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# 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 )

for xor.

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.

### 1 Answer

- 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?

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I do believe that A XOR B = NOT A XOR NOT B