# Use of pumping lemma for not regular languages. (Proof Verification)

$$L=\{w \in \{0,1,a\}^* | \#_0(w) = \#_1(w) \}$$

We show that L is not regular by pumping lemma.

We choose w=$$0^p 1^p a$$

|w| = 2p+1

Now |xy| has to be $$\leq p$$

So x and y could only contain zeros. And $$z=1^p a$$

$$xy^iz= 0^p 1^p a$$

Now let i = 0

$$xy^0z=0^{p-|y|} 1^p a$$

Now hence $$p-|y| \neq p$$ this choice of i would lead to a word that is not in L. So we can not pump y and stay in the language.

So L is not regular.

I’m trying to learn the usage of the pumping lemma. Is my proof correct?

Any suggestions are welcome. Thanks!