Converting a truth table into a boolean expression?

Question

I'm trying to convert a truth table into a boolean expression, but it has two output values. I know that if you only have one output then you can take the expressions that produce a 1 and "add" them together, but I'm not quite sure how it works for two output values.
For further clarification, here are the first few lines...
input output
A B C D x y
0 0 0 0 0 0
0 0 0 1 0 0
0 0 1 0 1 1
0 0 1 1 1 1
0 1 0 0 0 0
0 1 0 1 0 1
0 1 1 0 1 1

Answer 1

A boolean expression MUST have one output. So if you are manipulating the x and y outputs to get a final expression we have to know how they are related ... do you add them, do you AND them, do you or them, xor.... etc



There is not enough information to complete this question. You cannot guess on what you do with those, it must be in the question.

Answer 2

The technique you're meant to be utilizing is "canonical sum of goods." All you do is look on the output column, and then convey mutually in elementary words the lines that go back real. then you definitely record the situations which convey about those truths through expressing each and each and every human being line through logical conjunction ( often times written like ' A·B ' similar as logical AND ) and then once you've finished all this, you're taking the fellow boolean words and connect all of them through dysjunction ( often times written like ' A+B ' similar as logical OR ). So 0 0 a million has a real output. We see that interior the inputs, both A and B are off. So the first expression is going: (A')·(B') this can be written in many diverse strategies like ' notA AND notB ' or ' ~A+~B ' or ' ¬A¬B ' yet they are all an identical. the precise syntax is determined by your instructor, or they'd no longer care that you take advantage of as long as you're consistent. The truthy line: a million 0 a million has a real output. sounds like A is on and B is off so we get: A·(B') Inotherwords: A and notB finally we only smush the (in our case only 2) real expressions mutually through dysjunction to get: (A')·(B') + A·(B') there's a thanks to more beneficial simplify the expression, even if it would no longer be needed.

Answer 3

each output value needs its own logical expression