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I am stuck on a problem that I know the logical answer to, yet I cannot seem to simplify properly to get there.

  • I want to simplify $$F(x,y,z)=y(x′z + xz′) + x(yz + yz′)$$
  • I know the simplest form (using a truth table) is $$xy + yz$$
  • Yet my simplification yields $$xy + x’yz$$

My steps are below with the identities used

$y(x′z + xz′) + x(yz + yz′)$
$y(x′z + xz′) + x(y(z + z’))$ Distributive
$y(x′z + xz′) + x(y(1))$ Inverse
$y(x′z + xz′) + xy$ Identity
$yx’z + yxz’ + xy$ Distributive
$y(x’z + xz’ + x)$ Distributive
$y(x’z + x)$ Absoption
$yx’z + yx$ Distributive
$\mathbf {xy + x’yz}$ Commutative x 2

Where did I go wrong?

Elliott
  • 123

3 Answers3

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I would probably just write $$ \begin{eqnarray} y(x′z+xz′)+x(yz+yz′) &=& y(x'z+xz')+xyz+xyz' \\ &=&y(x'z+xz')+yxz+yxz' \\ &=&y(x'z+xz'+xz+xz') \\ &=&y(x'z+xz+xz') \\ &=&y(x'z+xz+xz+xz') \\ &=&y((x'+x)z+x(z+z')) \\ &=&y(z+x). \end{eqnarray} $$

mjqxxxx
  • 41,358
  • This is an upworthy answer because it showed me that I did not approach the problem in the simplest terms from the start. Distributing the second OR part to find that there is a common y in all parts is elegant. – Elliott Oct 16 '14 at 20:41
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$x+z$ is equivalent to $x+x'z$. I don't remember the law. Absorption may be? The proof can be obtained from the true/false table or as follows: $$ (x+x'z)'=x' (x+z')=x'z'=(x+z)' $$

Alexander Vigodner
  • 2,237
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Continuing from your simplification:

$$\begin{array}{ccl} xy+x'yz&=&xy(z+z')+x'yz\\ &=& xyz+xyz'+x'yz\\ &=& xyz+xyz+xyz'+x'yz\\ &=& (xyz+xyz')+(xyz+x'yz)\\ &=& xy+yz \end{array}$$

paw88789
  • 40,402