Typed Regular Expressions, Complete Bounded Languages and Membership
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batch-20261015-advance-20261118-r06-regular-expression-foundations · 2026-10-15 · 100 questions
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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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Regular expression · qf-mod49-regex-boundary · candidate
Canonical question JSONL · English JSONL · Validation report