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.
Here is a very careful application of the pumping lemma for regular languages, showing all of the relevant details. It uses pumping down with your second choice of word to be pumped.
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.