I am attempting to simplify the following expression: (!A && !B) || (!B && !C) || (C && !A). It should simplify to only two terms: (!A and C) || (!B and !C)

I have tried applying almost all of the laws and tried different combinations of factoring to see if anything will reduce but it does not lead to the required answer.

2 Answers

-1
DigitalJedi On

Here you go:

(! a && c) || (! b && ! c)

from: https://www.dcode.fr/boolean-expressions-calculator

EDIT: Sorry I didnt know you were looking for HOT to solve it.

In this case I would suggest using a truth table. This will probably help you alot:

https://www.wolframalpha.com/input/?i=(!A+%26%26+!B)+%7C%7C+(!B+%26%26+!C)+%7C%7C+(C+%26%26+!A)

0
Joseph Sible On

You know C || !C is true by the law of excluded middle. Combine that with !A && !B to get !A && !B && (C || !C). Distribute to get (!A && !B && C) || (!A && !B && !C). Substitute this back into the original expression to get (!A && !B && C) || (!A && !B && !C) || (!B && !C) || (C && !A).

For any expressions P and Q, if P -> Q, then P || Q is the same as just Q. You should see that !A && !B && C implies C && !A, so you can remove the former. Same for !A && !B && !C and !B && !C. You can now easily reorder the remainder into (!A && C) || (!B && !C).