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Note

The wording of the title may be a bit vague, but I'm not asking if CFLs are closed under reversal. Please see below.

Problem Description

Given a word $w$, define $w^{r}$ to be its reversal.

Let $L=\{ G \vert G \text{ is a } CFG \text{ and for every } w \in L(G), w^{r} \in L(G) \}$

Prove that $L$ is undecidable.

My Attempt

I am aware that I should reduce a known-to-be-undecidable language to L, but by looking at the four undecidable languages here (Equivalence, Disjointness, Containment, Universality), I still failed to determine which language I can use. Please guide me a direction, thank you.

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Let $G_1,G_2$ be two context-free grammars. We can construct a context-free grammar $G$ such that $$ L(G) = \#L(G_1) \cup L(G_2)^r\#, $$ where $\#$ is a new symbol. The language $L(G)$ is closed under reverse iff $L(G_1) = L(G_2)$.

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  • $\begingroup$ Thank you for the response. But what if $L(G_{1})=\{a\}$ and $L(G_{2})=\emptyset$? In this case, $L(G)=\{\#a\#\}$ is closed under reversal, but $L(G_{1}) \neq L(G_{2})$. $\endgroup$ – David Chen Dec 26 '19 at 9:35
  • $\begingroup$ In your case, $L(G) = \{\#a\}$, which isn’t closed under reversal. $\endgroup$ – Yuval Filmus Dec 26 '19 at 9:37
  • $\begingroup$ Just to make sure, so you mean $$ L(G) = (\#L(G_1)) \cup (L(G_2)^r\#) $$ $\endgroup$ – David Chen Dec 26 '19 at 9:41
  • $\begingroup$ What do you think? What would make sense? $\endgroup$ – Yuval Filmus Dec 26 '19 at 10:09
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    $\begingroup$ I think when there might be ambiguity, it doesn't hurt to make things clear. $\endgroup$ – David Chen Dec 26 '19 at 10:14

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