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

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: $(x'+y')+z$

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?

  • 1
    $\begingroup$ It is the sixth similar question in two days: what is the "real" issue ? $\endgroup$ Commented Mar 8 at 7:43
  • $\begingroup$ @MauroALLEGRANZA, is it OK to post here? $\endgroup$ Commented Mar 8 at 7:44
  • $\begingroup$ @MauroALLEGRANZA, I am self-studying all on my own. And as soon as I get stuck, I google or, go back and read in chapter or visit YouTube. Finally, as an option I come in here and decide to post as a last resort to try to get hints to possible explanations. $\endgroup$ Commented Mar 8 at 7:48
  • 1
    $\begingroup$ Sorry, now are seven. Please, review the first ones and accept the answers received, if you are satisfied with them. $\endgroup$ Commented Mar 8 at 7:48
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    $\begingroup$ Not sure how you got line 4 Rewriting. $y'+yz$ on line 3 should become $\left(y'+y\right) \left(y'+z\right)$ by distribution. $\endgroup$
    – peterwhy
    Commented Mar 13 at 21:34

2 Answers 2


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:


$(x’ +y’)+yz$ (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!

  • $\begingroup$ @AlixBlaine Well, excuse me for trying to correct your mistakes and also pointing out that the book’s answer is wrong. Check my counterexample or do a full truth-table, but you will find that the statement you start out with is not equivalent to the book’s answer. So either the book’s answer is wrong, or you didn’t copy the starting statement correctly into your post here. So yes, something’s definitely wrong here, but it’s not me. So if you are waiting for an answer that correctly gets you from your post’s original statement to the book’s answer, you can wait for a very, very long time :P $\endgroup$
    – Bram28
    Commented Mar 14 at 11:36
  • $\begingroup$ Bram, added the original problem from the book, see my EDIT. $\endgroup$ Commented Mar 14 at 11:56
  • $\begingroup$ Bram, you are CORRECT! My BIG MISTAKE, I LOOKED AT A DIFFERENT ANSWER IN THE BACK AND WENT CONFUSED FOR DAYS. I AM IN BIG STRESS FOR UP-Coming Written Exam NEXT WEEK! $\endgroup$ Commented Mar 14 at 12:02
  • 1
    $\begingroup$ @AlixBlaine Aha! :) Glad you figured that out. Good luck on your exam! Do be careful in exactly following the patterns … I have noted you regularly make mistakes in that. So this is not a skill issue, but it’s an issue of being calm and composed. I know, easy for me to say when you’re the one with a big exam coming up. But it’s good you’re working on this now. Indeed, the night before don’t stress yourself further and just make sure you get in some good sleep! $\endgroup$
    – Bram28
    Commented Mar 14 at 12:09
  • $\begingroup$ Bram28, thanks. How would you have solved the steps I did, more correctly and precise? I really want to know, so that, I learn from examples too by others. $\endgroup$ Commented Mar 14 at 12:14

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.


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