# In Pumpin Lemma, let's say we have a decomposition of the string s = xyz, is it necessary for x to take the automaton through distinct states?

I understand what Pumping Lemma is, how it works and also understand it's proof. The gist is that it uses "Pigeon Hole" principle to guarentee us a repeatation in the sequence of states and we use that repeatation as an exploit to prove non regularity using contradiction

There is a condition $$|xy| <= p$$, this condition guarentees us that we focus on first repeatation, which makes sense.

Now, we know that to use pumping lemma we need to "prove that a language is non regular" we need to assume that the language is regular and pick a specific word ~ 'w' from the language, and decompose that 'w' into w = xyz

My question is that since we need to pump only one certain loop, the very first loop, can x can a loop? or does it need to have only those symbols that take the finite automata through a distinct set of states? similarly for z

In the following, by distinct I mean ~ the mentioned part does not have any loops

This first link which proves pumping lemma say's 'x' part of decomposition must be distinct: https://cs.stackexchange.com/a/133970/150275

Whilst in this example the user didn't take the 'x' part of decomposition as distinct: Pumping Lemma: 1^n^2 is not regular and even in this one https://math.stackexchange.com/questions/1209631/proving-0n-1n-2n-n-0-is-not-regular

They didn't allow the x part to be distinct, now here is the thing, if they don't take the x part as distinct ~ and choose other loop part in y, then they aren't considering the very first loop guareented by pumping lemma, they are using some other loop

Tldr: Should the x part of the decomposition be distinct or not?