When does $p$ break the pumping lemma
Let's start with stating the pumping lemma, in its contrapositive form:


  Suppose that $L$ is a language such that for all $p&gt;0$, 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 $p &gt; 0$, 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.