# Pumping Lemma for CFL

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.
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