Prove not context free

Question

How can we prove that:

$$ L = \{ w_1\#w_2 \mid w_1 \in w_2;\; |w_2| > |w_1|;\; w_1 , w_2 \in \{0, 1\}^*\} $$

is not context-free?

The language defines $w_1$ as a sub-string of $w_2$, and they are separated by a $\#$. This is easy with the CFG pumping-lemma for a slightly different language with $|w_2| \ge |w_1|$ by using the special case of $|w_2| = |w_1|$ (i.e. $w_1 = w_2$).

But here, $w_1$ is a proper sub-string of $w_2$ so I can't do the same. I fail to push the string out since we can always pump, for example the first symbol of $w_2$.

Answer 1

For large enough $p$, consider the word $w = 0^{p+1}1^{p+1}\#0^{p+1}1^{p+2} \in L$. Mark the part $1^{p+1}\#0^{p+1}$. According to Ogden's lemma, we can write $w = uxyzv$ such that $xyz$ contains at least $1$ and at most $p$ marked symbols, and $ux^iyz^iv \in L$ for all $i \geq 0$. The pumped part $xyz$ cannot lie all to the right of $\#$ since then pumping it down would result in a word not in the language (here we crucially use the fact that only the $0^{p+1}$ part to the right of $\#$ is marked). It also cannot lie all to the left of $\#$ since then pumping it up would result in a word not in the language. It follows that the part of $xyz$ to the right of $\#$ is of the form $0^k$, and the part of $xyz$ to the left of $\#$ is of the form $1^\ell$ (otherwise, there would be more than $p$ marked symbols). However, pumping up, the resulting word is not in the language. This contradiction shows that $L$ is not context-free.