I have to show that the following expression are equal using the simplification rules:
$$x' y z + x y' z + x y z' + x' y z' + x y' z' + x' y' z = x y' + y z' + z x'$$
But I can only get to:
$$y'z + y'x + x' y z$$
Please let me know what I am doing wrong here. I don't know how to proceed from here.
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ Hint: $ab+ab'=a(b+b')=a$. $\endgroup$– PoypoyanCommented Mar 24, 2018 at 23:36
-
1$\begingroup$ The biggest hint is in the RHS of your expression. Collect expressions that have $xy'$, $yz'$ and $zx'$ in common. Then take a look at @Poypoyan's hint. $\endgroup$– an4sCommented Mar 24, 2018 at 23:39
-
$\begingroup$ Thank you! it worked. For future reference, is there a rule for which inputs should I group together before simplifying? $\endgroup$– johnny laymanCommented Mar 25, 2018 at 1:27
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Use:
Adjacency
$pq + pq'=p$
In other words: look for pairs of terms that have all literals in common except one.
Thus, the first and sixth term simplify to:
$x'yz+x'y'z=x'z$
The second and fifth and also be combined:
$xy'z+xy'z'=xy'$
And finally, the third and fourth:
$xyz'+x'yz'=yz'$
So, the whole thing simplifies to:
$x'z+xy'+yz'$
