Using pumping lemma to prove {a^ib^j | i > j} non-regular
Now if an $i$ is chosen to be 0, this would conflict with $|y|&gt;0$?


Note it's $|y|&gt;0$, not $|y^i|&gt;0$, so there's no problem.

All your other confusions come from the typos in the proof. I rewrite the proof (the fixed typos are colored red) as follows.

... Clearly $w \in L_2$ and $|w| &gt; p$, so according to the pumping lemma there exist $x, y, z \in \Sigma^∗$ such that $w = xyz,$ $ |xy| \le p,$ $ |y| &gt;
 0$ and for all $i \ge 0$ it holds that $xy^iz \in L$. Since $|xy| \le p$, then $x = a^n, y = a^m$, and $z = a^kb^\color{red}p$ such that $m + n + k = \color{red}{p+1}$, and $m &gt; 0$. We pump with $i = 0$ and get the word $xz = a^nz = a^na^kb^\color{red}p$. Since $m+n+k = p+1$ and $m &gt;0$, then $n+k \color{red}{\le p}$ ...