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I am breaking my head over this. Let the alphabet $A$ be given by $A = \{a,b,c\}$ and let

$$L = \{ww^R \mid w \in A^* \}.$$

Prove that the language $L$ is not regular without using the pumping lemma, but using:

  • The fact that $\{a^nb^n \in A^* \mid n \geq 0 \}$ is not regular.
  • The fact that $\{w \in A^* \mid w \text{ is a palindrome} \}$ is not regular.
  • Closure properties of the class of regular languages (such as if $L_1$ and $L_2$ are regular, then $L_1 \cup L_2$ is also regular, etc.).
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  • $\begingroup$ $L$ consists of the even length palindromes. $\endgroup$ – Hendrik Jan Dec 2 '16 at 16:31
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    $\begingroup$ Welcome to Computer Science! What have you tried? Where did you get stuck? We do not want to just hand you the solution; we want you to gain understanding. However, as it is we do not know what your underlying problem is, so we can not begin to help. See here for tips on asking questions about exercise problems. If you are uncertain how to improve your question, why not ask around in Computer Science Chat? $\endgroup$ – Raphael Dec 2 '16 at 18:37
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Let

$$ L_1 = \{\ \omega\omega^r \ | \ \omega \in \Sigma^* \} $$ $$ L_2 = \{\ a^nb^n \ | \ n \in \mathbb{N}\ \}$$

Because $\Sigma = \{a, b, c\}$ is finite, $\Sigma^*$ must be countable.

This means we can define a function $f : \Sigma^* \rightarrow \mathbb{N}$ which returns the position of a word $\omega \in \Sigma^*$ in an arbitrary ordering defined by $f$. $f$ is invertible, we define $f^{-1} : \mathbb{N} \rightarrow \Sigma^*$, which returns the word $\omega$ at position $n$ in the enumeration of $\Sigma^*$ as defined by $f$.

Now, since every word $\omega$ is uniquely defined by a natural number $n$, and every word $\omega$ has exactly one unique reverse $\omega^r$, we can define an isomorphism between $L_1$ and $L_2$:

$$ g \colon\ L_1 \rightarrow L_2; \quad \omega\omega^r \mapsto a^{f(\omega)}b^{f(\omega)} $$

$$ g^{-1} \colon \ L_2 \rightarrow L_1; \quad a^nb^n \mapsto f^{-1}(n) f^{-1}(n)^r $$

Since $L_2$ is not regular and we have a bijection between $L_1$ and $L_2$, $L_1$ must not be regular.

This proof works for any finite alphabet $\Sigma$ because the transitive closure of a finite alphabet is always countable.

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  • $\begingroup$ "isomorphism"? You meant bijection. It does not work; otherwise we can show $\{\ a^n \mid n \in \Bbb{N}\}$ is not regular since we can build an "isomorphism" between it and $L_2$ easily. $\endgroup$ – Apass.Jack Dec 21 '18 at 9:07
  • $\begingroup$ @Apass.Jack Yes, I meant bijection. Thanks! $\endgroup$ – ThreeFx Dec 21 '18 at 9:11
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Suppose that $L$ is regular. Then $L' = L \cap (a+b)^*c^2(a+b)^*$ is also regular. Another description for $L'$ is $$ L' = \{ wccw^R : w \in (a+b)^* \}. $$ Consider now the regular substitution $\sigma$ given by $$ \sigma(a) = \{a\}, \sigma(b) = \{b\}, \sigma(c) = \{a,b,\epsilon\}. $$ Then $L'' = \sigma(L')$ is also regular. But $L''$ is the language of palindromes.

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