Why is {$0^ 1^ : ≥0$} context free?

Question

This Language {$0^ 1^ : ≥0$} is context-free as per another answer I read on this site.

Link: What does it mean to say a language is context-free?

For this example take S = $0^p 1^p$ However, if I take a split of:

u = ε

v = $0^p$

x = ε

y = ε

z = $1^p$

And pump it up for some i = 2 such that uvxyz becomes uvvxyyz, I get:

uvvxyyz = $0^{2p}$$1^p$

For 1 Doesn't that mean that this condition is not satisfied: $|vxy|<=p$ and therefore is not context free since: v = $0^{2p}$

Also after a pump up wouldn't the language break since the number of 0s are more than 1s?

Answer 1

The pumping lemma says that there is some split that satisfies the conditions of the lemma, but you have only shown that there is a particular kind of split that doesn't satisfy the conditions. Hence, in order for your proof to work, you would have to argue that any split would violate the conditions, but you cannot, because it is indeed context-free.