I'm trying to prove this language is not context-free: {a^x b^y c^z | where x=z and x<y}
The pumping lemma for context-free languages says the following: 
For every context-free-language $L$ there exists a constant $p$ such that for every word $s \in L$ of length at least $p$, we can write $s = uvwxy$ and:

$|vx| \ge 1$
$|vwx| \le p$
$uv^nwx^ny \in L$ for $n \ge 0$

To prove that $L = \{a^xb^yc^z: x &lt; y \land x = z \}$ is not context-free we will do the following: 
Assume language $L$ i context-free and let $p$ be the constant from the pumping lemma. 
Take the word $s = a^pb^{p+1}c^p$, which is clearly in $L$, and express it as $s = uvwxy$ as is stated in the lemma. 
We know that $|uvw| \le p$ so $v$ consists only of $a$'s. 
Denote $s_n = uv^nwx^ny$ 
Now consider the possible cases:

If either $v$ or $x$ contain more than one letter, than $s_2$ will not be of the form $a^*b^*c^*$, so we know that $v$ and $x$ will consist of at most one letter.
If $v$ is empty and $x$ consists of $b$'s then $s_0$ will not have more $a$'s than $b$'s. If $x$ consists of $c$'s then $s_2$ will have more $c$'s than $a$'s.
If both $v$ and $x$ are non-empty then if $v$ consists of $a$'s and $x$ of $b$'s then we in $s_2$ we will have more $a$'s than $c$'s. If $v$ consists of $a$'s and $x$ of $c$'s than for large enough $m$ in $s_m$ we will have more $a$'s than $b$'s. If $v$ consists of $b$'s and $x$ of $c$'s then in $s_2$ we will have more $c$'s than $a$'s.
If $v$ and $x$ consist of the same letter (i.e $w$ is empty) then following the same type of reasoning we will also reach a contradiction.

Thus we have reached a contradiction for every possible case and so $L$ is not context-free.