# Pumping Lemma - Clarification of Usage

I'd like to make sure my understanding of the Pumping Lemma is correct.
Consider $L=\{ 0^n1^m2^{n-m}:\, n \ge m \ge 0\}$

I'm going to give 2 solutions to prove that $L$ is not regular. One using "pumping down" and the other using a "cheap trick". I'm not sure whether either solution is correct, so any comments would help clarify my understanding.

Solution 1: Pumping Down
Let p be the pumping length. Let $S= 0^{p+1}1^p2 = xy^iz$ for $i \ge 0$
Now, $|xy| \le p$ implies $y$ contains only $0's$
So, for $i=0,$ $S=xy^0z=xz$ and since $xy$ had only one more $0$ than the number of $1's, \, x$ will not have more 0's than the number of 1's. Hence, $S \notin L$ giving the required contradiction.

Solution 2: Cheap Trick
As before, let p be the pumping length. But this time, $S= 0^p1^p2^0 = 0^p1^p$ which is generally the example proven in most textbooks. However, I'm not sure this is a valid approach, so some clarification would be helpful.

Suppose that $L$ is regular; then it has a pumping length $p$. Let $w=0^p1^p2^0$; the pumping lemma for regular languages says that we can decompose $w$ as $w=xyz$, where $|xy|\le p$, $|y|\ge 1$, and $xy^kz\in L$ for each $k\ge 0$. Clearly $xy$ is contained in the initial $0^p$ of $w$, so there are $r\ge 0$ and $s>0$ such that $x=0^r$ and $y=0^s$, and therefore $z=0^{p-r-s}1^p$. Then
$$xy^0z=xz=0^r0^{p-r-s}1^p=0^{p-s}1^p\in L\;,$$
but $p-s<p$, so $xy^0z\notin L$. This contradiction shows that $L$ is not regular after all.
• @EggHead: No, you can never choose the pumping length: you just know that if the language is regular, there is one. I didn’t choose one here: the only thing that I chose was the word $w$, which I chose based on the supposed pumping length (which in fact doesn’t exist, since $L$ isn’t regular). – Brian M. Scott Nov 18 '13 at 3:03
• @EggHead: Ah, okay. Yes, as long as $|w|\ge p$, $w$ can be as long as you need to make the proof work. – Brian M. Scott Nov 18 '13 at 17:11