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I have a truth table as follows

X Y Z A B C

0 0 0 0 0 1

0 0 1 0 1 0

0 1 0 1 0 1

0 1 1 1 0 0

1 0 0 0 1 1

1 0 1 1 0 0

1 1 0 1 0 1

1 1 1 1 1 0

where x, y and z are inputs and a, b and c are outputs. Normally, simplifying this into an expression would be easy with one output. I would just use k-maps and get the simplified Boolean expression from it. But with three outputs, I don't know how.

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  • $\begingroup$ What are you trying to do with the outputs, are you trying to construct logic gates ? $\endgroup$
    – asddf
    Oct 11, 2017 at 21:34
  • $\begingroup$ I need to get the simplified boolean expression for the outputs so I can draw a circuit diagram $\endgroup$
    – Shoaib Ahmed
    Oct 11, 2017 at 21:41
  • $\begingroup$ with 3 outputs you need 3 expressions. $\endgroup$
    – asddf
    Oct 11, 2017 at 21:43
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    $\begingroup$ So would I have to make 3 k-maps, one for each output? $\endgroup$
    – Shoaib Ahmed
    Oct 11, 2017 at 21:50
  • $\begingroup$ yes or you can just take the dnf. $\endgroup$
    – asddf
    Oct 11, 2017 at 21:53

1 Answer 1

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It's pretty clear by inspection that $c = \neg z$, so we can get that out of the way.

The simplest way I can see to specify $a$ is that $a = y \vee (x \wedge z)$. Again by inspection.

$b$ has no simple expression that I can see; you're not doing much better than just saying "x and not y and not z, or z and not y and not x, or x and y and z".

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