Can someone please show me how to simplify this expression? I have no clue of how it's done so any help would be appreciated. Thank you!

x'y'+xz'+yz+x'yz'+xy'z

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- Apr 16th 2013, 05:52 PMelorabeesCan someone help me simplify this boolean expression?
Can someone please show me how to simplify this expression? I have no clue of how it's done so any help would be appreciated. Thank you!

x'y'+xz'+yz+x'yz'+xy'z - Apr 16th 2013, 06:17 PMchiroRe: Can someone help me simplify this boolean expression?
Hey elorabees.

Hint: Try using De-morgans laws on a few of the terms. - Apr 16th 2013, 06:21 PMelorabeesRe: Can someone help me simplify this boolean expression?
- Apr 16th 2013, 06:41 PMchiroRe: Can someone help me simplify this boolean expression?
Demorgans theorems are as follows:

(a+b)' = a'b' and

(ab)' = a' + b'

Try using these to convert terms to common ones and simplify using the other laws (namely distributive). - Apr 16th 2013, 07:00 PMelorabeesRe: Can someone help me simplify this boolean expression?
Okay, but what can I do with xz'? I don't know what to do when we have a variable with a complement and a variable without a complement together?

I just reduced this part:

x'y'+xz'+yz=x'+y'+xz'+yz=(x'+x)(x'+z')+(y'+y)(y'+z )=(x'+z')+(y'+z)=x'z'+(y'+z) , Can we get y'z from that? and can you tell me if that is right? - Apr 16th 2013, 07:48 PMchiroRe: Can someone help me simplify this boolean expression?
Just as a hint for these problems, you can double check your work by creating a truth table consisting of 0 and 1 for all variables to check whether two statements are equivalent. I recommend you do this to check your own answers.

- Apr 16th 2013, 07:51 PMelorabeesRe: Can someone help me simplify this boolean expression?