show that p ↔ q and (p ∧ q) ∨ (¬p ∧ ¬q) are logically equivalent
Let us show that p↔qp ↔ qp↔q and (p∧q)∨(¬p∧¬q)(p ∧ q) ∨ (¬p ∧ ¬q)(p∧q)∨(¬p∧¬q) are logically equivalent. It follow that
p↔q=(p→q)∧(q→p)=(¬p∨q)∧(¬q∨p)=(¬p∧¬q)∨(¬p∧p)∨(q∧¬q)∨(q∧p)=(¬p∧¬q)∨F∨F∨(q∧p)=(¬p∧¬q)∨(p∧q)=(p∧q)∨(¬p∧¬q).p ↔ q=(p\to q)\land (q\to p) \\=(\neg p\lor q)\land (\neg q\lor p) \\=(\neg p\land\neg q)\lor(\neg p\land p)\lor(q\land\neg q)\lor(q\land p) \\=(\neg p\land\neg q)\lor F\lor F\lor(q\land p) \\=(\neg p\land\neg q)\lor (p\land q) \\= (p\land q)\lor (\neg p\land\neg q).p↔q=(p→q)∧(q→p)=(¬p∨q)∧(¬q∨p)=(¬p∧¬q)∨(¬p∧p)∨(q∧¬q)∨(q∧p)=(¬p∧¬q)∨F∨F∨(q∧p)=(¬p∧¬q)∨(p∧q)=(p∧q)∨(¬p∧¬q).
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