Is $\{ w_1cw_2 \mid w_1 ≠ w_2 \}$ a context-free language?
The idea is that $w_1 \neq w_2$ if and only if either (1) $|w_1| \neq |w_2|$ or (2) the $i$th letters of $w_1,w_2$ are different. This leads to the following partition of $L_1$:
$$
\begin{align*}
L_1 &amp;= (a+b)^+(a+b)^nc(a+b)^n \\ &amp;\cup (a+b)^nc(a+b)^n(a+b)^+ \\ &amp;\cup (a+b)^na(a+b)^*c(a+b)^nb(a+b)^* \\ &amp;\cup (a+b)^nb(a+b)^*c(a+b)^na(a+b)^*
\end{align*}
$$
Each of the summands is clearly context-free, hence so is their union.