(~ ABC) + (A ~ B ~ C) + (AB ~ C) + (ABC)

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(~ ABC) + (A ~ B ~ C) + (AB ~ C) + (ABC)

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#1 by (6 votes)

2

I'm trying to solve this formula:

(~ABC)+(A~B~C)+(AB~C)+(ABC)
(~ABC)+(A~B~C)+AB
(~ABC)+A(B+~B~C)

But I do not know how to get out of this last part. I know the end result has to be a~c + bc . But I do not know how to get there.

booleano

asked by anonymous 20.10.2018 / 06:36

1 answer

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score 6 · Accepted Answer

Let's start with this:

(~ABC)+(A~B~C)+(AB~C)+(ABC)

Let's reorder the expressions:

(~ABC)+(ABC)+(A~B~C)+(AB~C)

Let's put BC and A~C in evidence:

BC(~A+A)+(A~C)(~B+B)

Every expression in the form X+~X is true. Logo:

BC+(A~C)

Note that its original expression (~ABC)+(A~B~C)+(AB~C)+(ABC) has an interesting property: It says exactly which are the four rows of the truth table in which the expression is true, since each subexpression in parentheses has all three variables A , B , and C exactly once each.