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Equivalence of CFGs and PDAs

Equivalence of CFGs and PDAs. Section 3.4 Mon, Oct 24, 2005. Every CFG is Equivalent to a PDA. Theorem: Let G = {V, ?, R , S } be a context-free grammar. There exists a PDA M such that L ( M ) = L ( G ). Proof: Let M = { K , ?, ?, ?, s , F }, where K = { s , f }. ? = V.

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Equivalence of CFGs and PDAs