Can someone help me simplify this in Boolean algebra? It should be one step at a time so I can understand it. The expression is:
$(x+y+z)(x+z)(x'+y+z)$
I tried doing this: (it's probably wrong, because I think it should simplify to just z?)
$(x+y+z)(x+z)(x'+y+z)$
$(x+y+z)(z+(x+x'+y))$ //distributive property
$z+(x+y+1+y)$ //distributive again
$z+x+y+(x+x')$ //+ identity
$z+x+y+x'$ //+ identity (Edit I know this is wrong, x+x' = 1 not 0, my mistake)
$z+y$ //+ identity
I feel this is wrong, because 'y' and not z wouldn't make the initial expression false.
Best Answer
I think this is right:
$(x+y+z)(x+z)(x'+y+z)$
$(y+z+xx')(x+z)$
$(y+z)(x+z)$
$z+yx$
Can someone confirm this please?