simplifying $(XY' + YZ)(X'Y' + XZ + YZ)$ into a polynomial with few terms as possible

Question

I expand out the brackets and through the law of complementation, I'm left with: $$ XY'Z + XYZ + YZ $$

which through law of absorption, simplifies to: $$ XY'Z + YZ $$

However the answer from the book is: $$YZ + XZ$$

What am I missing here?

Thanks

Answer 1

Note that $YZ=XYZ+YZ$, so

$$\begin{align*} XY'Z+YZ&=XY'Z+XYZ+YZ\\ &=X(Y'+Y)Z+YZ\\ &=XZ+YZ\;. \end{align*}$$

Of course this probably seems like an after the fact justification, and you might wonder how you might have spotted that further simplification is possible. You might have noticed that $XY'Z+YZ=(XY'+Y)Z$, and the $Y'$ in the $XY'$ term really is redundant given the $Y$ term, because $XY'+Y$ is true when $X$ is true whether or not $Y$ is true: if $Y$ is true, it’s true simply because $Y$ is true, and if $Y$ is false, it’s true because $X$ is true.

Answer 2

I don't see how you use the absorption law to get your answer, but anyway one has $$ XY'Z + XYZ = X(Y + Y')Z = XZ $$ which gives the answer to your question.

Answer 3

Calculate : $(XY'+YZ)(X'Y'+XZ+YZ)=$

$=XY'X'Y'+XY'XZ+XY'YZ+YZX'Y'+YZXZ+YZYZ=$

$=(XX')Y'Y'+(XX)Y'Z+(YY')XZ+(YY')ZX'+YX(ZZ)+(YZ)(YZ)=$

$=(0)Y'Y'+(X)Y'Z+(0)XZ+(0)ZX'+YX(Z)+(YZ)=$

$=0+XY'Z+0+0+YXZ+YZ=$

$=XY'Z+XYZ+YZ=$

$=X(Y+Y')Z+YZ=$

$=X(1)Z+YZ=$

$=XZ+YZ.$

Answer 4

Use a Karnaugh map.

  |0011 X
Z |0110 Y
--+------
0 |0000
1 |0111

From this it's immediate that your function is equivalent to $XZ+YZ$

You just have to find a sum of monomials that covers exactly the 1s in the table: The part .11. is covered by $YZ$, and the part ..11 is covered by $XZ$.

Answer 5

Two equivalence principles apply here.

First, we have:

Reduction

$P +P'Q = P + Q$

This is because:

$P + P'Q = (P + P')(P+Q)=1(P+Q) = P+Q$

That is: in the context of $P$, the term $P'Q$ can be reduced to just $Q$

This can be generalized to:

Contextualized Reduction

$PR+P'QR=PR+QR$

And that is because:

$PR+P'QR=(P + P'Q)R \overset{Reduction}{=} (P+Q)R=PR+QR$

That is, in the context of $R$, we can yet again reduce the $P'Q$ in the context of $P$

Notice that this principle applies immediately to your problem:

$XY'Z+YZ\overset{Contextualized \ Reduction}{=}XZ+YZ$

Few textbooks will actually lists these principles, but in my experience I have found plenty of occasion to make use of them: it quickly simplifies expressions!

In fact, here are a few others that are not always listed but that are also super useful:

Absorption

$P + PQ = P$

Adjacency

$PQ +PQ'=P$

Consensus

$PQ+P'R+QR=PQ+P'R$

Add these to your list and you can thank me later :)