How would I go about proving that the language accepted by a regular expression is subset of the language...
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Recursive cases: Let A, B be arbitrary RegExps and Let C$_{A}$, C$_{B}$ be cfg(A) and cfg(B) where the properties of cfg can be defined as:
- cfg($phi$) = the CFG with no productions
- cfg($epsilon$) = the CFG with no productions
- cfg(a) = the CFG with just the prouction s $to$ a for any a $in$ $sum$
- cfg(AB) = C$_{A}$, C$_{B}$, S$to$S$_{A}$S$_{B}$
- cfg(A$bigcup$B) = C$_{A}$, C$_{B}$, S$to$S$_{A}$ | S$_{B}$
- cfg(A*) = C$_{A}$, S$toepsilon$ | S$_{A}$S$_{B}$
The goal: Prove that for any R $in$ RegExp, the language accepted by R is a subset of the language accepted by cfg(R).
For my base cases, I think I should probably make A, B be empty sets... but where do I go from there? I'm still really confused about the structure of induction proofs and what my inductive hypothesis should be.
discrete-mathematics proof-writing recursion context-free-grammar regular-expressions
asked Dec 6 at 1:01
digiHarmonious
211
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Recursive cases: Let A, B be arbitrary RegExps and Let C$_{A}$, C$_{B}$ be cfg(A) and cfg(B) where the properties of cfg can be defined as:
- cfg($phi$) = the CFG with no productions
- cfg($epsilon$) = the CFG with no productions
- cfg(a) = the CFG with just the prouction s $to$ a for any a $in$ $sum$
- cfg(AB) = C$_{A}$, C$_{B}$, S$to$S$_{A}$S$_{B}$
- cfg(A$bigcup$B) = C$_{A}$, C$_{B}$, S$to$S$_{A}$ | S$_{B}$
- cfg(A*) = C$_{A}$, S$toepsilon$ | S$_{A}$S$_{B}$
The goal: Prove that for any R $in$ RegExp, the language accepted by R is a subset of the language accepted by cfg(R).
For my base cases, I think I should probably make A, B be empty sets... but where do I go from there? I'm still really confused about the structure of induction proofs and what my inductive hypothesis should be.
discrete-mathematics proof-writing recursion context-free-grammar regular-expressions
asked Dec 6 at 1:01
digiHarmonious
211
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Recursive cases: Let A, B be arbitrary RegExps and Let C$_{A}$, C$_{B}$ be cfg(A) and cfg(B) where the properties of cfg can be defined as:
- cfg($phi$) = the CFG with no productions
- cfg($epsilon$) = the CFG with no productions
- cfg(a) = the CFG with just the prouction s $to$ a for any a $in$ $sum$
- cfg(AB) = C$_{A}$, C$_{B}$, S$to$S$_{A}$S$_{B}$
- cfg(A$bigcup$B) = C$_{A}$, C$_{B}$, S$to$S$_{A}$ | S$_{B}$
- cfg(A*) = C$_{A}$, S$toepsilon$ | S$_{A}$S$_{B}$
The goal: Prove that for any R $in$ RegExp, the language accepted by R is a subset of the language accepted by cfg(R).
For my base cases, I think I should probably make A, B be empty sets... but where do I go from there? I'm still really confused about the structure of induction proofs and what my inductive hypothesis should be.
discrete-mathematics proof-writing recursion context-free-grammar regular-expressions
asked Dec 6 at 1:01
digiHarmonious
211
Recursive cases: Let A, B be arbitrary RegExps and Let C$_{A}$, C$_{B}$ be cfg(A) and cfg(B) where the properties of cfg can be defined as:
- cfg($phi$) = the CFG with no productions
- cfg($epsilon$) = the CFG with no productions
- cfg(a) = the CFG with just the prouction s $to$ a for any a $in$ $sum$
- cfg(AB) = C$_{A}$, C$_{B}$, S$to$S$_{A}$S$_{B}$
- cfg(A$bigcup$B) = C$_{A}$, C$_{B}$, S$to$S$_{A}$ | S$_{B}$
- cfg(A*) = C$_{A}$, S$toepsilon$ | S$_{A}$S$_{B}$
The goal: Prove that for any R $in$ RegExp, the language accepted by R is a subset of the language accepted by cfg(R).
For my base cases, I think I should probably make A, B be empty sets... but where do I go from there? I'm still really confused about the structure of induction proofs and what my inductive hypothesis should be.
discrete-mathematics proof-writing recursion context-free-grammar regular-expressions
discrete-mathematics proof-writing recursion context-free-grammar regular-expressions
asked Dec 6 at 1:01
digiHarmonious
211
asked Dec 6 at 1:01
digiHarmonious
211
asked Dec 6 at 1:01
digiHarmonious
211
asked Dec 6 at 1:01
digiHarmonious
211
asked Dec 6 at 1:01
digiHarmonious
211
211
add a comment |
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