(10 pt., 5 pt. each) Describe the error in the following proofs that incorrectly conclude that A={0m1nm,n0} is not a regular language. a. Proof (by contradiction). Assume that A is a regular language. Let p be the pumping length for A. Let s be the string 0p1p. s is a valid string because sA and sp. Consider all ways of dividing s into x,y, and z. If y consists only of 0 's, then condition 3 of the pumping lemma does not hold because xy2z=0p+n1p/A (for n>0 ). If y contains one or more 1 's, then condition 2 of the pumping lemma does not hold because xy=0p1n>p (for n>0 ). Therefore, the pumping lemma does not hold and A is not a regular language. b. Proof (by contradiction). Assume that A is a regular language with pumping length p. Let s be the string 0p1pA with length at least p. By the pumping lemma, s can be divided into x,y, and z such that y>0,xyp, and xylzA for any i0. Let x=0p,y=1, and z=1p1. Condition 2 of the pumping lemma does not hold because xy=0p1>p. Therefore, the pumping lemma does not hold and A is not a regular language