MHB Proving L is not Regular using Pumping Lemma

[mathmari]

Hey!

I want to prove that the language $$L=\{ww^R \mid w \in \{0, 1\}^{\star} \}$$ is not regular using the pumping lemma. I have done the following:

We suppose that $L$ is regular and is recognized by a DFA $M$ with $k$ states.
So, $M$ accepts the string $0^k110^k$.
Since $|0^k110^k|=2k+2>k$, there is a state $q$ through which $0^k110^k$ passes at least twice.
So, $0^k110^k$ is of the form $0^k110^k=xyz$, where $x$ leads $M$ from $q_0$ to $q$, $y$ leads $M$ from $q$ to $q$ (and $y \neq \varepsilon$) and $z$ leads $M$ from $q$ to an accepting state.
We suppose that $M$ accepts every string of the form $xy^iz, i \geq 1$.

Is the formulation until this point correct? Could I improve something? (Wondering)

To get a contradiction do we have to take cases for what $y$ contains? (Wondering)

 

[Evgeny.Makarov]

We suppose that $L$ is regular and is recognized by a DFA $M$ with $k$ states.
So, $M$ accepts the string $0^k110^k$.
Since $|0^k110^k|=2k+2>k$, there is a state $q$ through which $0^k110^k$ passes at least twice.
So, $0^k110^k$ is of the form $0^k110^k=xyz$, where $x$ leads $M$ from $q_0$ to $q$, $y$ leads $M$ from $q$ to $q$ (and $y \neq \varepsilon$) and $z$ leads $M$ from $q$ to an accepting state.

This is correct, but it is not necessary to invoke automata during an application of the pumping lemma. You don't want to prove it all over again.

We suppose that $M$ accepts every string of the form $xy^iz, i \geq 1$.

It's not "suppose", it follows that it is the case. You could say, "Then the pumping lemma asserts that every string of the form $xy^iz, i \geq 1$ is in $L$".

To get a contradiction do we have to take cases for what $y$ contains?

Yes, you need to consider the cases when $y$ is contained in the first zeros portion, second zeros portion and when it contains a 1. However, one variant of the pumping lemma says in addition that $|xy|\le k$ where $k$ is the pumping length, i.e., the constant whose existence is guaranteed by the lemma (it is equal to the number of states in the automaton, but, again, it is not necessary to mention it). Then it follows that $y$ is in the first zeros portion, which makes considering other cases unnecessary.