# (Pumping Lemma for Regular Languages) Is this proof that L is not regular?

I have a language $$L$$:

$$L = \{w : a^ib^j; i > j \}$$

I need to prove this language is not regular using Pumping Lemma.

I need to find a suitable $$w$$, where $$|w| \ge$$ some $$p$$

$$w = a^{p+1}b^{p}$$

$$w$$ makes sense because it is in $$L$$ and has a length greater than $$p$$

We must be able to break it up into 3 substrings $$xyz$$ where:

$$|xy| \le p$$,

$$|y| \ge 1$$, and

$$xy^iz$$ is in $$L$$ for all $$i \ge 0$$

For all possible choices, which also satisfy conditions 1 and 2, we have the following cases:

Case 1: $$xy$$ is only composed of just $$a$$'s. If we pump $$y$$ with $$i=0$$, then we will end up with and equal amount of a's and b'stherefore not in the language.

Case 2: $$xy$$ is in both a and b parts. Therefore $$y$$ is some length of b's. Then if we pump $$y$$ with $$i=2$$, the number of b's will be greater than or equal to the number of a's, therefore not in the language.

Case 3: $$xy$$ consists of just b's. Then if we pump $$y$$ with $$i=2$$, then the number of b's will be greater than or equal to the number of $$a's$$, and not be part of the language.

Since we have shown that the string $$w$$ from $$L$$ cannot satisfy all 3 conditions at once for all pumping length $$p \ge 1$$, then $$L$$ is not a regular language.

If $$w = a^{p+1} b^p$$ and $$|xy|\le p$$, then $$y$$ cannot contain any $$b$$'s. If it did, then $$|xy|>p$$ by the construction of $$w$$.
So how do we fix this? We know that $$y=a^k$$ for some $$1 \le k \le p$$, because it cannot be empty and it cannot contain any $$b$$'s. Now you should be able to argue that $$xy^0z \notin L$$, which will contradict that $$L$$ is regular.