Edit: I was totally leading you down the wrong track. That's what happens when I try to help out when I haven't completely solved the problem myself.
Ogden's Lemma
Suppose L is context free. By Ogden's lemma, there exists an integer p that has the following properties:
Given a string w in L at least p symbols long, where at least p of those symbols are "marked", w can be represented as uvxyz, which satisfy:
- x has at least one marked symbol,
- either u and v both have marked symbols or y and z both have marked symbols,
- vxy has at most p marked symbols, and
- u vi x yi z is in L for i >= 0
That's Ogden's lemma. Now, let q be an integer divisible by every positive integer no greater than p. Let w = ap+q bp+q cp. Mark every c. By #2, u or v must contain at least one c. If either u or v contains any other symbol, then #4 fails, so u and v must contain only c. But then #4 fails when i = q/|uv|. We know q is divisible by |uv| because p > |uv| > 0, and q is divisible by all positive integers less than p.
Note that Ogden's lemma turns into the pumping lemma when you mark all symbols.
Pumping Lemma
Suppose L is context free. By the pumping lemma, there is a length p (not necessarily the same p as above) such that any string w in L can be represented as uvxyz, where
- |vxy| <= p,
- |vy| >= 1, and
- u vi x yi z is in L for i >= 0.
Given a string w in L, either m > n or m < n. Suppose p = 2.
Suppose that m > n. (Note that Λ denotes the empty string.)
- Let u = an bn cm-1
- Let v = c
- Let x = Λ
- Let y = Λ
- Let z = Λ
Suppose that n > m.
- Let u = an-1
- Let v = a
- Let x = Λ
- Let y = b
- Let z = bn-1 cm
This demonstrates that no string from L provides a counterexample using the pumping lemma to the supposition that L is a context free language (even though it is context sensitive).