As a follow up to this question
Given an alphabet $\{a, b\}$.
Why are palindromes not regular? Could you not select $x=z=(a|b)$ and $y=$ the remaining characters in the word.
For example given $aba$ could you not select $x=z=a$ and $y = b$ and create a valid pumpible solution?
Or $abaaba$ $x=z=a$ $y=baab$ ?
I thought that as long as you could find a way to assign $x, y, z$ such that a repeating $y$ created a word in the language, the language couldn't be proven irregular.
The pumping lemma for regular languages says that if a language $L$ is regular, there is a constant $N \ge 1$ such that all strings $\sigma \in L$ with $\lvert \sigma \rvert \ge N$ can be written $\alpha \beta \gamma = \sigma$, where $\lvert \alpha \beta \rvert \le N$, $\beta \ne \epsilon$, so that for all $k \ge 0$ it is $\alpha \beta^k \gamma \in L$.
To prove the language of even length palindromes over $\Sigma = \{a, b\}$, say, is not regular we use a proof by contradiction.
Suppose $L$ is regular, so the lemma applies. Call the constant of the lemma $N$, and select $\sigma = a^N b b a^N$. Clearly $\lvert \sigma \rvert = 2 N + 2 \ge N$, so we can divide $\sigma = \alpha \beta \gamma$. As $\lvert \alpha \beta \rvert \le N$, they are formed only by $a$s. If you leave out $\beta$ (take $k = 0$), there will be less $a$ before $bb$ than after, i.e., $\alpha \beta^0 \gamma \notin L$, contradiction.
Note that we want to contradict the lemma. It says there is $N$, so we can't assume anything about it, just that it is a positive integer; it talks about all long enough strings, we can select any $\sigma$ that is easy to work with; there is some division into $\alpha, \beta, \gamma$, we have no control over that, just know it has to respect $\lvert \alpha \beta \rvert \le N$ and $\beta \ne \epsilon$; for all $k$ it is $\alpha \beta^k \gamma \in L$, we can select a $k$ at will so $\alpha \beta^k \gamma \notin L$ and have a contradiction (but if you manage to prove that $\alpha \beta^1 \gamma \notin L$, you are way off base ;-).
