Question

I need to show that the expression:

~(A XOR B)

is equivilant to

(~A XOR B)

using boolean algebra.

I really have no idea how to start, any help would be appreciated.

Answer 1

In order to show that two logical expressions are equivalent, you may proceed in two different ways.

1. Write a truth table for each of the expressions and then, if the resulting functional truth values are the same, then the expressions are equivalent;

2. Equivalence is the same as implication in both directions;

   A <=> B is equivalent to (A => B) AND (B => A)

   So, what you need is try to get (~A xor B) from ~(A xor B) and vice versa.

   • ~(A xor B) =
   • by definition of xor + negation = ~ ( (~A and B) or (A and ~B) ) =
   • by De Morgan law = ~ (~A and B) and ~(A and ~B) =
   • by De Morgan law again = (A or ~B) and (~A or B) =
   • by applying distributive law = (A and ~A) or (A and B) or (~B and ~A) or (~B and B) =
   • ignore contradictions = (A and B) or (~B and ~A) =
   • apply definition of xor in another direction = ~A xor B.

   the end

   The same procedure must be done in the other direction ( get ~(A xor B) from (~A xor B) ). Then the proof will be complete.