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{⟨M,N⟩ | All strings in L(M)∩L(N) begin with 110.}

I think that this language is decidable. We can make a Turing Machine TM, which takes as input . For every string that is in L(M)∩L(N), if the string starts with 110, after the first 3 digits, we halt and accept. If the first three digits are not 110, we halt and reject. I am unsure what we do if the string is not in L(M)∩L(N).

Also overall I am unsure if my Turing Machine is actually working or not. Could I get some feedback on this?

jPrime
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1 Answers1

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If M and N are Turing machines, then this language is not decidable. If it were, we could make N a TM that accepts all strings, and then we'd have a decider for {M | all strings in M begin with 110}. We can recognize this is not decidable since the condition is true for some TMs, false for others, and it's semantic in that it deals with the strings in the language; so Rice's theorem applies.

Patrick87
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