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I've been trying to perform Boolean simplification on the following expression: $y(x' + (x+y)')$

So far, my steps have been:

$y(x' + (x+y)')$

$y(x' + x'y')$

$y(x' + x')(x' + y')$

$(yx')(x' + y') $

I have no idea where to go from here. I feel as if I've complicated this. I know the end result is supposed to be $x'y$.

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  • $\begingroup$ I believe your third step there is where you are going off target. $\endgroup$ – IntegrateThis Oct 8 '17 at 1:40
  • $\begingroup$ @IntegrateThis The third step is valid (it's an instance of Distibution) ... but indeed not helpful $\endgroup$ – Bram28 Oct 8 '17 at 2:51
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$x' + x'y'$ simplifies to $x'$.

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  • $\begingroup$ Out of curiosity, what rule is used to simplify x' + x'y' to x'? $\endgroup$ – Derrin Oct 8 '17 at 1:54
  • $\begingroup$ x' + x'y' = x' ( 1 + y') , and 1+ y' is always true $\endgroup$ – IntegrateThis Oct 8 '17 at 1:59
  • $\begingroup$ But also, consider the truth table. It's obvious I think. $\endgroup$ – IntegrateThis Oct 8 '17 at 2:00
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    $\begingroup$ @Derrin The rule for this is called Absorption. $\endgroup$ – Bram28 Oct 8 '17 at 2:50

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