Question

simplify the following using Boolean algebra.
(xy)'+xz'+yz

a)y+z'
b)x+y
c)X'+y
d)xz'

Answer 1

c) x'+y

Answer 2

$$
\Large{
(xy)'+xz'+yz\\
=[(xy)(xz')'(yz)']'\\
=[(xy)(x'+z)(y'+z')]'\\
=[((xyx')+(xyz))(y'+z')]'\\
=[0+xyzy'+xyzz']'\\
=[0+0+0]'\\
=[0]'\\
=1
}
$$
This is a tautology, because it is always true.

Here is another method using a Truth Table

This also shows a tautology