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Verify the following using Boolean lows x’ + y’z = x’y’z’ + x’yz’ + x'yz + x’y’z + xy’z
Answer:
To prove:
x’ + y’z = x’y’z’ + x’yz’ + x'yz + x’y’z + xy’z
Proof:
RHS
= x’y’z’ + x’yz’ + x'yz + x’y’z + xy’z
= x’y’z + x’y’z’ + x’yz’ + x’yz + xy’z (rearranging the terms)
= x’y’(z+z’) + x’y(z+z’) + xy’z (using distributive law)
= x’y’.1 + x’y.1 + xy’z (using complement law)
= x’y’ + x’y + xy’z (using identity law)
= x’(y’+y) + xy’z (using distributive law)
= x’.1 + xy’z (using complement law)
= x’ + xy’z (using identity law)
= (x+x’).(x’+y’z) (using distributive law)
= 1.(x’+ y’z) (using complement law)
= x’ + y’z (using identity law)
= LHS
Hence, the expression x’ + y’z = x’y’z’ + x’yz’ + x'yz + x’y’z + xy’z is verified.
