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I have the following Boolean expression:

$$w'x'y'z + wx'y'z + xz + xyz'\tag{1}$$

Upon doing my own work, I can only get as far as:

$$zx + xy + zy'\tag{2}$$

Now, when I put the original equation into the following webpage (http://calculator.tutorvista.com/math/582/boolean-algebra-calculator.html, I've been using it to double check myself), it keeps saying the full simplification is:

$$xy + zy'\tag{3}$$

Using Truth Tables, I've compared all three equations to each other and all prove equal that I can tell, and what confuses me most is how (2) apparently simplifies to (3).

Does (2) simplify to (3)? Or am I missing some step between getting from (1) to (3)? Or is (3) simply incorrect (perhaps a glitch in the online software)?

Asaf Karagila
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skippr
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1 Answers1

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You've already answered part of your question yourself: Since the truth tables are the same, the expressions are equivalent, so it's not a glitch in the software.

To see that $(2)$ is equivalent to $(3)$, note that for $zx$ to have an effect, both $x$ and $z$ would have to be true; but in that case $xy$ is $y$ and $zy'$ is $y'$, and $y+y'$ is true; thus $zx$ is redundant. More formally:

$$ \begin{align} xy+zy'+zx&=xy+zy'+zx(xy+zy')' \\ &=xy+zy'+zx(xy)'(zy')' \\ &=xy+zy'+zx(x'+y')(z'+y) \\ &=xy+zy'+zxy'y \\ &=xy+zy' \end{align}$$

(where I moved $zx$ to the right to make it typographically apparent that the rest stays the same).

joriki
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  • See, that's where my problem is. When you add the negation of (xy + zy') at the end there, what exactly is that "saying?" and why do you add it? I was just introduced to Boolean Algebra and only have basic identities at my disposal. That last step seems to go beyond that though? – skippr Feb 14 '13 at 06:30
  • @sunday: I'm confused. First, I added $(xy+zy')'$, not $(xy+zy')$. Second, that's the first step, not the last step. So which step do you mean, that first step, or the last step? – joriki Feb 14 '13 at 06:52
  • Your first step. And yes, you negated (xy+zy'). I'm confused why you added that there. – skippr Feb 14 '13 at 06:54
  • @sunday: That's a formalization of the argument I gave in words. I'm using $a+b=a+ba'$, which can be proved in terms of more basic identities:

    $$a+b=a+b\cdot1=a+b(a+a')=1\cdot a+ba+ba'=(1+b)a+ba'=1\cdot a+ba'=a+ba';.$$

    For a question like this where you're only allowed to use certain basic identities because others haven't been proved yet, it's usually a good idea to state in the question which identities you're allowed to use.

     – joriki Feb 14 '13 at 07:01
  • @joriki, is there some kind of law we can apply to simplify $(xy+zy′+zx)$ to $(xy+zy′)$ ? – Pacerier May 07 '13 at 10:34
  • @Pacerier: I don't understand the question. I showed in the answer how to simplify $xy+zy'+zx$ to $xy+zy'$ using laws of logic. What other kind of law do you have in mind? – joriki May 07 '13 at 10:56
  • @joriki, I mean not using logic, but using algebra working. – Pacerier May 07 '13 at 15:01
  • @Pacerier: I don't understand the difference. Isn't what I did in the answer a proof using the algebraic laws of logic? Could you give an example of something that you'd consider more algebraic than this? – joriki May 07 '13 at 15:11