differs_at_next_bound, unbounded_equivalence_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(bounded_equality, 6, [[]], [[]], Yes, 7, [[]], [[], [c, c, c, c, c, c, c]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (bounded_equality, 6, [[]], [[]], Yes, 7, [[]], [[], [c, c, c, c, c, c, c]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0089/) Regular expression · qf-mod49-regex-boundary · candidate ## 90. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=concat_order, complete model={"alphabet":["a","b","c"],"left":{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},"right":{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"c"},{"type":"symbol","value":"c"}]},"max_length":3}. Give (mode, left_AST, right_AST, left_bounded_language, right_bounded_language, bounded_languages_equal, order_irrelevance_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(concat_order, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"c"},{type:"symbol",value:"c"}],type:"concat"}, [[c, c, a]], [[a, c, c]], No, No, concatenation_order_is_semantic) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (concat_order, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"c"},{type:"symbol",value:"c"}],type:"concat"}, [[c, c, a]], [[a, c, c]], No, No, concatenation_order_is_semantic). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0090/) Regular expression · qf-mod49-regex-boundary · candidate ## 91. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=bounded_equality, complete model={"alphabet":["a"],"left":{"type":"epsilon"},"right":{"type":"union","children":[{"type":"epsilon"},{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"}]}]},"max_length":7}. Give (mode, bound, left_bounded_language, right_bounded_language, equal_through_bound, next_bound, left_next_language, right_next_language, differs_at_next_bound, unbounded_equivalence_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(bounded_equality, 7, [[]], [[]], Yes, 8, [[]], [[], [a, a, a, a, a, a, a, a]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (bounded_equality, 7, [[]], [[]], Yes, 8, [[]], [[], [a, a, a, a, a, a, a, a]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0091/) Regular expression · qf-mod49-regex-boundary · candidate ## 92. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=concat_order, complete model={"alphabet":["a","b","c"],"left":{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},"right":{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"}]},"max_length":4}. Give (mode, left_AST, right_AST, left_bounded_language, right_bounded_language, bounded_languages_equal, order_irrelevance_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(concat_order, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"a"}],type:"concat"}, [[a, a, a, b]], [[b, a, a, a]], No, No, concatenation_order_is_semantic) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (concat_order, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"a"}],type:"concat"}, [[a, a, a, b]], [[b, a, a, a]], No, No, concatenation_order_is_semantic). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0092/) Regular expression · qf-mod49-regex-boundary · candidate ## 93. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=bounded_equality, complete model={"alphabet":["b"],"left":{"type":"epsilon"},"right":{"type":"union","children":[{"type":"epsilon"},{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"b"}]}]},"max_length":1}. Give (mode, bound, left_bounded_language, right_bounded_language, equal_through_bound, next_bound, left_next_language, right_next_language, differs_at_next_bound, unbounded_equivalence_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(bounded_equality, 1, [[]], [[]], Yes, 2, [[]], [[], [b, b]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (bounded_equality, 1, [[]], [[]], Yes, 2, [[]], [[], [b, b]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0093/) Regular expression · qf-mod49-regex-boundary · candidate ## 94. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=concat_order, complete model={"alphabet":["a","b","c"],"left":{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},"right":{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"}]},"max_length":5}. Give (mode, left_AST, right_AST, left_bounded_language, right_bounded_language, bounded_languages_equal, order_irrelevance_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(concat_order, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"b"}],type:"concat"}, [[b, b, b, b, c]], [[c, b, b, b, b]], No, No, concatenation_order_is_semantic) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (concat_order, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"b"}],type:"concat"}, [[b, b, b, b, c]], [[c, b, b, b, b]], No, No, concatenation_order_is_semantic). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0094/) Regular expression · qf-mod49-regex-boundary · candidate ## 95. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=bounded_equality, complete model={"alphabet":["c"],"left":{"type":"epsilon"},"right":{"type":"union","children":[{"type":"epsilon"},{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"c"},{"type":"symbol","value":"c"}]}]},"max_length":2}. Give (mode, bound, left_bounded_language, right_bounded_language, equal_through_bound, next_bound, left_next_language, right_next_language, differs_at_next_bound, unbounded_equivalence_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(bounded_equality, 2, [[]], [[]], Yes, 3, [[]], [[], [c, c, c]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (bounded_equality, 2, [[]], [[]], Yes, 3, [[]], [[], [c, c, c]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0095/) Regular expression · qf-mod49-regex-boundary · candidate ## 96. