Proving that $L=\lbrace{ab^{n}ba^{n}|n\geq1}\rbrace$ is not regular with pumping lemma I'm trying to understand the pumping lemma for regular languages and would like to prove that $L=\lbrace{ab^{n}ba^{n}|n\geq1}\rbrace$ is not regular.

My suggestion is as follows:

Assuming $L\in{}REG$, let $p$ be the number that exists for $L$ according to the pumping lemma.

Let $w=ab^{p}ba^{p}$ with $|w|=2p+2&gt;p$.

Since $w\in{}L$, $w$ can be written as $w=xyz=ab^{p}ba^{p}$, so that:


$|xy|\leq{}p$, in this case $xy=ab^{q}$ with $q\leq{}n-1\leq{}p$
$|y|\geq{}1$, in this case $y=b^{n}$
$\forall{}i\geq{}0,xy^{i}z\in{}L$


But for $i=0$ we have $xy^{0}z=xz=aba^{n}\in{}L$ which violates the definition of $L$, so that $L\notin{}REG$.

Does this constitute a sufficient proof with the pumping lemma or am I at least on the right track? I would appreciate any feedback.
It's not sufficient proof because you only proved that if you choose the decomposition $x=ab^q$ and $z$ the rest does not meet the three points of the pumping lemma but remember that the lemma says: 

If $L$ is regular then there is a constant $N&gt;0$ associated to $L$ such that  $\forall \sigma \in L$ with $|\sigma|\geq N$ $\textbf{exist}$ a decomposition $\sigma = xyz$ ... 

Summarizing. you proved that your decomposition doesn't work. Now you have to try all other possible decompositions.