Since
¬(P↔Q)≡¬((P→Q)∧(Q→P))≡¬((¬P∨Q)∧(¬Q∨P))≡¬(¬P∨Q)∨¬(¬Q∨P)≡(P∧¬Q)∨(Q∧¬P)
Then
¬(P↔Q)↔(P∧¬Q)∨(¬P∧Q)
Since
(P∧¬Q)∨(¬P∧Q)≡(P∨¬P)∧(¬Q∨¬P)∧(P∨Q)∧(¬Q∨Q)≡1∧(¬Q∨¬P)∧(P∨Q)∧1≡(¬Q∨¬P)∧(P∨Q)≡(P∨Q)∧¬(P∧Q)
Then
(P∧¬Q)∨(¬P∧Q)↔(P∨Q)∧¬(P∧Q)
But then
¬(P↔Q)↔(P∨Q)∧¬(P∧Q)↔(P∧¬Q)∨(¬P∧Q)
Q. E. D.
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