# What is the importance of the condition "| xy | < p" in pumping lemma? [duplicate]

Let $$L$$ be a language. $$w \in\ L$$ , and $$w$$ could be broken in $$xyz$$.

Then if $$L$$ is regular, there exists a pumping length $$p$$ such that:

1. $$\vert y\vert>0$$
2. $$\vert xy\vert \leq p$$
3. $$\forall i \geq 0,\ xy^i z\ \in L$$

I understand that this lemma is

• based on pigeon hole principle
• we use it to prove that a language is not regular by contradiction.

For condition (1), I understand that the string we are pumping should be $$\geq 1$$.

But I don't understand the significance of condition (2).

Why should the length of $$xy$$ be equal to or smaller than $$p$$ . Won't it work otherwise?

$$p$$ is the number of states in the automaton. As you say, the pumping lemma is about the pigeonhole principle. Suppose that $$q$$ is the first state that's repeated when you read input $$w$$. Then $$x$$ is the string that you read before you reach $$q$$ for the first time, and $$y$$ is the string you read between the first and second visits to $$q$$. Since there are only $$p$$ states including the start state, you must repeat a state after doing at most $$p$$ transitions, i.e., reading a string of length at most $$p$$. So the condition $$|xy|\leq p$$ comes from the proof.
In reality, when an automaton reads a particular string $$w$$, there might be many times when it returns to a state that has already been visited. If you take $$w\in L$$ any split $$w=xyz$$ with $$|y|\geq 1$$, such that the automaton is in the same state at the beginning and end of $$y$$ then $$xy^iz$$ will be in $$L$$ for all $$i$$, regardless of how long $$xy$$ is.