I've encountered many examples which its format is like: a^n b^n. For this I understand that w = 2n and is pretty straightforward, but what happens in my case? Is w = 1 + 2n? And in this case would |xy| be a subword of just a, or a subword of a b^n?
I'm struggling to pump the language in any meaningful way to get my proof going, any help is appreciated. Thanks.
Suppose that $L$ is regular and let $p$ be its pumping length. Consider $w = a b^p c^p \in L$. From the pumping lemma you know that $w$ can be written as $w=xyz$ with $|xy| \le p$ and $|y| \ge1$ such that $xy^ip \in L$ for all $i \ge 0$.
Notice that $y$ must contain the initial $a$ or at least one $b$ (possibly both) and cannot contain any $c$. Then $xz = xy^0z \not\in L$ since it either contains no $a$s or it contains less than $p$ $b$s and $p$ $c$s. This provides the sought contradiction.
