Answer to Question #132871 in Electrical Engineering for JANARAKSHAA

Question

Minimize the following Boolean expression using Boolean identities
F(A,B,C,D,E) = A’B’C’D’+A’BC’D’+A’B’CD’+A’B’D’E+AB’CD+ABD’E’+ABD’E+AB’CDE

Answer 1

F(A,B,C,D,E) = A’B’C’D’+A’BC’D’+A’B’CD’+A’B’D’E+AB’CD+ABD’E’+ABD’E+AB’CDE    //Original expression

F(A,B,C,D,E) = A’C’D’(B’+B) + A’B’CD’(1+E) + ABD’(E+ E’)+ A’B’D’E+AB’CD      //Distributive rule

F(A,B,C,D,E) = A’C’D’+ A’B’CD’+ ABD’+ A’B’D’E+AB’CD         //Compliment, Annulment, compliment  respectively

F(A,B,C,D,E) = A’C’D’(B’+ B) + A’B’CD’+ ABD’+ A’B’D’E+AB’CD            // B’+ B=1 (Complement)

F(A,B,C,D,E) = A’C’D’B’ + A’C’D’B+ A’B’CD’+ ABD’+ A’B’D’E+AB’CD   // Distributive

F(A,B,C,D,E) = A’D’B’(C’+ C) + A’C’D’B+ ABD’+ A’B’D’E+AB’CD            // Distributive

F(A,B,C,D,E) = A’D’B’ + A’C’D’B+ ABD’+ A’B’D’E+AB’CD         // C’+ C=1

F(A,B,C,D,E) = A’D’B’(E+1) + A’C’D’B+ ABD’+ AB’CD                              // Distributive

F(A,B,C,D,E) = A’D’B’ + A’C’D’B+ ABD’+ AB’CD                        // Annulment

F(A,B,C,D,E) = A’D’(B’ + C’B) + ABD’+ AB’CD                            // Distributive

F(A,B,C,D,E) = A’D’((B’+ C’)( B’+ B)) + ABD’+ AB’CD                // consequence of distributivity, e.g. P+QR≡(P+Q)(P+R).

F(A,B,C,D,E) = A’D’(B’+ C’) + A(BD’+ B’CD)                // complement., Distributive