Question

Show that (p↔q) ⇔ (¬p↔¬q) without using the truth table.

Answer 1

Okay:) That's an inverse

Answer 2

I'm not 100% sure of these signs, just confirm for me that ¬ means negative?

Answer 3

Yup! ¬ means a negation.

Answer 4

Okay. Since p<->q and ¬p<->¬q we can assume ¬p<->¬q = ¬¬p<->¬¬q. Since a double negation has no logical effect, the inverse of the inverse is logically equivalent to the original conditional p<->q

Answer 5

Wait wait wait. I don't understand...

Answer 6

How do you show "¬p<->¬q ⇔ ¬¬p<->¬¬q" ?

Answer 7

Convert it all to and/or/not.

Answer 8

(¬p↔¬q)
⇔ (¬p→¬q) ∧ (¬q→¬p)
⇔ (p ∨ ¬q) ∧ (q ∨ ¬p) (Implication law)
⇔ [ (p ∨ ¬q) ∧ q ] ∨ [ (p ∨ ¬q) ∧¬p ] (Distributive law)
⇔ [ q ∧ (p ∨ ¬q) ] ∨ [ ¬p ∧ (p ∨ ¬q) ] (Commutative law)
⇔ [ (q ∧ p) ∨ (q ∧ ¬q) ] ∨ [ (¬p ∧ p) ∨ (¬p ∧¬q) ] (Distributive law)
⇔ (q ∧ p) ∨ (¬p ∧¬q) (Negation law)
⇔ [ (q ∧ p) ∨ ¬p ] ∧ [ (q ∧ p) ∨ ¬q) ] (Distributive law)
⇔ [ ¬p ∨ (q ∧ p) ] ∧ [ ¬q ∨ (q ∧ p) ] (Commutative law)
⇔ [ ¬p ∨ q ∧ (¬p ∨ p ) ] ∧ [ (¬q ∨ q) ∧ (¬q ∨ p) ] (Distributive law)
⇔ [ ¬p ∨ q ] ∧ [ (¬q ∨ p) ] (Negation law)
⇔ ( p → q ) ∧ (q → p) (Implication law)
⇔ p↔q
OMG :O
How come I couldn't do it... :(

Answer 9

Thanks!