Boolean Expression x'y + x(x + y')

Question

I'm still learning Boolean Algebra so I apologise if the question seems pretty straight forward. After working on this problem I thought the answer was 'x', my working out was:

x'y + x(x + y')

= x'y + xx + xy'

= x'y + x + xy'

= x + xy'

= x

but after runnng this through a logic circuit I realise if x = 0 and y = 1 then the ouput does not equal x. Can someone help me simplify this expression

Maybe you wanted to apply something similar as in your last step. There it works because $x+xy' = x(1+y') \stackrel{1+y' = 1}{=} x$. But from the 3rd to the 4th line the situation is different. — trancelocation, Sep 25 '19 at 11:57

score 1 · Answer 1 · answered Sep 25 '19 at 15:00

1

Two really useful principles are:

Absorption

$x(x+y)=x$

$x+xy=x$

Reduction

$x(x'+y)=xy$

$x+x'y=x+y$

With those:

$x'y+x(x+y')\overset{Absorption}{=} x'y+x\overset{Reduction}{=} y+x$

answered Sep 25 '19 at 15:00

Bram28

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score 0 · Accepted Answer · answered Sep 25 '19 at 12:06

You correct until the step

x'y + x + xy' = x + xy'

But you can go from there and see that x'y + xy' = x ⊕ y.

This follows since x'y = 1 ⇒ x=0, y=1 and xy' = 1 ⇒ x=1, y=0.

Following this x ⊕ y + x can be simplified.

x ⊕ y + x = x + y

With your final solution x'y + x(x + y') = x + y.

Boolean Expression x'y + x(x + y')

2 Answers2