Thread: boolean algebra to simplify

hey
i've got this expression:
[x'+(yz)'](x+z')'
and i'm supposed to simplify it. I've gotten this far:
[x' + y' + z')(x'z)

I have no clue on what to do next. i've not been taught how to do this and i'm expected to hand in an assignment today about this.
any sort of help will be much appreciated.

thanks

Well now you can distribute getting: (x'x'z+y'x'z+z'x'z) = x'z + y'x'z + x'

cheers. helped me out

Originally Posted by wsldam

Well now you can distribute getting: (x'x'z+y'x'z+z'x'z) = x'z + y'x'z + x'

zz' = 0 (i.e., always false), so we have x'z + y'x'z, not x'z + y'x'z + x'. Further, x'z + y'x'z = (x' + x'y')z = x'z because x' + x'y' = x'(1 + y') = x'1 = x'.

i got that when i finished off
thanks though