# Proving that $L=\{ \langle M \rangle \colon L(M)=L(M)^R \}$ is undecidable [duplicate]

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I'm trying to show that $L=\{ \langle M \rangle \colon L(M)=L(M)^R\}$ is undecidable, but I don't even know where to begin. Google wasn't much of a help, maybe because it's hard describing the language in words. Any hints?

## marked as duplicate by Raphael♦Jul 12 '15 at 9:46

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• Use Rice's theorem. If you don't know what that is, look it up and read the proof. Now you can prove your special case in the very same way. – Yuval Filmus Jul 11 '15 at 19:49
• What do you define $L(M)^{R}$ to be? – Ryan Jul 11 '15 at 20:28
• Every string in $L(M)^R$ is a reverse of some string in $L(R)$ – Yotam Jul 12 '15 at 7:42
• "I don't even know where to begin" -- review your class material, textbooks, and our reference question. "it's hard describing the language in words" -- what about "the set of Turing machines whose languages are closed against reversal"? – Raphael Jul 12 '15 at 9:48