Showing that a language satisfies the pumping lemma 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| &lt;=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?
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.