Is this proof using the pumping lemma correct? I have this proof and it goes like this:

We have a language $L = \{\text{w element of } \{0,1\}^* \mid w = (00)^n1^m \text{ for } n > m \}$.
Then, the following proof is given:
There is a $p$ (pumping length) for $L$. Then we have a word $w = (00)^{p+1}1^p$ and $w$ element of $L$ and $|w| \le p$. $w$ can be written as $xyz$ with $y$ not empty and $|xy| \le p$.
This implies that $y=(00)^n$ for $0 < n \le p$.
Now for all $i \ge 1$ it means $xy^iz = (00)^{p+n(i-1)+1}1^p$ is an element of $L$.
Hence $L$ is regular.

I can clearly see that this proof is wrong (at least that's what I think). I can see that three things are wrong:


*

*The pumping lemma cannot be used to proof the regularity of a language!

*One thing bothers me: $y=(00)^n$ could never hold for $0 < n \le p$ because $n$ can be $p$ and therefore $y=(00)^p$ which means that the length of $y$ is $2p$! Condition no. 2 of the pumping lemma won't hold because $|y| = 2p$ and the 2nd condition of the pumping lemma is $|xy| \le p$.

*Another thing is that you should also consider $i = 0$. Oh, and if $i = 0$ and $n \ge 1$ we would get that $n \le m$ and not $n > m$!


My question is actually if my assumptions are correct because it didn't take me much time to figure this out. There is a chance that I missed something. Can someone clarify?
I'm not asking for a proof, so please don't give one. I just want to know if my assumptions are correct and if not what you would think is wrong?

Yes, the proof is bogus, as Brian M. Scott's answer expertly explains.
It is easier to prove that the reverse of your language isn't regular, or (by minor adjusting the proof of the lemma to place the pumped string near the end, not the start) that it isn't regular.
