# requirement for pumping lemma in regular language

I am a bit confused on the theory of the pumping lemma. As I know is used to decide if a language is regular or not.

This is what I have understood so far though For a regular language $L$, there exists a $p > 0$ such that for all $w ∈ L$ where $|w| ≥ p$, there exists some split $w = vxu$, for which the following holds:

$|vx| ≤ p$

$|x| > 0$

$vx^iu ∈ L$ for all $i ≥ 0$

but what is the rationale behind the requirement of $|vx| ≤ p$ what happens if we drop that requirement??

The constraint $|vx| \leq p$, which is necessary for the effectiveness of the lemma as showed by the answers, comes up directly out of the proof. Have you seen the proof of the pumping lemma? –  Yuval Filmus May 25 at 18:02
$|vx| < p$ mean that $x$ is short and it is at the beginning of $w$. If you were to remove that constraint, all regular languages would still satisfy the lemma, but more irregular ones would too. For example $L = \{a^n \mid n ~\text {is not prime}\}$, when $n = m \cdot k$ can be pumped by taking $x = a^k$.