Can someone simplified this boolean

[tonyzgolfer]

Homework Statement

here are my function.

s = xy'z + x'y'z + xyz

Homework Equations

boolean algebra

The Attempt at a Solution

after i try many boolean law. i got this
c. F(x,y,z) = xy’z+x’y’z+xyz
= xy’z + x’y’z + x’y’z + xyz
= y’z(x + x’) + x’y’z + xyz
= y’z(x + x’) + z(x’y’+xy)
= y’z(1) + z(x’y’+xy)
but i stuck in this line
i know x'y'+xy is (x xor y )'



what is the simplest ?

thank in advance

 

[berkeman]

Homework Statement

here are my function.

s = xy'z + x'y'z + xyz

Homework Equations

boolean algebra

The Attempt at a Solution

after i try many boolean law. i got this
c. F(x,y,z) = xy’z+x’y’z+xyz
= xy’z + x’y’z + x’y’z + xyz
= y’z(x + x’) + x’y’z + xyz
= y’z(x + x’) + z(x’y’+xy)
= y’z(1) + z(x’y’+xy)
but i stuck in this line
i know x'y'+xy is (x xor y )'



what is the simplest ?

thank in advance

First, you have two terms in the initial equation that have y'z in them. You have them correctly combined in your last equation above.

Second, if you use a Karnaugh map to help you see what to do in the logic, that may help. Draw a K-map of the 3 terms in the equation. See how the y'z combination becomes obvious? Do you see another combination that will get you to the minimim implementation?

 

[zgozvrm]

Without using a Karnaugh map:

F(x,y,z) = xy'z + x'y'z + xyz
= (xy'z + x'y'z) + xyz
<< middle steps edited out by berkeman >>
= (y' + x)z
= y'z + xz

 

[berkeman]

Without using a Karnaugh map:

F(x,y,z) = xy'z + x'y'z + xyz
= (xy'z + x'y'z) + xyz
<< middle steps edited out by berkeman >>
= (y' + x)z
= y'z + xz

Please do not do all the work for the OP. They must be doing the bulk of the work on their homework/coursework problems.

 

[Robbert]

I always prefer to use Karnaugh maps with these sort of problems since I'm more visual, so give that a try.
Otherwise, zgozvrm has posted a useful start. I'd continue using De Morgan's law.

 

[zgozvrm]

I, too, prefer using Karnaugh maps for this sort of problem. I was simply showing how to solve the problem using Boolean algebra, since the OP was going that route. Perhaps, the original problem stated that Boolean algebra was required, or that Karnaugh maps were not to be used.