Consider the language L = {(aa)^m c(bb)^n: m, n greaterthanorequalto 0}. This language contains the strings c, aacbb, aaaacbb, et cetera. Consider the following theorem and proof: Theorem: L is regular. Proof: We show that in the regular expression game, A (the opponent) can always win. Suppose A picks the integer N 50, B picks any string (aa)^m c(bb)^n of length larger than 50, then if m greaterthanorequalto 1 A picks x = e, y = aa, z = (aa)^m-1 c(bb)^n. Now whatever Value of i B picks, the string xy^iz is in L because xy^iz is (aa)^i (aa)^m-1 c(bb)^n. If m = 0 then (because the string has length larger than 50), n greaterthanorequalto 1 and A picks x = c, y = bb, and z = (bb)^n-1. This is possible because n greaterthanorequalto 1. Whatever value of i B picks, the string xy^iz is in L because xy^iz is c (bb)^i (bb)^n-1. Therefore the opponent (A) can always win, so L is not regular. Is the theorem correct? Is the proof correct? Justify both of your answers. Consider the language L = {(aa)^m c(bb)^n: m, n greaterthanorequalto 0}. This language contains the strings c, aacbb, aaaacbb, et cetera. Consider the following theorem and proof: Theorem: L is regular. Proof: We show that in the regular expression game, A (the opponent) can always win. Suppose A picks the integer N 50, B picks any string (aa)^m c(bb)^n of length larger than 50, then if m greaterthanorequalto 1 A picks x = e, y = aa, z = (aa)^m-1 c(bb)^n. Now whatever Value of i B picks, the string xy^iz is in L because xy^iz is (aa)^i (aa)^m-1 c(bb)^n. If m = 0 then (because the string has length larger than 50), n greaterthanorequalto 1 and A picks x = c, y = bb, and z = (bb)^n-1. This is possible because n greaterthanorequalto 1. Whatever value of i B picks, the string xy^iz is in L because xy^iz is c (bb)^i (bb)^n-1. Therefore the opponent (A) can always win, so L is not regular. Is the theorem correct? Is the proof correct? Justify both of your answers