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a b c Y
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 1
1 1 1 0
1 1 0 0

How would you put this into a logical expression that shows Y's behaviour?

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closed as unclear what you're asking by David Richerby, Evil, Juho, Yuval Filmus, hengxin Dec 6 '16 at 3:40

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ What did you try? Where did you get stuck? We're happy to help with conceptual problems but just solving homework-style exercises for you is unlikely to really help you. $\endgroup$ – David Richerby Nov 27 '16 at 13:36
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    $\begingroup$ try karnaugh table $\endgroup$ – yoyo_fun Nov 27 '16 at 13:42
4
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Giving an example of what Eugene has said, but letting you do your own homework. Let's look at another Truth Table:

$ \begin{array}{|c|c|c|c|} \hline a& b & c & \varphi \\ \hline 1 & 1& 1& 1\\ \hline 1 & 1 & 0 & 1 \\ \hline 1 & 0 & 1 & 0 \\ \hline 1 & 0 & 0 & 1 \\ \hline 0 & 1 & 1 & 0 \\ \hline 0 & 1 & 0 & 1\\ \hline 0 & 0 & 1 & 0 \\ \hline 0 & 0 & 0 & 1 \\ \hline \end{array} $

To put it on DNF just look at the rows which evaluate to True.

$ \begin{array}{|c|c|c|c|c|} \hline a& b & c & \varphi & \\ \hline 1 & 1& 1& 1 & a \wedge b \wedge c \\ \hline 1 & 1 & 0 & 1& a \wedge b \wedge \lnot c \\ \hline 1 & 0 & 0 & 1& a \wedge \lnot b \wedge \lnot c \\ \hline 0 & 1 & 0 & 1& \lnot a \wedge b \wedge \lnot c \\ \hline 0 & 0 & 0 & 1& \lnot a \wedge \lnot b \wedge \lnot c \\ \hline \end{array} $

So the DNF of $\varphi$ is : $$ (a \wedge b \wedge c) \vee (a \wedge b \wedge \lnot c) \vee (a \wedge \lnot b \wedge \lnot c) \vee (\lnot a \wedge b \wedge \lnot c) \vee (\lnot a \wedge \lnot b \wedge \lnot c) $$

To put it on CNF just look at the rows which evaluate to False. Now just negate those rows to get True.

$ \begin{array}{|c|c|c|c|c|c|c|c|} \hline a& b & c & \varphi & & & \lnot \varphi & \\ \hline 1 & 0 & 1 & 0 & a \wedge \lnot b \wedge c & & 1 & \lnot a \vee b \vee \lnot c\\ \hline 0 & 1 & 1 & 0 & \lnot a \wedge b \wedge c & & 1 & a \vee \lnot b \vee \lnot c\\ \hline 0 & 0 & 1 & 0 & \lnot a \wedge \lnot b \wedge c & & 1 & a \vee b \vee \lnot c\\ \hline \end{array} $

So the CNF of $\varphi$ is:

$$ (\lnot a \vee b \vee \lnot c) \wedge (a \vee \lnot b \vee \lnot c) \wedge (a \vee b \vee \lnot c) $$

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1
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You can write either CNF or DNF from the truth table directly and then transform it to any form you like opening brackets and possibly using absorption rule.

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0
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Define Y as:

|&A!B&C!A

If you can't understand it, then here is a more readable version:

Y = (((A) AND (NOT B)) OR ((C) AND (NOT A)))

Explanation

If either:

  • A = 1 and B = 0
  • C = 1 and A = 0

Then Y = 1. Else, Y = 0.

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  • 2
    $\begingroup$ I'd describe infix as "more readable". Calling it "more basic" suggests that it's only needed by people who aren't smart enough to parse Polish notation. Frankly, life's too short to unpick that string of symbols. $\endgroup$ – David Richerby Nov 28 '16 at 10:18
  • $\begingroup$ @DavidRicherby "more basic" was referring to the complexity of the polish-notation-mess-that-I-might-have-alerady-forgot-so-you-need-to-make-a-parser-to-even-get-anything-out-of-it. But, it is a shorter notation. Ikr, life's so short. $\endgroup$ – EKons Nov 28 '16 at 11:43

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