4. (10 points) Consider the language L tu 1's^. Explain and correct the error below: t and u are strings over 0,1 with the same number of Proof that L is not regular using the Pumping Lemma: Assume (towards a contradiction) that L is regular. Then the Pumping Lemma applies to L. Let p be the pumping length of L. Choose 8 to be the string 1p0p1P0, which is in L because t = 1p0p and u = 1POP each have p 1's. Since this string is in L and has length greater than or equal to p, the Pumping Lemma guarantees that s can be divided into parts s-xyz such that for anyi 2 0, xy'z is in L, and that lyl> 0 and ryl p. Since the first p characters of s are all 1's and we have lyl >0 and lzyl 3 p, we know that y repeating y twice adds 1's to t but not to u, and strings in L are required to have the same number of 1's in t as in u. This is a contradiction. Therefore the assumption is false, and L is not regular. Created by Paint X