How to prove that a language is not context-free using pumping lemma?

Question

How could I realize this using the pumping lemma? What if the pumping part is between b and c or between a and b so that after pumping the word is still in L?

Answer 1

Let $p$ be our constant from pumping lemma, and consider word $a^{p + 1} b^{p + 1} c^{p + 1}$. Pumping lemma says this word is $uvwxy$ for some words s.t. $|vwx| \leq p$ and for any $k$, $uv^kwx^ky \in L$.

Of course, if $v$ or $x$ contains two different letters, taking $n = 2$ already gives us a word not from $L$. So each of them consists of the same letter. Also, from length restriction, it can't be that $v$ consists of $a$ and $x$ consists of $c$.

Now, to cases.

1. If both $u$ and $v$ consists of the same letter (with one of them may be empty), then taking $n = 2$ again gives us word not from $L$.
2. If $u$ consists of $a$ and $v$ consists of $b$, taking $n = 0$ gives us word not from $L$ (as we have fewer $b$s then $c$s).
3. If $u$ consists of $b$ and $v$ consists of $c$ - similar to 2.