When does $p$ break the pumping lemma

I want to prove that $$\{A^iB^jC^k \mid i=j \text{ or } j=k\}$$ is a not a regular language using the pumping lemma.

I've found that the only way to obtain a contradiction is when $$x \in A^*$$, $$y \in B^*$$, $$z \in C^*$$, as $$y$$ is unable to be pumped up and still fit in the language. However if I have any combination of $$y \in B^*C^*$$ or $$y \in A^*B^*$$ and either $$x$$ or $$z$$ is empty, it appears as if $$y$$ can be pumped and thus this would be a regular language. The only way I can think that this would violate the rules of pumping lemma is if $$|xy| ≤ P$$. Assuming that $$P$$ is $$2$$ (which may be completely wrong), then the above still wouldn't violate the pumping lemma and thus this would be a regular language, right?

• I cannot make heads or tails of your question. – Yuval Filmus Feb 7 '20 at 3:56

Suppose that $$L$$ is a language such that for all $$p>0$$, the following holds. There exists a word $$w \in L$$ of length at least $$p$$ such that for every decomposition $$w=xyz$$ in which $$|xy| \leq p$$ and $$y \neq \epsilon$$, there exists $$t \geq 0$$ such that $$xy^tz \notin L$$. Then $$L$$ is not regular.
Now let us prove that your language is not regular. Given $$p > 0$$, let $$w = a^pb^p \in L$$, which certainly has length at least $$p$$. Given a decomposition $$w = xyz$$ with $$|xy| \leq p$$ and $$y \neq \epsilon$$, we see that $$y = a^i$$ for some $$i \neq 0$$. Therefore $$xy^0z = a^{p-i}b^p \notin L$$. Hence the pumping lemma applies, and shows that $$L$$ is not regular.