Question

. Simplify the following functional expressions using Boolean algebra and its identities. List the identity used at each step.

y'(x'z' + xz) + z (x + y)'

Answer 1

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

The Simplified functional expressions is x'y' + y'z