How to simplify the boolean expression $(x\times y)'+(y\times z)$?

Question

Expression:

$$(x\times y)'+(y\times z)$$

My attempt:

1. $(xy)' + (yz)$
2. $(x'+y') + (yz) \quad \textit{After applying de Morgan's Axiom}$
3. $x' + (y' + yz) \quad \textit{After applying 1st Distributive Axiom}$
4. $x'+ (y' + y) + (y'z) \quad \textit{Rewriting}$
5. $x' + 1 + (y'z) \quad \textit{After applying Inverse Axiom}$
6. $x' + y' + z \quad \textit{After applying second Identity Axiom}$

The answer is supposed to be: $(x'+y')+z$

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N.B. I am still new to this and learning, I am using the following book:

Discrete Mathematics for Computing / Edition 3
by Peter Grossman

Question: What am I missing or have I done wrong?

Answer 1

There is a mistake on line 4. Applying Distribution on $y’ + (yz)$ will get you $(y’+y)(y’+z)$, and so line 4 should be $x’ + ((y’+ y)(y’+z))$

Accordingly, line 5 should be $x’+ 1(y’+z)$. In fact, if line 5 was what you have, $x’+1+y’z$, then your line 6 should be just $1$, because $1+ [anything]$ is always just $1$. (I don’t know if your book has an equivalence principle for that, but many books call it Annihilation)

I also note that between your line 5 and 6, the $y’z$ term suddenly changes to $y’+z$, which isn’t right either, so you made two mistakes in going from line 5 to line 6

Despite these mistakes, you somehow managed to reach the correct answer. As indicated, the correct line 5 should have been $x’+1(y’+z)$ which by Identity becomes $x’+y’+z$. So, you happened to get to the right answer, but you made several mistakes getting there.

Here is how you would have done this correctly:

$(xy)’+yz$

$(x’ +y’)+yz$ (DeMorgan)

$x’ + (y’ +yz)$ (Association)

$x’+ (y’+y)(y’+z)$ (Distribution)

$x’ + 1(y’+z)$ (Complement)

$x’+ (y’+z)$ (Identity)

$(x’+y’)+z$ (Association)

So … You were very close! It’s just a matter of staying calm and composed and carefully applying those patterns correctly!

Answer 2

We want to simplify $(x\times y)'+(y\times z)$.

I will first show the way how I would solve this task, after that I want to take a look onto your try to solve this task.

Simplifying the boolean expression

I will use the $\equiv$ symbol to show, that two boolean expressions $A$ and $B$ have the same value. I will use the symbol $\top$ to indicate the value true (or one).

\begin{align*} &(x\times y)' + (y\times z)&\overset{(1)}{\equiv} \\ &(x' + y') + (y\times z) &\overset{(2)}{\equiv} \\ &((x'+y')+y)\times((x'+y')+z) &\overset{(3)}{\equiv}\\ &(x'+(y'+y))\times(x'+y'+z) &\overset{(4)}{\equiv} \\ &(x'+\top)\times(x'+y'+z)&\overset{(5)}{\equiv} \\ &\top\times(x'+y'+z)&\overset{(6)}{\equiv} \\ &(x'+y'+z)&\overset{(7)}{\equiv} \\ &(x'+y')+z \end{align*}

We see that we are ending up with the solution presented in the book. Now I will shortly present the single steps. $A, B$ and $C$ are all boolean expressions:

1. Here I used the de Morgan Axioms.
2. After that I applied the distributivity rule $A+(B\times C)\equiv (A+B)\times(A+C)$. Where $A\equiv(x'+y')$, $B\equiv y$ and $C\equiv z$.
3. Now I used the associative law, which states that I can change the parantheses.
4. Here I used that $A+A'\equiv\top$ always holds. You can see that with a truth table.
5. Here I used that $A+\top\equiv\top$ always holds. As in step 4 you can also see this with a truth table.
6. Now I used that $\top$ is a so called neutral element for the $\times$ operation. So $\top\times A\equiv A$ always holds. This is again proofable with a truth table.
7. Now I used the associative law to match the parantheses to the parantheses in the solution. This step is in my opinion not necessary. I would end my calculations with the sixth step.

Your solution

Your first step and the second step is completly correct. But your third step was a bit hastly. You state the following: $$x'+(y'+(y\times z))\equiv x'+(y'+y)+(y'\times z))$$ But that is where you went wrong. It should be: $$x'+(y'+(y\times z))\equiv x'+(y'+y)+(y'\boldsymbol{+}z))$$

You come up with the correct answer because you just exchange $y'+z$ with $y'\times z$, and vice versa. But you can't do this.

I would recommend to first do the application of the rules as exact and small stepped as possible. After you are a bit trained you can do more than one rule in one step.