1

xyz' + x'y + x'z

the answer from boolean expression calculator on the internet is

x'z + yz'

I know the end

x'z

same like the answer. but I can't find a way to simplify

xyz' + x'y = yz'

sai-kartik
  • 2,090
mas bro
  • 113
  • 1
    Here's a hint. Don't try to show $xyz' + z'y = yz'$ in isolation, it is probably easier to work with the entire expression the whole way through. Relations such as $x+x'=1$ and $x+1=1$ are useful for this problem. You can use $x+x'=1$ to add in terms that can then factorise with other terms and cancel. – Joz Sep 23 '20 at 14:13
  • thank you for your reply. – mas bro Sep 23 '20 at 15:06

1 Answers1

1

We can only derive $$xyz'+x'y=y(xz'+x')=y(xz'+x'(z+z'))=\\ =y((x+x')z'+x'z)=y(z'+x'z)=yz'+yx'z$$ But it's enough to get $$xyz'+x'y+x'z=yz'+yx'z+x'z=yz'+(1+y)x'z=yz'+x'z$$

Berci
  • 90,745