So I solved some exercises where I have to use the pumping lemma for contextfree languages but this one is a problem for me:
Consider:
$ L = $ { $w_1£w_2£w_3 \in$ { $0,1,£$}$^*$ | $w_1, w_2, w_3 \in$ {0,1}$^*$ and $\exists i \neq j \in ${$1,2,3$} : $ w_i = w_j$ }.
Show that: L is not contextfree.
So L contains words, which are separated with £ and atleast two subwords are the same. My idea is that we create a word ( with the help of the pumping lemma ) that has three different subwords $ w_1, w_2, w_3 $. But I don't find a "start-word". My first idea was to consider z:= 0.....0£0.....0£1.....1. But the problem is this part $ uvwxy = 0.....0£0.....0£1.....1 $ . If I pumped only the 1....1 part,then I wouldnt have a counterexample. So can you please tell me which word is a possible start-word?
The definition of $L$ allows $w_1$, $w_2$ and $w_3$ to be empty. By chosing one of them to be empty you avoid the problem you have with being able to pump in the "$1\dots 1$ part".
But this is not enough. You cold pump zeroes in both $w_1$ and $w_2$ and thus stay within the language.
For $n$ the constant from the pumping lemma, take for example $$0^n1^n£0^n1^n£.$$
If you want to pump in both $w_1$ and $w_2$, it can only add $1$s to the former $0$s to the latter destroying the equality.