The only statement among the following  that is a tautology is:

Note thatA∧(A∨B) is F when A=F, A∨(A∧B) is F when A=F,B=F, and B→[A∧(A→B)] is F when A=F,B=T∴ We check only (c) [A∧(A→B)]→B≡[A∧(~A∨B)]→B≡[(A∧(~A))]∨(A∧B)]→B≡A∧B→B≡~(A∧B)∨B≡∼[(A∧B)∧(~B)]≡~[A∧(B∧~B)]≡∼[A∧F]≡∼F≡TThus,[A∧(A→B)]→B is a tautology.