A neat identity I happened upon: the negation of an equivalence is equivalent to an exclusive or.
Equivalence: A <--> B
Or (A --> B) & (B --> A)
Or (~A v B) & (~B v A)
Negating this last
~[(~A v B) & (~B v A)]
and then applying DeMorgan's law, we get:
~(~A v B) v ~(~B v A)
Apply DeMorgan again to the parenthesized symbols for:
(A&~B) v (B&~A)
This last offers only mutually exclusive options.
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Thursday, February 13, 2020
Trivial but neat
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