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)?


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).

  • $\begingroup$ 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? $\endgroup$ – skippr Feb 14 '13 at 6:30
  • $\begingroup$ @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? $\endgroup$ – joriki Feb 14 '13 at 6:52
  • $\begingroup$ Your first step. And yes, you negated (xy+zy'). I'm confused why you added that there. $\endgroup$ – skippr Feb 14 '13 at 6:54
  • $\begingroup$ @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. $\endgroup$ – joriki Feb 14 '13 at 7:01
  • $\begingroup$ @joriki, is there some kind of law we can apply to simplify $(xy+zy′+zx)$ to $(xy+zy′)$ ? $\endgroup$ – Pacerier May 7 '13 at 10:34

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