# Using pumping lemma to prove $L2 = \{a^ib^j |i > j \}$ non-regular

I'm having issues using the pumping lemma to prove $$L2 = \{a^ib^j |i > j \}$$ is non-regular. It's obvious to know that the language is non-regular as there is no way of tracking $$a^{i's}$$ and $$b^{j's}$$

So I've done some digging. On page 2 here I found a proof. However, it doesn't make sense to me.

Here is the proof, I will list my understandings as we go through the proof:

Assume by way of contradiction that $$L2 ∈ REG$$, then $$L2$$ satisfies the conditions of the pumping lemma. Let $$p > 0$$ be the pumping constant. Consider the word $$w = a^{p+1}b^p$$.

Makes complete sense.

Clearly $$w ∈ L2$$ and $$|w| > p$$, so according to the pumping lemma there exist $$x, y, z ∈ Σ^∗$$ such that $$w = xyz,$$ $$|xy| ≤ p,$$ $$|y| > 0$$ and for all $$i ≥ 0$$ it holds that $$xy^iz ∈ L$$.

Now if an $$i$$ is chosen to be $$0$$, this would conflict with $$|y| > 0$$? It seems it's using $$L$$ instead of $$L2$$. Is there a reasoning?

Since $$|xy| ≤ p$$, then $$x = a^n, y = a^m$$, and $$z = a^kb^{p+1}$$ such that $$m + n + k = p$$, and $$m > 0$$.

This makes sense to me, as $$z$$ is equal to the rest of the $$a's$$ not in $$x$$ and $$y$$ plus all the $$b's$$.

We pump with $$i = 0$$ and get the word $$xz = a^nz = a^na^kb^{p+1}$$. Since $$m+n+k = p+1$$ and $$m > 0$$, then $$n+k < p+ 1$$.

Now this doesn't make sense to me. Pumping down ridding of $$y$$ contradicts the $$|y| > 0$$ rule and how does $$m+n+k$$ change to now equal $$p+1$$

Thus, $$xz ∉ L1$$, in contradiction to the pumping lemma. So $$L2$$ is not regular. Note that it is crucial to “pump down” for this language.

If you guys can help me understand this proof, that would be great thanks

Now if an $$i$$ is chosen to be 0, this would conflict with $$|y|>0$$?
Note it's $$|y|>0$$, not $$|y^i|>0$$, so there's no problem.
... Clearly $$w \in L_2$$ and $$|w| > p$$, so according to the pumping lemma there exist $$x, y, z \in \Sigma^∗$$ such that $$w = xyz,$$ $$|xy| \le p,$$ $$|y| > 0$$ 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 $$m > 0$$. 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 $$m >0$$, then $$n+k \color{red}{\le p}$$ ...