Question

Digital Logic

Simplify:

F = (x’∙ y’∙ z’) + (x’∙ y ∙ z’) + (x ∙ y’ ∙ z’) + (x ∙ y ∙ z)

F = (x + y + z’) (x + y’ + z’) (x’ + y + z’) (x’ + y’ + z)

Answer 1

F = (x’ y’ z’) + (x’ y z’) + (x y’ z’) + (x y z)
F = x’z’(y’ + y) + (x y’ z’) + (x y z)  { By Distributive law PQ+PR= P(Q+R) }
F = x’z’(1) + (x y’ z’) + (x y z)  { We know that P+P'=1 }
F = x’z’ + x y’ z’ + (x y z)
F = z’(x’ + x y’) + (x y z)
F = z’[(x’ + x)(x’ + y’)[ + (x y z)  { By Distributive law P+QR = (P+Q)(P+R) }
F = z’[(1)(x’ + y’)[ + (x y z)  { We know that P+P'=1 }
F =(x’ z’ + y’ z’) + (x y z) { By Distributive law P(Q+R) = PQ+PR }
F =x’z’ + y’z’ + xyz

The Simplified Function is F =x’z’ + y’z’ + xyz

By Using K-map:

F = (x’∙ y’∙ z’) + (x’∙ y ∙ z’) + (x ∙ y’ ∙ z’) + (x ∙ y ∙ z)

Given

Simplified K-map

The Simplified Function is F =x’z’ + y’z’ + xyz

2)

F = (x + y + z’) (x + y’ + z’) (x’ + y + z’) (x’ + y’ + z)
F = (x + z’+ y) (x + z’+ y’) (x’ + y + z’) (x’ + y’ + z)
F = (x + z’+ yy’) (x’ + y + z’) (x’ + y’ + z) { By Distributive law (P+Q)(P+R) = P+QR }
F = (x + z’+ 0) (x’ + y + z’) (x’ + y’ + z)  { We know that PP'=0 }
F = (x + z’) (x’ + y + z’) (x’ + y’ + z)
F = z’+ x(x’ + y) (x’ + y’ + z)  { By Distributive law (P+Q)(P+R) = P+QR }
F = z’+ x(x’ + y) (x’ + y’ + z)
F = z’+ (xx’ + xy) (x’ + y’ + z)  { By Distributive law P(Q+R) = PQ+PR }
F = z’+ (0 + xy) (x’ + y’ + z)  { We know that PP'=0 }
F = (z’+ xy) (x’ + y’ + z)
F = (z’+ x)(z’+ y) (x’ + y’ + z) { By Distributive law P+QR = (P+Q)(P+R) }
F = (x+z’)(y+z’) (x’ + y’ + z)

The Simplified Function is  F = (x+z’)(y+z’) (x’ + y’ + z)

By Using K-map:

Given

Simplified K-map

The Simplified Function is F = (x+z’)(y+z’) (x’ + y’ + z)