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

Question

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

Answer 1

$L_1 = \{a^ib^jc^k: i,j,k \ge0\} = a^*b^*c^*$ 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.

Answer 2

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