Simplify the following Boolean expression.

Question

Can someone explain to me how this step done? I got a different answer than what the solution said.

Simplify $x(y+z)(\bar{x} + y)(\bar{y} + x + z)$

what the solution got

$x(y+z)(\bar{x} + y)(\bar{y} + x + z)$ = $x(y + z\bar{x})(\bar{y} + x + z)$ (Using distrubitive)

What I got

$x(y+z)(\bar{x} + y)(\bar{y} + x + z)$ = $x(y\bar{x} + y + z\bar{x} + zy)(\bar{y} + x + z)$ (Using distrubitive)

Answer 1

$y\bar{x} + y + z\bar{x} + zy$

= $y(\bar{x} + 1) + z\bar{x} + zy$

= $y + z\bar{x} + zy$

= $y(1 + z) + z\bar{x}$

= $y + z\bar{x}$

Direct rule -

X + YZ = (X+Y)(X+Z)

So we have -

$(y + z)(y + \bar x)$

= $(y + z \bar x)$

Answer 2

Hint : $ab + ac = a(b+c) = a$ and $a+a=a$.
Do you see the '$b$' and '$c$' (and also the '$a$') in your case?

Answer 3

Put the following equivalence into your boolean algebra toolkit:

Absorption

$x +xy = x$

Using Absorption twice in one step we get:

$y\bar{x}+y+z\bar{x}+zy = y+z\bar{x}$

Done!