I am looking over pumping lemma and the author is using it to show that the language is non-regular.
{a^n b^n a^n} = {aba aabbaa aaabbbaaa........}
Is there anyway that you can show this in a simpler way ? Without using math expressions cause it is really hard to read...
{a^n b^n a^n} = {aba aabbaa ........}
A = {a^n b^n a^n}
take the word a^p b^p a^p ∈ A
a^p b^p a^p = xyz
|xy| ≤ p
The xy^2 ∉ A
The above language is not regular
The given solution is more like a sketch of the solution. Here's the same solution, with some more sentences filled in to explain the logic at each step.
We want to show that the language: $$ A = \{a^nb^na^n \mid n \in \mathbb N\} $$ is non-regular. To do this, we argue by contradiction. Suppose instead that $A$ is regular. Then let $p \geq 1$ be the pumping length given by the Pumping Lemma and consider the word: $$ w = a^pb^pa^p $$ Certainly, we know that $w \in A$ (just take $n = p$) and that $|w| = 3p \geq p$. Hence, by the Pumping Lemma, we have that $w = xyz$, where $|xy| \leq p$ and $|y| \geq 1$ and $xy^i z \in A$ for all $i \geq 0$. But then we know that $y=a^k$ for some $k \in \{1,\ldots,p\}$. Now take $i=2≥0$ and consider the string: $$ xy^2z = a^{n+k}b^na^n $$ which should be in $A$. But this is impossible, since $xy^2z$ contains $k \geq 1$ more $a$'s to the left of the $b$'s than to the right of the $b$'s. So $A$ is non-regular, as desired. $~~\blacksquare$