Pumping Lemma's Condition 3 concept

Question

I'm following one of the examples from my textbook on the Pumping Lemma:

Let C = {w | w has an equal number of 0s and 1s}

Condition 3 stipulates: |xy| <= p



If |xy| <= p, then y must consist only of 0s, so xyyz is not in C. 
Therefore s cannot be pumped

I'm having trouble understanding how condition 3 leads to the conclusion that "y must only consist of 0s, so xyyz is not in C"

If you could add a few preceding lines from the book, it would help — uba, Mar 08 '13 at 04:46

score 0 · Accepted Answer · answered Mar 08 '13 at 04:57

0

I guess the string chosen is 0^p1^p. Since |xy| <= p, and xyz = 0^p1^p, the string xy will be 0^k where k <= p since the first p symbols of 0^p1^p are all 0's. Since xy consists of only 0's, y must also consist of only 0's

And learn to put your question in a proper manner. You cannot expect others to "predict" your question while you put half of the information

answered Mar 08 '13 at 04:57

uba

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Hi the string chosen was indeed 0p1p thank you I understand it now – Locke McDonnell Mar 08 '13 at 05:35

Pumping Lemma's Condition 3 concept

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