# Confused about 3rd rule of CFG pumping lemma

Let $$L = \{\space ww \space | \space w \in \{0,1\}^*$$} (need to prove that $$L$$ is not CFL)

Assuming $$L$$ is CFL we can use the PL and split $$s=uvxyz$$ and we choose $$s = 0^p1^p0^p1^p$$ where $$p$$ is the pumping length.

Using the 3$$^{rd}$$ rule, namely $$|vxy| \le p$$, can I say that $$vxy$$ can consist only of $$0's$$ and either pumping up(e.g. $$uv^2xy^2z)$$ or down ($$uxz$$) will throw us out of $$L$$ ?

Remember that the PL is stated for any partition $$s=uvxyz$$, so it is not enough to show that for one specific partition pumping $$v$$ and $$y$$ will exclude the resulting string from $$L$$. While it is true that $$xyz$$ can consist only of 0′s, it could also be the case that $$xyz$$ is of the form $$0^i1^j$$ for some $$i, j < p$$, for instance. You must show that pumping excludes the resulting string from $$L$$ in this (and all possible other) cases as well.
Here is a decomposition $$s = uvxyz$$ such that $$|vxy| \leq p$$ but $$xyz$$ doesn't consist only of zeroes: $$u = 0^p \\ v = 1^p \\ x = \epsilon \\ y = \epsilon \\ z = 0^p 1^p$$