Answer:
L is not a regular language with formal proofs
Explanation:
(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails that L is a regular language. Then by the Pumping Lemma for Regular Languages,
there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p,
s = xyz subject to the following conditions:
(a) |y| > 0
(b) |xy| ≤ p, and
(c) ∀i > 0, xyi
z ∈ L
(b) To determine that L is not a regular language, we mke use of proof by contradiction. lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,
The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject to the condtions as follows :
(a) |y| > 0
(b) |xy| ≤ p, and
(c) ∀i > 0, xyi
z ∈ L.
Choose s = 0p10p
. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.
for some k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus, xy0
z should be in L. xy0
z = xz = 0(p−k)10p
It is shown that is is not in L. This is a contraption with the pumping lemma. our assumption that L is regular is incorrect, and L is not a regular language
