How can ws with |w| = |s| and w ≠ s be context-free while w#s is not?

Why does (if so) the seperator $ \#$ is making a difference between the two languages ?

Let say:

$ L=\{ws : |w|=|s|\, w,s\in \{0,1\}^{*}, w \neq s \}$

$ L_{\#}=\{w\#s : |w|=|s|\, w,s\in \{0,1\}^{*}, w \neq s \}$

Here is a proof and a grammer representing $ L$ as a $ CFL$

And below Im adding a proof for $ L_{\#} \notin CFL$ :

Does the $ \#$ sign truly make a difference? if so, why is that ? and if not, which one of the proofs is wrong and where?

Proof that $ L_{\#} \notin CFL$ :

Assume by way of contradiction that $ L ∈ CFL$ . Let $ p > 0$ be the pumping constant for $ L$ guaranteed by the pumping lemma for context-free languages. We consider the word $ s = 0^{m}1^{p}\#0^{p}1^{m}$ where $ m=p!+p$ so $ s ∈ L$ . Since $ |s| > p$ , according to the pumping lemma there exists a representation $ s = uvxyz$ , such that $ |vy| > 0$ , $ |vxy| ≤ p$ , and $ uv^{j}xy^{j} z ∈ L$ for each $ j ≥ 0$ .

We get a contradiction by cases:

  • If $ v$ or $ y$ contain $ \#$ : Then for $ i = 0$ , we get that $ uxz$ does not contain $ \#$ , so $ uxz \notin L$ in contradiction.
  • If both $ v$ and $ y$ are left to $ \#$ : Then for $ i = 0$ , we get that $ uxz$ is of the form $ w\#x$ , where $ |w| < |x|$ , so $ uxz \notin L$ .

  • If both $ v$ and $ y$ are right to $ \#$ : Similar to the last case.

  • If $ v$ is left to $ \#$ , $ y$ is right to it, and $ |v| < |y|$ : Then for $ i = 0$ , we get that $ uxz$ is of the form $ w\#x$ , where $ |w| > |x|$ , so $ uxz \notin L$ .

  • If $ v$ is left to $ \#$ , $ y$ is right to it, and $ |v| > |y|$ : Similar to the last case.

  • If $ v$ is left to $ \#$ , $ y$ is right to it, and $ |v| = |y|$ : This is the most interesting case. Since $ |vxy| ≤ p$ , $ v$ must be contained in the $ 1^{p}$ part of $ s$ , and $ y$ in the $ 0^{p}$ part. So it holds that $ v = 1^{k}$ and $ y = 0^{k}$ for the same $ 1 ≤ k ≤ p$ (in fact, it must be that $ k < p/2$ ). For each $ j ≥ 0$ , it holds that $ uv^{j+1}xy^{j+1}z = 0^{m}1^{p+j·k}\#0^{p+j·k}1^{m}$ , so if it happens that $ m = p + j · k$ , then it holds that $ uv^{j+1}xy^{j+1}z \notin L$ in contradiction. To achieve this, we must take $ j = (m-p)/k$ , which is valid only if $ m − p$ is divisible by $ k$ . Recall that we chose $ m = p + p!$ , so $ m − p = p!$ , and $ p!$ is divisible by any $ 1 ≤ k ≤ p$ as wanted.