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

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    Why don't you set up a truth-table and find out??? – 2017-01-08
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    I don't know how to do that, sir... – 2017-01-08

3 Answers 3

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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.

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    umm.. I never heard of creating a copy method? is it one of the theorem? – 2017-01-08
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    @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 $x'yz'$. – 2017-01-08
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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.

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    so... I'm correct? – 2017-01-08
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    @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. – 2017-01-08
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    Yeah I think I should. Thanks! – 2017-01-08
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    @DonnyAvaris: You're welcome :) – 2017-01-08
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One further simplification: x'y'+y(x+z')

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    That seems incorrect. I got $xy + x'(y'+z')$ – 2017-01-08
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    Nevertheless, I persist. Both are correct. – 2017-01-08
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    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. – 2017-01-08