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Prove that the class of Turing decidable languages is strictly larger than the class of context free languages. (Give a language that is Turing decidable, but which violates the pumping lemma for context free languages).

I don't know how to solve this. Please help

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HINT: Can you think of a set $A$ of finite strings which is

  • Easy to describe (this is informal, but it's better to start informally than to look for something which you know fits the definition of "decidable" - in practice, most things which are "easily describable" will be decidable), but

  • Doesn't satisfy the pumping lemma?

In fact, probably any example you've seen of a set of strings not satisfying the pumping lemma is Turing decidable . . .

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  • $\begingroup$ Sorry, I'm pretty new to this. I don't even know where to start. $\endgroup$ – Brice Petty Nov 21 '15 at 1:31
  • $\begingroup$ @BricePetty Have you had any exercises of the form "Show that --- is not a CFL by showing it fails the pumping lemma?" $\endgroup$ – Noah Schweber Nov 21 '15 at 1:37
  • $\begingroup$ yes, I have. I just don't understand the connection $\endgroup$ – Brice Petty Nov 21 '15 at 2:58
  • $\begingroup$ @BricePetty OK, so take your favorite example of a language L which is not a CFL, since it fails the pumping lemma. If you can show L is Turing decidable, then you're done. $\endgroup$ – Noah Schweber Nov 21 '15 at 3:01
  • $\begingroup$ Okay, I have chosen the language L = even length binary string where the first half of the string is the same sequence as he second half of the string. By pumping lemma, I have proven that this language is not a CFL. Now, how would I go about showing this string is turing decidable? $\endgroup$ – Brice Petty Nov 21 '15 at 4:16

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