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May 1st, 2016, 08:33 AM   #1
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Question 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?
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May 1st, 2016, 08:54 AM   #2
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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
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May 1st, 2016, 02:04 PM   #3
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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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