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→q)∧(q→p)=(¬p∨q)∧(¬q∨p)=
=(¬p∧(¬q∨p))∨(q∧(¬q∨p))=((¬p∧¬q)∨(¬p∧p))∨((q∧¬q)∨(p∧q))
(¬p∧p) is always false, as well as (q∧¬q), since (x∨False) is true only if x is true, (x∨False)=x. Using this we can remove (¬p∧p) and (¬q∧q), which gives as
p↔q=(p∧q)∨(¬p∧¬q)
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