Is the language with decreasing numbers of a, b and c context-free by pumping lemma? So I've been given the following language on an assignment.  It is the only question I have left of 10, and I've been racking my brains out trying to solve it for hours.  

$$L=\{w:w\in(a+b+c)^*, n_a(w) &gt; n_b(w)&gt;n_c(w)\}$$

where $n_x(w)$ represents the number of character $x$ found in $w$.  The problem statement is: prove or disprove that the language is context-free.  Context-free grammars or pushdown automata are acceptable proofs.  Use pumping lemma to disprove.

I've extensively explored both possibilities and I'm fairly certain that it is context-free.  

The approach I've taken in finding a context-free grammar for the language involves using rules which preserve the constraint $n_a(w) &gt; n_b(w) &gt; n_c(w)$ (ie. whenever a $b$ is added, add an $a$; whenever a $c$ is added, add a $b$).  Then, I've attempted to enforce that there are at least two $a$s and at least one $b$ (base case for the constraint).

The grammar I've used is:

$S\implies XaXaXbX | XaXbXaX | XbXaXaX$

$X \implies XX| A | B | C | \lambda$

$A \implies a$

$B \implies ab | ba$

$C \implies abc | acb | bac | bca | cab | cba$

(where $\lambda$ is the empty string)

My grammar fails for strings like $cccaaaaabbbb$.

I'm confused as to where to go from here.  

I would really like a push in the right direction, not an answer.
 Any help is greatly appreciated!
If your language is context free, so is its intersection with the regular language $a^* b^* c^*$. And your proof in the answer that this isn't context free does go through. Your proof omits the cases where one of the strings straddles the $a/b$ or $b/c$ boundaries.