Boolean algebra Simplification of "xy + x'z + yz"

Question

I'd like to simplify the following expression "xy + x'z + yz":

         xy + x'z + yz = xy + z(x' +y)
                       = (xy + z)(xy + x' + y)
                       = (xy + z)(y(x + 1) + x') 
                       = (xy + z) ( y + x')

What do I do after the last step?

It is hard to hit a target you can't see. What do you want the end result to be like? — rschwieb, Jan 17 '15 at 19:44
i want to simplify in detail as with lower expressions it can get..do you understand my dude ? — Pantelis10, Jan 17 '15 at 19:46

score 7 · Accepted Answer · answered Jan 17 '15 at 19:57

By either drawing a Karnaugh map or recognizing that this is just the Consensus Theorem, observe that: \begin{align*} xy + x'z + yz &= xy + x'z + (1)yz \\ &= xy + x'z + (x + x')yz \\ &= xy + x'z + (xyz + x'yz) \\ &= (xy + xyz) + (x'z + x'yz) \\ &= xy(1 + z) + x'z(1 + y) \\ &= xy(1) + x'z(1) \\ &= xy + x'z \end{align*}

Boolean algebra Simplification of "xy + x'z + yz"

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