Pumming Lemma Question -Not Context Free
I understand the general concept of pumping lemma but I don't quite understand how to write proofs formally. In this particular case (see image attached),I understand that:
if we divide it into $uvxyz$ and $v$ gets all the $b$'s and then the power of $b$ is 2, the number of $A$'s will no longer be equal to the number of $B$'s. But how can I write a formal "mathematical" proof for it? After writing the basic 3 conditions?
Assume that $A=\{a^n b^n c^i | i \leq n\}$ is a context-free language. That means $\exists p >0$ where if some $s \in A$ has $|s| \geq p$ then we can split $s = uvxyz$ with the following properties $$ uv^ixy^iz \in A \forall i \geq 1 $$ $$ |vy| > 0 $$ $$ |vxy| \leq p $$
Choose the following string in $s = a^pb^pc^p \in A$. Since $|vy| > 0$ and $|vxy|\leq p$ we are guaranteed that either $v, y$ must be uniform all $a$'s, $b$'s, or $c$'s. If both are uniform both $v,y$ must all be $b$'s, so then all times you pump them there will be an unequal amount of $b$'s compared to $a$. If there is a non-uniform $v,y$ then, there will be sections of the string that do not fit into the form $a^*b^*c^*$ ($b$'s come before $a$'s).
From here we know there are no ways to split the string such that all properties are satisfied. Therefore $A$ must not be context-free.
