Got a final coming up, so I gotta get this concept down.

Given $A= \{(G_1,G_2 \mid G_1$ and $G_2$ are two CFG's and $L(G_1)=L(G_2) \}$. Is $A$ decidable?

Would it be sufficient to take an undecidable language such as $B= \{x \mid L(M_x)$ is a CFG $ \}$. Then reduce it to A with function $f(x) = G'_{M_x}$?


I don't really understand your reduction, but there is a very easy reduction from $ALL_{CFG}=\{G| G \text{ is a CFG and } L(G)=\Sigma^*\}$. Given $G$, Simply output $(G,G_{\Sigma^*})$, where $G_{\Sigma^*}$ is a (concrete) grammar that generates every word.

And the problem $ALL_{CFG}$ can be shown to be undecidable with a very nice (but elaborate) reduction from the word problem for TMs.

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