Pumping lemma $L=\{a^ib^j | i \neq j ; i,j \ge 0\}$ So, let's have language $L=\{a^ib^j | i \neq j ; i,j \ge 0\}$
I have to prove that it's not regular.
\begin{align}
\omega=a^nb^{n+1}=a^{n-1}ab^{n+1}
\end{align}
\begin{align}
x&=a^k\\
y&=a\\
z&=a^{n-k-1}b^{n+1}\\
\\
i=2:xy^iz&=a^ka^ia^{n-k-1}b^{n+1}\\
&=a^ka^2a^{n-k-1}b^{n+1}\\
&=a^{n+1}b^{n+1}\\
\end{align}
$\#_a(\omega)=\#_b(\omega)\implies a^{n+1}b^{n+1} \not\in L $
I'd like to ask if this is a correct proof?
 A: The problem with your approach is that you are determining the substring $y$ that you intend to pump.  Recall the statement of the Pumping Lemma:

If $L$ is regular, then there is an integer $p \geq 1$ such that for every string $w$ in $L$ of length $\geq p$ you can decompose $w$ into $xyz$ such that etc.....

To show that a language $L$ is not regular you must show that no such integer $p$ exists.  That is, for every $p \geq 1$ you must exhibit (at least) one string $w$ in $L$ of length $\geq p$ such that there is no decomposition of $w$ into $xyz$ such that etc.....
You have essentially only provided a single decomposition of your $w = \mathtt{a}^{n} \mathtt{b}^{n+1}$ which doesn't allow you to pump and remain in your language.  However, there might be other decompositions that do work.  For example, if you take $y = \mathtt{a}^2$ (so that $x = \mathtt{a}^{k}$ and $z = \mathtt{a}^{n-k-2} \mathtt{b}^{n+1}$), you will find that this can be pumped.  (The number of $\mathtt{a}$s in $xy^iz$ will always be of opposite parity to the number of $\mathtt{b}$s.)