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=concat_order, complete model={"alphabet":["a","b","c"],"left":{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"a"}]},"right":{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"c"}]},"max_length":2}. Give (mode, left_AST, right_AST, left_bounded_language, right_bounded_language, bounded_languages_equal, order_irrelevance_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(concat_order, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"c"}],type:"concat"}, [[c, a]], [[a, c]], No, No, concatenation_order_is_semantic) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (concat_order, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"a"}],type:"concat"}, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"c"}],type:"concat"}, [[c, a]], [[a, c]], No, No, concatenation_order_is_semantic). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0096/) Regular expression · qf-mod49-regex-boundary · candidate ## 97. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=bounded_equality, complete model={"alphabet":["a"],"left":{"type":"epsilon"},"right":{"type":"union","children":[{"type":"epsilon"},{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"}]}]},"max_length":3}. Give (mode, bound, left_bounded_language, right_bounded_language, equal_through_bound, next_bound, left_next_language, right_next_language, differs_at_next_bound, unbounded_equivalence_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(bounded_equality, 3, [[]], [[]], Yes, 4, [[]], [[], [a, a, a, a]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (bounded_equality, 3, [[]], [[]], Yes, 4, [[]], [[], [a, a, a, a]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0097/) Regular expression · qf-mod49-regex-boundary · candidate ## 98. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=concat_order, complete model={"alphabet":["a","b","c"],"left":{"type":"concat","children":[{"type":"symbol","value":"a"},{"type":"symbol","value":"a"},{"type":"symbol","value":"b"}]},"right":{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"a"},{"type":"symbol","value":"a"}]},"max_length":3}. Give (mode, left_AST, right_AST, left_bounded_language, right_bounded_language, bounded_languages_equal, order_irrelevance_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(concat_order, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"a"},{type:"symbol",value:"a"}],type:"concat"}, [[a, a, b]], [[b, a, a]], No, No, concatenation_order_is_semantic) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (concat_order, {children:[{type:"symbol",value:"a"},{type:"symbol",value:"a"},{type:"symbol",value:"b"}],type:"concat"}, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"a"},{type:"symbol",value:"a"}],type:"concat"}, [[a, a, b]], [[b, a, a]], No, No, concatenation_order_is_semantic). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0098/) Regular expression · qf-mod49-regex-boundary · candidate ## 99. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=bounded_equality, complete model={"alphabet":["b"],"left":{"type":"epsilon"},"right":{"type":"union","children":[{"type":"epsilon"},{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"}]}]},"max_length":4}. Give (mode, bound, left_bounded_language, right_bounded_language, equal_through_bound, next_bound, left_next_language, right_next_language, differs_at_next_bound, unbounded_equivalence_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(bounded_equality, 4, [[]], [[]], Yes, 5, [[]], [[], [b, b, b, b, b]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (bounded_equality, 4, [[]], [[]], Yes, 5, [[]], [[], [b, b, b, b, b]], Yes, No, bounded_language_equality_does_not_prove_regex_equivalence). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0099/) Regular expression · qf-mod49-regex-boundary · candidate ## 100. Empty Language, Epsilon, Wildcards and Bounded Equality Are Not Interchangeable Use mode=concat_order, complete model={"alphabet":["a","b","c"],"left":{"type":"concat","children":[{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"c"}]},"right":{"type":"concat","children":[{"type":"symbol","value":"c"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"},{"type":"symbol","value":"b"}]},"max_length":4}. Give (mode, left_AST, right_AST, left_bounded_language, right_bounded_language, bounded_languages_equal, order_irrelevance_claim, boundary). The four modes have6/5/11/8 fields. [] is the empty language;[[]] contains the empty word. bounded_equality supplies a genuine next-length distinction; concatenation retains order. View answer and explanation(concat_order, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"b"}],type:"concat"}, [[b, b, b, c]], [[c, b, b, b]], No, No, concatenation_order_is_semantic) empty_vs_epsilon separates [] from[[]]. A wildcard token is not a core typed AST node. bounded_equality retains equality through the bound and a genuine distinguishing next-length slice, so unbounded equivalence is No. concat_order retains the ordered ASTs and their genuinely different bounded languages. The exact result, retaining the requested order, is (concat_order, {children:[{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"c"}],type:"concat"}, {children:[{type:"symbol",value:"c"},{type:"symbol",value:"b"},{type:"symbol",value:"b"},{type:"symbol",value:"b"}],type:"concat"}, [[b, b, b, c]], [[c, b, b, b]], No, No, concatenation_order_is_semantic). [Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r06-regular-expression-0100/) Regular expression · qf-mod49-regex-boundary · candidate [Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261118-r06-regular-expression-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261015-advance-20261118-r06-regular-expression-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261118-r06-regular-expression-foundations/validation-report.json)