logic: How could I form something like De Morgan for xor?
How could I get something like De Morgan ( ~(A and B) <=> ~A or ~B )
How to get there, and how to get it in a simple form.
Any help is greatly appreciated. Thanks.
De Morgan for xor meaning rules to express
~(A xor B) as something else, and same for
~A xor ~B.
- Anonymous1 decade agoFavorite Answer
I dont get your question
The Boolean "XOR" is equivalent to the logical "Negation of a Biconditional"... these are rules of implications
DeMorgans Rule is a rule of replacement. Completely different entity.
Rules of Implications are designed to evaluate two statements for a logical conclusion from them.
Rules of Replacement are designed to rearrange a given statement for same meaning but a different way of saying it.
If you are asking for a means of replacing a negated biconditional (xor) for a new set of equivalent statements... then there is a way.
A ≡ B :: (A • B) \/ (~A • ~B)
~(A ≡ B) :: ~[ (A • B) \/ (~A • ~B) ]
A XOR B = NOT [(A AND B) OR (NOT A AND NOT B]
Hows that work for you?
I do believe that A XOR B = NOT A XOR NOT B