My Math Forum Simplifying this Boolean expression

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 May 1st, 2016, 08:33 AM #1 Newbie   Joined: Jan 2016 From: Naples Posts: 4 Thanks: 0 Simplifying this Boolean expression This is the function: $(x + y)(x' + z)$ I did this: $(x + y)(x' + z) =$ $xx' + xz + x'y + yz =$ $xz + x'y + yz =$ $x(y + y')z + x'y(z + z') + (x + x')yz =$ $xyz + xy'z + x'yz + x'yz' + xyz + x'yz =$ $xyz + xy'z + x'yz + x'yz' + x'yz =$ $xy'z + x'y(z + z') + (x + x')yz =$ $xy'z + x'y + yz$ Is this correct? Was there a better and faster way to do it?
 May 1st, 2016, 08:54 AM #2 Newbie   Joined: May 2016 From: Kolkata Posts: 1 Thanks: 1 I think the steps are correct uptil 3rd step , no need to drag along the process after that ! Its a sum relating to simplification and not proof ! Thanks from Kernul
 May 1st, 2016, 02:04 PM #3 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,790 Thanks: 629 Math Focus: Yet to find out. It depends what you are trying to do, as hellosuman has said. Do you need to simplify, or convert your POS expression into a SOP expression?

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