Final answer:
To prove that F = {ww l w is a string from {0,1}*} is not regular, we can use the pumping lemma for regular languages. By assuming that F is regular and applying the pumping lemma, we can prove that F is not regular. We consider the specific string s=0^p10^p1, split it into three parts and demonstrate that the pumped string is not in F, thus contradicting the assumption.
Step-by-step explanation:
To prove that F = {ww l w is a string from {0,1}*} is not regular, we can use the pumping lemma for regular languages. Let's assume that F is regular. According to the pumping lemma, for any regular language F, there exists a pumping length p such that any string s of length greater than or equal to p can be split into three parts, s = xyz, satisfying the conditions: |xy| ≤ p, |y| > 0, and xy^iz ∈ F for all i ≥ 0.
Now, consider the string s = 0^p10^p1 where p is a pumping length. We can split this string into three parts: x = 0^k, y = 0^l (where 1 ≤ l ≤ p), and z = 0^(p-l)10^p1. Applying the pumping lemma, we can pump the y part by repeating it, i.e., xy^2z = 0^k0^l0^l0^(p-l)10^p1 = 0^k0^(2l)0^(p-l)10^p1. This pumped string is not in F, as the number of 0s after the first set of 0s is no longer equal to the number of 0s before the 1. Hence, F is not regular as it fails the pumping lemma condition, contradicting our assumption.