showing Boolean algebra equality

Question

I have this exercise in my worksheet :

Show that x (z ⊕ y) = xz ⊕ xy

I reached this in solving it , but didn't reach the final equation
x(z'y + zy')
xz'y + xzy'
please can someone show how

What does it mean that you "reached this" but "didn't reach the final equation"? What is the final equation? — hardmath, Apr 10 '14 at 12:13
Isn't that an axiom of boolean algebras? What's your definition of a boolean algebra? — fgp, Apr 10 '14 at 12:14
Isn't it now an application of a DeMorgan law distributing AND over OR? — hardmath, Apr 10 '14 at 13:35

score 0 · Answer 1 · answered Apr 12 '14 at 20:13

0

The truth table shows that both expressions are in fact equivalent:

enter image description here

answered Apr 12 '14 at 20:13

Axel Kemper

4,943

turtle · Answer 2 · 2019-02-26T05:07:03.180

0

Checking the above expression with the truth table is the best option. There is another crude way to prove the same.

xz ⊕ xy = xz (xy)' +(xz) 'xy

      = xz (x'+ y') + (x'+ z') xy
      = xx'z + xzy' + x'xy + xyz'       xx'=0 and yy'=0
      = xzy' + xyz'
      = x ( zy' + yz')
      = x (y ⊕ z)

edited Feb 26 '19 at 05:07

answered Aug 23 '14 at 10:05

turtle

45

showing Boolean algebra equality

2 Answers2