# Is this enough to prove that the language L is not context-free? (Pumping lemma for CFL's)

The language $$L = a^nb^nc^n | n>=1$$

We assume that the language $$L$$ is context-free. Then it must satisfy these conditions:

We can break any string $$Z$$, where $$|Z| >= p$$ into 5 substrings: $$uvwxy$$

So I choose the string:

$$Z = a^pb^pc^p$$

Now I need to show that the string $$uvwxy$$ satisfies these conditions:

$$|vx| >= 1$$, $$|vwx| <= p$$, and $$uv^iwx^iy$$ is in $$L$$, for all $$i>=0$$

For all the cases that satisfies the first 2 conditions, we have:

Case 1: $$vx$$ is composed of a mix of $$a$$'s and $$b$$'s (in that order). If we choose $$i=0$$, then the number of $$c$$'s will be greater than the number of $$a$$'s or $$b$$'s, thus not being part of the language

Case 2: $$vx$$ is composed of a mix of $$b$$'s and $$c$$'s (in that order). If we choose $$i=0$$, then the number of $$a$$'s will be greater than the number of $$b$$'s or $$c$$'s, thus not being part of the language

Case 3: $$vx$$ is composed of just $$a$$'s, $$b$$'s or $$c$$'s. With $$i = 0$$, the letter chosen becomes less than the number of both unchosen letters, which is not part of the language.

Since $$Z$$ cannot satisfy these conditions, $$L$$ is not a CFL.

In general, for pumping lemma proofs by contradiction, you need to show that for all ($$\forall$$) pumping lengths there exists ($$\exists$$) a string $$s$$ (with $$|s|\ge p$$) that cannot be pumped.
In your case, you chose a generic pumping length $$p$$. And you found a counter-example string $$s$$ that absolutely cannot be pumped (i.e. you showed that all possible cases cannot be pumped). Therefore, you've showed that for any arbitrary pumping length $$p$$, you can construct a string $$s$$ that cannot be pumped by the Pumping Lemma for CFL's. Thus, you've contradicted that $$L$$ is a CFL.