Show that { xy ∣ |x| = |y|, x ≠ y } is context-free
Claim: $L$ is context-free.

Proof Idea: There has to be at least one difference between the first and second half; we give a grammar that makes sure to generate one and leaves the rest arbitrary.

Proof: For sake of simplicity, assume a binary alphabet $\Sigma = \{a,b\}$. The proof readily extends to other sizes. Consider the grammar $G$:

$\qquad\begin{align} 
  S &amp;\to AB \mid BA \\
  A &amp;\to a \mid aAa \mid aAb \mid bAa \mid bAb \\
  B &amp;\to b \mid aBa \mid aBb \mid bBa \mid bBb 
\end{align}$

It is quite clear that it generates

$\qquad \mathcal{L}(G) = \{ \underbrace{w_1}_k x \underbrace{w_2v_1}_{k+l}y\underbrace{v_2}_l \mid |w_1|=|w_2|=k, |v_1|=|v_2|=l, x\neq y \} \subseteq \Sigma^*;$

the suspicious may perform a nested induction over $k$ and $l$ with case distinction over pairs $(x,y)$. Now, $w_2$ and $v_1$ commute (intuitively speaking, $w_2$ and $v_1$ can exchange symbols because both contain symbols chosen independently from the rest of the word). Therefore, $x$ and $y$ have the same position (in their respective half), which implies $\mathcal{L}(G) = L$ because $G$ imposes no other restrictions on its language.



The interested reader may enjoy two follow-up problems:

Exercise 1: Come up with a PDA for $L$!

Exercise 2: What about $\{xyz \mid |x|=|y|=|z|, x\neq y \lor y \neq z \lor x \neq z\}$?