The pumping lemma for regular languages

Choose the DFA $D$ recognizing $L$ supplied by [L0], and let $p$ be its number of states, which is finite by [L1]. Fix a word $w=a_1\cdots a_m\in L$ with $m\ge p$. As $D$ reads the prefixes $\epsilon ,a_1,a_1{a}_2,\dots ,a_1\cdots a_p$, [L3] gives $p+1$ visited states.

By [L4], two of those first $p+1$ visited states are equal; write them as the states after the prefixes of lengths $i<j\le p$, and set $x:=a_1\cdots a_i$, $y:=a_{i+1}\cdots a_j$, and $z:=a_{j+1}\cdots a_m$. Then $w=xyz$, $\mathrm{\mid }xy\mathrm{\mid }=j\le p$, and $\mathrm{\mid }y\mathrm{\mid }=j-i>0$.

Because the run returns to the same state after reading $y$, repeating or deleting that loop does not change the state reached before the suffix $z$. Therefore [L2] implies $x{y}^{i}z\in L$ for every $i\ge 0$, so the factorization from step 2.1 satisfies all three required conditions.