Proving irregularity of $\{a^nb^k \mid n > k \text{ or } n \neq k-1\}$
If $n &gt; k$ then $n \neq k-1$ (since $k-1 \not&gt; k$), so your language is really $L = \{ a^n b^k \mid n \neq k-1 \}$. If $L$ were regular then so would be the following language $a(a^*b^* \setminus L) \cup \{\epsilon\} = \{ a^n b^n \mid n \geq 0 \}$, which is known not to be regular.