Simplifying this boolean function

Question

How can I completely simplify this equation using algebraic simplification rules?

$$x'y'z + x'yz + xyz$$

Answer 1

We use the distributive law twice:

$$\begin{align} x'y'z + x'yz + xyz & = x'z(\underbrace{y' + y}_{= 1}) + xyz \\ \\ & = x'z + xyz \\ \\ &= z(x' + xy)\end{align}$$

What counts as simplified depends on the context.

Answer 2

Hint:

The first two terms have two common factors you should factor out.

Answer 3

The first two terms have a common factor of $x'z$; after pulling it out, you have

$$x'z(y'+y)+xyz=x'z+xyz\;.$$

Once again there’s a common factor, this time of $z$; pulling it out may or may not lead to further simplification, but there’s no hardm in trying:

$$x'z+xyz=(x'+xy)z\;.$$

At this point it would be easy to think that no further simplification is possible: certainly there is no simplification of $x'+xy$ that is as obvious as $y'+y=1$ after the first step. However, you might try drawing a Venn diagram of $x'+xy$:

The blue is $x'$, the red is $xy$, and the white is $xy'$. Thus, $x'+xy$ is the union of the red and the blue, which is the complement of the white: $x'+xy=(xy')'=x'+y$. The Venn diagram doesn’t really constitute a proof, but now that we know that we should be able to simplify $x'+xy$ to $x'+y$, we can look for a computational proof. It’s a little tricky to discover the first times, but after you’ve seen it a time or two, it’s easier to spot: $x'=x'+x'y$ by one of the absorption laws, so

$$x'+xy=x'+x'y+xy=x'+(x'+x)y=x'+y\;.$$

So the original expression can be boiled down to $(x'+y)z$ or to $x'z+yz$, whichever you consider simpler.

Answer 4

We can further simplify by the distributive property:

$$z(x' + xy)=z((x'+x)(x'+y))=z(x'+y)$$ where $x'+x$ is always true.

Check out: http://en.wikipedia.org/wiki/Boolean_algebra_(structure)#Definition