Question

simplify using Boolean algebra: (a'+b+c')(a'+b++d+e)(c+d)​

Answer 1

Answer:

Solution

Given,F(A,B,C)=A′B+BC′+BC+AB′C′

Or,F(A,B,C)=A′B+(BC′+BC′)+BC+AB′C′

[By idempotent law, BC’ = BC’ + BC’]

Or,F(A,B,C)=A′B+(BC′+BC)+(BC′+AB′C′)

Or,F(A,B,C)=A′B+B(C′+C)+C′(B+AB′)

[By distributive laws]

Or,F(A,B,C)=A′B+B.1+C′(B+A)

[ (C' + C) = 1 and absorption law (B + AB')= (B + A)]

Or,F(A,B,C)=A′B+B+C′(B+A)

[ B.1 = B ]

Or,F(A,B,C)=B(A′+1)+C′(B+A)

Or,F(A,B,C)=B.1+C′(B+A)

[ (A' + 1) = 1 ]

Or,F(A,B,C)=B+C′(B+A)

[ As, B.1 = B ]

Or,F(A,B,C)=B+BC′+AC′

Or,F(A,B,C)=B(1+C′)+AC′

Or,F(A,B,C)=B.1+AC′

[As, (1 + C') = 1]

Or,F(A,B,C)=B+AC′

[As, B.1 = B]

So,F(A,B,C)=B+AC′is the minimized form.

Problem 2

Minimize the following Boolean expression using Boolean identities −

F(A,B,C)=(A+B)(A+C)