Theorem $[$Pumping Lemma for Regular Languages, contrapositive$]$:

Language $L$ is not regular if the following holds: for every positive integer $p,$ there

is a string $s$ in $L$ of length at least $p$ such that for all strings $x,y,$ and $z$ with $s=xyz$

and $|xy|p,$ and $|y|>0,$ there is an iinN such that $x{y}^{i}z!$inL.

The operative statement here has four nested quantifiers:

For every positive integer $p,$

there is a string $s$ in $L$ of length at least $p$ such that

for all strings $x,y,$ and $z$ with $s=xyz,|xy|p,$ and $|y|>0,$

there is an iinN such that $x{y}^{i}z!$inL.

Given a language $L,$ consider the following $2-$player game:

player $1$ picks a positive integer $p$

player $2$ picks a string $s$ in $L$ of length at least $p$

Player $1$ picks strings $x,y,$ and $z$ with $s=xyz,|xy|p,$ and $|y|>0,$

Player $2$ picks iinN

Player $2$ wins if $x{y}^{j}z!$inL. Otherwise, Player $1$ wins.

The pumping lemma can be interpreted to say that Player $2$ will have a winning

strategy for this game if and only if $L$ is not regular.

For example, let $L$ be the language of all strings of $a^{\text{'}}$ s and $b^{\text{'}}$ s that are palindromes

$($i$.$e$.,$ the same forward as backward$).$ Player $2\text{'}$s strategy is:

In Step $2,$ choose a string of all $a^{\text{'}}$ s of length p$,$ followed by a string of $b\text{'}$s of

length $p,$ followed again by a string of $p$ a$\text{'}$s$.$

In step $4,$ choose $i=2.$