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This question already has an answer here:

Are there any closure properties which I can use to come up with a contradiction?

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marked as duplicate by xskxzr, Yuval Filmus, David Richerby, Evil, Apass.Jack Nov 19 '18 at 1:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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 Jun 13 '17 at 17:32
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$L_1 = \{a^ib^jc^k: i,j,k \ge0\} = a^*b^*c^*$ is regular.

Assume that $L = \{a^ib^jc^k: (i=1) \Rightarrow (j=k)\}$ is also regular.

Then $L_2 = \{ab^jc^k: j\neq k\} = L_1 - L$. By closure properties (difference of regular sets is regular) $L_2$ is regular. But using Pumping lemma it is not difficult to prove that $L_2$ is not regular. So we have a contradiction.

In fact we could prove that $L$ is not regular directly using Pumping lemma.

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Let $h$ be the homomorphism defined by $h(a) = \epsilon$, $h(b) = b$, $h(c) = c$. Then $$ h(L \cap ab^*c^*) = \{ b^n c^n : n \geq 0 \}. $$

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