Well, basically we do not take any "m" as pumping length but just assumes "m" as some integer value which is supposed to be existing (if pumping lemma is violated no such $m$ will exist). Our only aim is to prove that some $w$ given by $xy^iz$ is not in $L$ which would make the language irregular.
So, we should find an appropriate "i" and also the split $x,y,z.$ Here, finding "i" is like a puzzle and I do not think there exist any systematic way which works for all languages. For example,
- $L = \{0^{n^2}\mid n \geq 0\},$ Taking $w = 0^{m^2}$ where $m$ is the pumping length so that the string length is greater than or equal to the pumping length as mandated by pumping lemma. $0^{m^2}= xyz \text{ or }0^X0^Y0^Z.$ Now, pumping lemma condition says that for "SOME" $\mid xy\mid < m, \mid y \mid >0, xy^iz \in L, \forall i \geq 0$.
- Now the trick here is that we have to prove "NO SUCH" $x,y,z$ exist such that $xy^iz \in L, \forall i \geq 0.$
- So, the split entirely depends on the property of the language. For our $L,$ we need a square number of $0's.$ Our $w = 0^{m^2} = xyz.$ Now without specifying what $x$ and $y$ are, if we can say $xy^iz,i \geq 0,$is not in $L$, then $L$ is irregular. To say this we just need to show for some $i$, $\mid xy^iz \mid$ is not a perfect square. We know that $\mid xyz \mid = m^2$ is a perfect square. Now, by pumping one extra $y$, we increase the string length by $\mid y \mid$. This will be minimum $1$ and maximum $m$ because of the conditions $\mid y \mid > 0$ and $\mid xy \mid \leq m.$ Now, to get the next perfect square after $m^2$ we need to add minimum $2m +1 \because (m+1)^2 = m^2 + 2m+1$, but we are adding maximum $m$ here. So, we proved that for no $x,y,z$ pumping lemma holds thus making the language irregular. Instead of pumping one extra $y$ we could have removed $y$ also (making $i = 0$).
Now, for
- $L = \left\{0^p \mid p \text{ is prime}\right\}$, we can take $w = 0^n,$ where $n$ is the first prime number after the pumping length $m$ to satisfy the minimum length condition. Again we do not know what is $m$, but just assuming it exists and has an integer value.
- $w = 0^{n} = xyz,\mid xyz \mid = n \text{ is prime}.$ Here we are trying to make $\mid xy^iz \mid$ not a prime number for some $i$. The easiest way is to take $i = n,$ which gives $\mid xy^nz\mid = n + n \times \mid y\mid = (n+1) \mid y \mid$ which can never be a prime number. So, pumping lemma is violated -- we did not fix any particular $x,y,z$ and showed for all such $x,y,z$ -- and $L$ is irregular.
