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Simplify the boolean expression $(xy')\cdot(x'+ y')'$.

Here is my work:

$(xy')\cdot(x'+ y')' = [(x' + y'') · (x' + y')]'$ de Morgans rule (b)

$= [(x' + y) · (x' + y')]'$ involution

$= (x' + y)' + (x' + y')'$ de Morgans rule (b)

$= (x''·y') + (x''·y'')$ de Morgans rule (a)

$= (x·y') + (x·y)$ involution

$= x(y' + y)$ distributive rule (b)

$= x(1)$ complement rule (a)

$= x$ identity rule (b)

darij grinberg
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DanS
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1 Answers1

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A bit more briefly:

$$[ (xy')' * (x'+y') ]' \equiv (xy') + (x'+y')'\tag{DeMorgan's}$$

$$\equiv (xy') + (xy)\tag{DeMorgan's}$$

$$\equiv x(y' + y)\tag{Distributive property}$$

$$\equiv x*1 \tag{complement rule}$$

$$\equiv x \tag {identity rule}$$

amWhy
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