Pumping Lemma clarification, and regularity of language of equal number of zeros and ones.

Question

All explanations and proofs I find about the Pumping Lemma are ambiguous. So if I understand this correctly, if we can find some $p>0$, then for any string $|w| \ge p$, we should be able to split it up into $xyz$ according to some conditions which are satisfied as shown below.

Here's the problem. Say $L$ is the language with equal number of zeros and ones. As far as I know, $L$ is considered to be irregular. But we could easily find $p$ that satisfies the conditions:

$$p=4 $$ $$xyz=0011$$ Let $y$ be the middle part $01$.

1. $y$ can be pumped.

2. $|y|>0$

3. $|xy|\le p$

But what is wrong with this reasoning? It seems to satisfy all the conditions yet still this language is considered irregular.

Answer 1

The Pumping Lemma is a statement true of all regular languages. However, it is not necessarily false for irregular languages. In general, the Pumping Lemma is used to prove something must be irregular. It cannot be used to prove something is regular, because of exactly this example.

This is analogous to saying "9 is odd and greater than 2! Why isn't it prime?" Just because all primes greater than 2 are odd doesn't make all odd numbers a prime greater than 2.