Question #132871

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

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

Expert's answer

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

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## Comments

Janarakshaa15.09.20, 18:06thanks for the problem

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