Irregularity of $\{a^iba^j : i \neq j\}$ using pumping lemma Proving the language $L=\{ a^i b a^j |\ i&lt;j,\text{ or }i&gt;j;\ i,j&gt;0 \} $ is not regular, I am trying to use the pumping lemma for regular languages:

pumping length $:= m$

case 1: $i&lt;j; j=m$ 

$w_1 = a^i b a^m = xyz$ 
$|xy| &lt; m$ 
$|y| &gt; 0$ 

$\Rightarrow$ 
$y$ can be:


$a$s only
$a$s + $b$
just $b$ 


However, this leads to $xy^hz\ \not \in L\ \forall h,\ h&gt;0$.

case 2: $i&gt;j; i=m$ 

$w_2 = a^m b a^j = xyz$ 
$|xy| &lt; m$ 
$|y| &gt; 0$ 

$\Rightarrow$ 
$y$ can be:


$a$s only


Pumping with this $y$:
 $xy^hz\ \in L\ \forall h$ holds true.

As this answer does not seem to even consider the second case, is the first case sufficient to proof the language is not regular? 
However, taking a look at this answer I think I'd have to considere the second case as well, in order to cover all possible $xyz$ combinations.
Here is the contrapositive form of the pumping lemma, which is what we use to prove that a language is not regular:


  Let $L$ be a language. Suppose that for all $p \geq 1$ there exists a word $w \in L$ of length at least $p$ such that for every decomposition $w=xyz$ in which $|xy| \leq p$ and $y \neq \epsilon$ there exists an integer $i \geq 0$ such that $xy^iz \notin L$. Then $L$ is not regular.


Using the pumping lemma directly for the language $L$ in the question is a bit tricky. Given $p \geq 1$, let $w = a^p b a^{p+p!} \in L$; note that $|w| \geq p$. We have to show that for every decomposition $w = xyz$ in which $|xy| \leq p$ and $y \neq \epsilon$ there exists an integer $i \geq 0$ such that $xy^iz \notin L$. Consider any such decomposition. It is easy to check that $y = a^\ell$ for some $1 \leq \ell \leq p$, and that $xy^ij = a^{p+(i-1)\ell}ba^{p+p!}$. For $i = p!/\ell+1$ (which is an integer!), $xy^ij = a^{p+p!}ba^{p+p!} \notin L$. Thus the condition in the pumping lemma is satisfied, and we conclude that $L$ is not regular.