Is $L = \{w_1cw_2 : w_1, w_2 \in \{a, b\}^* , w_1 \neq w_2 \}$ a CFL?
In my opinion it is not since if we want to know the inequality of $w_1$ and $w_2$ we must be aware of their equality and that is not a $CFG$.
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@G.Bach: I don't agree, you cannot check two words for equality with a DFA. Hadi_Amiri: be careful, CFG are not closed under complement – Denis Dec 09 '13 at 13:55
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as far as I know for $ww$, turing machine is the only accpter, but for this case I'm confused, anyway can anyone write a argument ? – Hadi Amiri Dec 09 '13 at 14:14
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What have you tried? We expect you to make a serious effort before asking and to show us what you've tried in the question. – D.W. Dec 09 '13 at 18:34
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I thought that with simple observation we can determine which class this language belong. – Hadi Amiri Dec 09 '13 at 19:16
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@HadiAmiri, No, we expect you to make a serious effort on your own, and to show in the question what you've tried. – D.W. Dec 10 '13 at 07:20
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(Very) Similar question. – Raphael Feb 23 '15 at 07:46
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This is a CFL, but the complement (with equality) is not. Notice that both the language and the complement become CFL when you mirror one of the two words.
To show that this is a CFL, you can design a PDA that guesses the letter at position $i$ that will be different in $w_1$ and $w_2$. It start by reading the first $i$ letters of $w_1$ and pushing $i$ symbols to the stack up to this letter, and remembers the letter $a$ in $w_1$ at position $i$. Then, it waits to read $c$, and pops the $i$ symbols off the stack. It accepts if the current letter (at position $i$ in $w_2$) is not $a$.
You have to also treat the case of different lengths, but it's not hard, I leave it as exercise ;)
Denis
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