Prove $x' + y' + xyz' = x' + y' + z'$

Question

I had this question in an exam this morning but was unable to solve it. How would I start approaching this proof? I've tried multiplying x by ($y+y'$) or ($z + z'$) and similarly with y, multiplying it by ($x+x'$) or ($z+z′$) and gotten stuck.

For notation's sake, our text uses "+" to represent an OR operation and products such as $xyz'$ to represent the AND operations.

Answer 1

A very useful principle is:

Reduction

$x+x'y=x+y $

Notice that with this principle, your equivalence is immediately proven with two applications of Reduction.

You were probably not provided with Reduction (most texts don't, which is unfortunate, since it is such a very handy principle to have around when doing Boolean algebra!). Here is how you perform reduction with more basic principles:

$$x + x'y= \text{ (Distribution)}$$

$$(x+x')(x+y)=\text{ (Complement)}$$

$$1(x+y)=\text{ (Identity)}$$

$$x+y$$

Applied to your statement:

$$x'+y'+xyz'=\text{ (Distribution)}$$

$$(x'+y'+x)(x'+y'+y)(x'+y'+z')=\text{ (Complement)}$$

$$(1+y')(1+x')(x'+y'+z')=\text{ (Annihilation)}$$

$$11(x'+y'+z')=\text{ (Identity)}$$

$$x'+y'+z'$$

Answer 2

If $x$ and $y$ are true then $xyz'=z'$.

If $x$ or/and $y$ is false then $x'$ or/and $y'$ is true and $x'+y'$ is true and so is $x'+y'+\text{anything}$.

So all in all the two expressions are equivalent.

Answer 3

Using the various axioms for Boolean algebra, we have

$$\begin{align} x'+y'+xyz' &=x'+y'+(x'+y'+z)'\\ &=(xy(x'+y'+z))'\\ &=(xx'y+yy'x+xyz)'\\ &=(0+0+xyz)'\\ &=(xyz)'\\ &=x'+y'+z' \end{align}$$