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Can someone help me with simplifying this Boolean algebra

Prove: (x * y') + (y' * z) + (x' * z) = (x * y') + (x' * z)

x'= not x

Can you please show me step by step and the laws which you have applied in proving?

Thank you very much!

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  • $\begingroup$ That's your second Boolean algebra question without showing us any effort! $\endgroup$
    – Bram28
    Jul 7, 2017 at 12:30
  • $\begingroup$ It is quite fast for you to come to that conclusion? How do you know I am not trying? The concept of the two questions I asked are totally different. I have spent at least 1 hour in trying to solve each question. If I can't think of the answer, why should I stop myself from asking for help and learn from that? $\endgroup$
    – Anlin yang
    Jul 7, 2017 at 13:16
  • $\begingroup$ I wasn't saying you weren't trying, for indeed I cannot know that, but I was commenting on the fact that you haven't shown us what you did try, and where you get stuck. $\endgroup$
    – Bram28
    Jul 7, 2017 at 13:31
  • $\begingroup$ Also, look up Consensus theorem for this one. $\endgroup$
    – Bram28
    Jul 7, 2017 at 13:32
  • $\begingroup$ @Bran28 don't be so hard on Anlin. After all writing 2 questions in a row is already very tiring :-) $\endgroup$
    – magma
    Jul 8, 2017 at 13:02

2 Answers 2

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It's worth your while to study this proof because this pattern shows up a lot. In fact, it's the proof of an identity known as Consensus:

\begin{array}{l} & xy' + y'z + x'z & \text{ Given }\\ & xy' + y'z1 + x'z & \text{ Identity }\\ & xy' + y'z(x + x') + x'z & \text{ Identity }\\ & xy' + y'zx + y'zx' + x'z & \text{ Distributive }\\ & xy' + y'zx + x'z + y'zx' & \text{ Associative }\\ & xy'(1 + z) + x'z(1+ y') & \text{ Distributive }\\ & xy'1 + x'z1 & \text{ Identity }\\ & xy' + x'z & \text{ Identity }\\ \end{array}

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  • $\begingroup$ Thank you so much. I totally did not think of multiplying by a 1 $\endgroup$
    – Anlin yang
    Jul 7, 2017 at 13:17
  • $\begingroup$ @Anlinyang. That's an important trick to keep in mind, not only with logic but in all areas of mathematics. It changes the form of the expression to something more manageable without changing its truth value. $\endgroup$
    – User4407
    Jul 7, 2017 at 13:19
  • $\begingroup$ @anlin you should also upvote the answer, since you like it, don't you think so? $\endgroup$
    – magma
    Jul 8, 2017 at 12:57
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The Consensus Theorem does this in 1 step:

Consensus Theorem

$XY'+Y'Z+X'Z \Leftrightarrow XY'+X'Z$

We can derive the Consensus Theorem (and thus your equivalence as well) from some more basic principles:

$XY'+Y'Z+X'Z= \text{ Adjacency}$ (Adjacency says $PQ+PQ'=P$)

$XY'+XY'Z+X'Y'Z+X'Z= \text{ Absorption}$ ($XY'$ absorbs $XY'Z$ and $X'Z$ absorbs $X'Y'Z$)

$XY+X'Z$

If you didn't already know of Absorption and Adjacency, then you should immediately add them to your Boolean algebra toolbox, because they are super useful!

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  • $\begingroup$ Thank you for your help. This method has the same concept of multiplying with (x' + x) $\endgroup$
    – Anlin yang
    Jul 9, 2017 at 3:23

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