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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?

StuckWithHomeworkProbably
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  • The pumping lemma says that such $x, y, z, u, v$ exist not that for every such $x, y, z, u, v$ it works, in particular it doesn't work for your choices. – kingW3 Dec 09 '21 at 08:59

1 Answers1

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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.

mrp
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  • So even if 1 split satisfies the conditions and a pump up or down exists within that language, the context freeness holds? – StuckWithHomeworkProbably Dec 09 '21 at 08:56
  • @StuckWithHomeworkProbably it's the other way around: context-freeness guarantess that one such split exists. You cannot use the pumping lemma to prove that a language is context-free. – mrp Dec 09 '21 at 09:17