# How can I prove that the language is not CFL?

I have this language: $$L=\{x^ny^jz^k:k=j\cdot n\}$$, I know I can use the pumping lemma for CFL, but I have managed to do it for $$k, j, n \leq 0$$, and with $$k = j \cdot n$$ I have not been able.

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– D.W.
May 29, 2021 at 7:22

We will replace the symbols x,y,z by a,b,c respectively , so we don't get confused when using the pumping lemma

Consider the string w = a^p b^p c^(p^2) , clearly w ∈ L , we will show that w cannot be pumped

Neither v or y can have 2 type of symbols or the order gets messed up when pumping , so both v and y must have one type of symbols ( so for ex v has only as , y has only bs )

Now if v and y don't have cs the string can't be pumped , if they have only cs the string can't be pumped too , since you can't maintain k=jn

So we know now that each of v and y must have only one type of symbol and that they must include cs but no only cs ,so y must have cs , and v must have bs (it can't have as or |vxy| > p)

The only way to pump the string now is for v = b and y = c^p , in this way whenever the string is pumped , when we add a b we add c^p , maintaining k=jn

But for this to happen |vxy| > p which is unacceptable

So w can't be pumped and thus L is not a CFL