Solution to Show that p ↔ q and (p ∧ q) ∨ (¬p ∧ ¬q) are logically … - Sikademy
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Archangel Macsika

Show that p ↔ q and (p ∧ q) ∨ (¬p ∧ ¬q) are logically equivalent.

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1)First method

Lets make truth value table of both statements. Lets denote p ↔ q as S1, (p ∧ q) as A, (¬p ∧ ¬q) as B, (p ∧ q) ∨ (¬p ∧ ¬q) as S2.

| p | q | A | B | S2 | S1 |

| T | T | T | F | T | T |

| T | F | F | F | F | F |

| F | T | F | F | F | F |

| F | F | F | T | T | T |

2)Second method

p ↔ q = (p \to q) \land (q \to p) = (\lnot p \lor q) \land (\lnot q \lor p) =

=(\lnot p \land( \lnot q \lor p)) \lor (q \land (\lnot q \lor p))=((\lnot p \land \lnot q)\lor(\lnot p \land p)) \lor ((q \land \lnot q)\lor(p \land q))

(\lnot p \land p) is always false, as well as (q \land \lnot q), since (x \lor False) is true only if x is true, (x \lor False) =x. Using this we can remove (\lnot p \land p) and (\lnot q \land q), which gives as

p ↔ q =(p \land q) \lor (\lnot p \land \lnot q)

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