Solution:
¬(P⇔Q)⇔¬((P→Q)∧(Q→P))⇔¬((P∨¬Q)∧(Q∨¬P))⇔¬(((P∨¬Q)∧Q)∨((P∨¬Q)∧¬P))⇔¬((P∧Q)∨(¬Q∧Q)∨((P∧¬P)∨(¬Q∧¬P))⇔¬((P∧Q)∨¬(P∨Q))⇔¬(P∧Q)∧¬(¬(P∨Q))⇔(P∨Q)∧¬(P∧Q)…(1)
Also,
¬(P⇔Q)⇔¬((P→Q)∧(Q→P))⇔¬((P∨¬Q)∧(Q∨¬P))⇔(¬P∧Q)∨(¬Q∧P)…(2)
Therefore the result follows from (1) and (2).
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