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# Figure 2. i'A for L] n L; Pruhlem 2 20% L1={I E {. b3t lam is nut 21 substring ufxi; L2 = {I E in. b}' II ends with abl. For the FA [Figure...

L1 = {x∈{a,b}∗|aa is not a substring of x}; L2={x∈{a,b}∗|x ends with ab}. For the FA (Figure6.1)that accepts L1∩L2, prove that there can not be any other FA with fewer states accepting the same language.

Figure 2.1: i—‘A for L] n L; Pruhlem 2 20% L1={I E {.51. b3"t lam is nut 21 substring ufxi; L2 = {I E in. b}' II ends with abl. For the FA [Figure 2.1} that accepts L. r1 L2. prove that than: cannut be any other FA with fewer statesaccepting the same language.