I’ll walk you through the argument.
Suppose that $L=\{ba^nbc^n:n\ge 1\}$ is regular. Then the pumping lemma for regular languages says that it has a pumping length $p$ such that if $w$ is any word of $L$ whose length is at least $p$ (i.e., such that $|w|\ge p$), then $w$ can be decomposed as $w=xyz$ in such a way that $|xy|\le p$, $|y|\ge 1$, and $xy^kz\in L$ for every $k\ge 0$. In order to use the pumping lemma to show that $L$ is not in fact regular, we must find a word $w$ for which this yields some contradiction. Finding such a $w$ is largely a matter of practice and experience with similar problems.
Here we can take $w=ba^pbc^p$. No matter how we split this as $w=xyz$, if $|xy|\le p$, then the $xy$ part is some initial segment of the first $p$ letters of $w$, i.e., of $ba^{p-1}$. There are now two possibilities that have to be distinguished.
- If $|x|\ge 1$, then the initial $b$ of $w$ is part of $x$, and $y=a^r$ for some $r$ such that $1\le r\le p-1$.
If this is the case, then $x=ba^s$ for some $s\ge 0$ such that $r+s\le p-1$, and $z=a^{p-r-s}bc^p$; why? Then for any $k\ge 0$ we have
$$xy^kz=ba^sa^{kr}a^{p-r-s}c^p=ba^{p-r+(k-1)r}bc^p\;.$$
You can easily check that this is in $L$ if and only if $k=1$. If $k=0$, the block of $a$s is shorter than the block of $c$s, and if $k>1$, the block of $a$s is longer than the block of $c$s; in both cases the word is not in $L$.
- If $x$ is empty, then $y=ba^r$ for some $r\le p-1$.
In this case $z=a^{p-r}bc^p$, and
$$xy^2z=y^2z=ba^rba^pbc^p\;,$$
which is plainly not in $L$. This is sufficient, but we can go further and note that in fact $xy^kz\in L$ if and only if $k=1$, so that we just have the original $w$: $xy^kz$ has $k+1$ $b$s, while every member of $L$ has exactly two $b$s.
No, this isn't enough to prove that the language is not regular. You have to take into account all the possible options for $y$. As I can see you consider $y$ contains only $b$'s. You should add the cases:
i) $y$ contains both $a$'s and $b$'s then in the word $xy^kz$ for any $k\geq 2$ there would be switching of $a$'s and $b$'s. For instance, if $x=aa,y=ab,z=b$ then $xy^2z=aaababb$ which isn't in $L$.
ii) $y$ contains only $a$'s, then for $k=0$ the number of $a$'s is less than the number of $b$'s in the word $xz$. So $xz$ isn't in $L$.
Best Answer
Just consider the string $111\cdots111 @ 111\cdots 111$, where the number of $1$'s is the same on both sides and is large: longer than the pumping length. Then the decomposition into $xyz$ must be such that $y$ contains only $1$'s, and so repeating it any number of times (other than exactly once) will make $\#_1(w_1) \neq \#_1(w_2)$.