I am wanting to show that this language fails to show that it is not context-free. So, in essence, it satisfies the pumping lemma
If L = {ambncndn | m,n >= 1 }
Should I have n be the constant of the pumping lemma? Is z=ambncndn and |z|= m+3n>n ?
By the pumping lemma, there exist u,v,w,x,y such that z=uvwxy, |vwx| <=n, |vx|>=1 . I'm just a little thrown off by the two different letters for the exponents, can I treat m and as another n?
You want to show that the language $L = \{a^mb^nc^nd^n \mid m,n\ge 1 \}$ does satisfy the Pumping Lemma (for context-free languages) while it is not context-free.
In order to do so, you must show that every string can be pumped within the language, so for every string $z=a^mb^nc^nd^n$ (for large enough $m+3n$) you must find a decomposition $z=uvwxy$ that can be pumped. The problem however is that your language $L$ does not satisfy the Pumping Lemma. Possibly?) you wanted to pump the first $a$ in a string (taking $u=x=y=\lambda$, $v=a$, $w=a^{m-1}b^nc^nd^n$). Generally that works, but not if the string contains a single $a$, like $ab^nc^nd^n$. Then we can pump the letter $a$ `down' choosing $i=0$ which results in a string outside of $L$. All other decompositions will fail (they will destroy the $abcd$ order or the equal numbers of $b,c,d$).
An example of a non-regular language that satisfies the Pumping Lemma (for regular languages) can be found here: Languages that satisfy the pumping lemma but aren't regular? .
It is of a very specific form $\$\cdot K \cup \{\$^k \mid k\neq 1\}\cdot \Sigma^*$ (for a complicated language $K$) that can be used for context-free pumping examples too.
Assume that pumping lemma is satisfied for $L$. Let $N$ be the pumping length.
Now take $z = ab^Nc^Nd^N \in L$. $|z| = 3N+1$ and hence this string can be pumped.
For this string, every choice of $u,v,w,x$ and $y$ that satisfies the conditions of pumping lemma will have a choice of $i \in \mathbb{N}$ such that $uv^iwx^iy \not\in L$. This is a contradiction. Hence $L$ is not context-free.
