# Prove that $L=\{a^i b^j c^k \mid i=1$ implies $j=k\}$ is NOT regular [duplicate]

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

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