Prove or disprove: Complement of language $L=\left\{baba^2ba^3b...ba^{n-1}ba^nb \, | \, n \geq 1\right\}$ is context-free
No, reasoning with the complement will not work as context-free languages are not closed under complement.

Consider an alternative description of $L$: it consists of all strings $ba^{a_1}ba^{a_2} \dots ba^{a_n}b$ such that (1) $a_1 =1$ and (2) for all $1\le k&lt;n$, $a_{k+1} = a_k+1$.