Help on Boolean Expression Simplification

Question

I got a question and I'm wondering is the way I solve it correct? $$xyz + x'z' + xyz' + x'y'z + x'yz'$$ $$= xyz + xyz' + x'y'z + x'yz' + x'yz' + x'y'z'$$ $$= xyz + xyz' + x'y'z + x'yz' + x'y'z'$$ $$= xy + x'y' + x'yz'$$

Why don't you set up a truth-table and find out??? – barak manos Jan 08 '17 at 15:32 — barak manos, Jan 08 '17 at 15:32
I don't know how to do that, sir... – Donny Avaris Jan 08 '17 at 15:35 — Donny Avaris, Jan 08 '17 at 15:35

Bram28 · Accepted Answer · 2017-01-08T17:57:56.617

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You did everything correct, but there is one more simplification you can make from your third step on:

$$xyz + xyz' + x'y'z + x'yz' + x'y'z'$$ (and now create a copy of $x'y'z'$)

$$xyz + xyz' + x'y'z + x'y'z' + x'yz' + x'y'z'$$

$$xy + x'y' + x'z'$$

You can also combine the last two terms to $x'(y'+z')$, and you can do some DeMorgan's as well to further minimise the number of operations, but if it has to be in CNF (SOP), then stick to the last line as is.

edited Jan 08 '17 at 17:57

answered Jan 08 '17 at 16:57

Bram28

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umm.. I never heard of creating a copy method? is it one of the theorem? – Donny Avaris Jan 08 '17 at 17:06
@DonnyAvaris By 'copying' I meant that I used the equivalence $x=x+x$, so you can always create or remove duplicates. In fact, you did this yourself when you went from step 2 to 3 as you deleted one of the two $x'yz'$. – Bram28 Jan 08 '17 at 17:52

score 0 · Answer 2 · answered Jan 08 '17 at 17:01

Your input expression gives:

 x | y | z | xyz | x'z' | xyz' | x'y'z | x'yz' | xyz+x'z'+xyz'+x'y'z+x'yz'
---|---|---|-----|------|------|-------|-------|---------------------------
 0 | 0 | 0 |  0  |  1   |  0   |   0   |   0   |              1
 0 | 0 | 1 |  0  |  0   |  0   |   1   |   0   |              1
 0 | 1 | 0 |  0  |  1   |  0   |   0   |   1   |              1
 0 | 1 | 1 |  0  |  0   |  0   |   0   |   0   |              0
 1 | 0 | 0 |  0  |  0   |  0   |   0   |   0   |              0
 1 | 0 | 1 |  0  |  0   |  0   |   0   |   0   |              0
 1 | 1 | 0 |  0  |  0   |  1   |   0   |   0   |              1
 1 | 1 | 1 |  1  |  0   |  0   |   0   |   0   |              1

Your output expression gives:

 x | y | z | xy | x'y' | x'yz'| xy+x'y'+x'yz'
---|---|---|----|------|------|---------------
 0 | 0 | 0 | 0  |  1   |  0   |       1
 0 | 0 | 1 | 0  |  1   |  0   |       1
 0 | 1 | 0 | 0  |  0   |  1   |       1
 0 | 1 | 1 | 0  |  0   |  0   |       0
 1 | 0 | 0 | 0  |  0   |  0   |       0
 1 | 0 | 1 | 0  |  0   |  0   |       0
 1 | 1 | 0 | 1  |  0   |  0   |       1
 1 | 1 | 1 | 1  |  0   |  0   |       1

As you can see, they (the right-most columns) are identical.

@DonnyAvaris: Yes. But I've noticed that the other answer even suggests a slight simplification of your answer (which I have verified BTW), that you might want to use. — barak manos, Jan 08 '17 at 17:05

score 0 · Answer 3 · answered Jan 08 '17 at 17:48

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One further simplification: x'y'+y(x+z')

answered Jan 08 '17 at 17:48

Senex Ægypti Parvi

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That seems incorrect. I got $xy + x'(y'+z')$ – Bram28 Jan 08 '17 at 18:46
Nevertheless, I persist. Both are correct. – Senex Ægypti Parvi Jan 08 '17 at 21:54
Yes, it is indeed equivalent ... I just didn't see how it was a 'further simplification' from what we were working with as it will take a different route to get what you have. – Bram28 Jan 08 '17 at 22:05

Help on Boolean Expression Simplification

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