Using rules in logic, prove the following : (i) p ↔ q ≡ ~ (p ∧

1.	Using rules in logic, prove the following : (i) p ↔ q ≡ ~ (p ∧ ~q) ∧ ~(q ∧ ~p)(ii) ~p ∧ q ≡ (p ∨ q) ∧ ~p(iii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Answer» (i) p ↔ q ≡ ~ (p ∧ ~q) ∧ ~(q ∧ ~p) By the rules of negation of biconditional, ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p) ∴ ~ [(p ∧ ~ q) ∨ (q ∧ ~p)] ≡ p ↔ q ∴ ~(p ∧ ~q) ∧ ~(q ∧ ~p) ≡ p ↔ q …(Negation of disjunction) ≡ p ↔ q ≡ ~(p ∧ ~ q) ∧ ~ (q ∧ ~p). (ii) ~p ∧ q ≡ (p ∨ q) ∧ ~p (p ∨ q) ∧ ~ p ≡ (p ∧ ~p) ∨ (q ∧ ~p) … (Distributive Law) ≡ F ∨ (q ∧ ~p) … (Complement Law) ≡ q ∧ ~ p … (Identity Law) ≡ ~p ∧ q …(Commutative Law) ∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p. (iii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p ~ (p ∨ q) ∨ (~p ∧ q) ≡ (~p ∧ ~q) ∨ (~p ∧ q) … (Negation of disjunction) ≡ ~p ∧ (~q ∨ q) … (Distributive Law) ≡ ~ p ∧ T … (Complement Law) ≡ ~ p … (Identity Law) ∴ ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.

Using rules in logic, prove the following : (i) p ↔ q ≡ ~ (p ∧ ~q) ∧ ~(q ∧ ~p)(ii) ~p ∧ q ≡ (p ∨ q) ∧ ~p(iii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p

Answer»

(i) p ↔ q ≡ ~ (p ∧ ~q) ∧ ~(q ∧ ~p)

By the rules of negation of biconditional,

~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)

∴ ~ [(p ∧ ~ q) ∨ (q ∧ ~p)] ≡ p ↔ q

∴ ~(p ∧ ~q) ∧ ~(q ∧ ~p) ≡ p ↔ q …(Negation of disjunction)

≡ p ↔ q ≡ ~(p ∧ ~ q) ∧ ~ (q ∧ ~p).

(ii) ~p ∧ q ≡ (p ∨ q) ∧ ~p

(p ∨ q) ∧ ~ p

≡ (p ∧ ~p) ∨ (q ∧ ~p) … (Distributive Law)

≡ F ∨ (q ∧ ~p) … (Complement Law)

≡ q ∧ ~ p … (Identity Law)

≡ ~p ∧ q …(Commutative Law)

∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

(iii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p

~ (p ∨ q) ∨ (~p ∧ q)

≡ (~p ∧ ~q) ∨ (~p ∧ q) … (Negation of disjunction)

≡ ~p ∧ (~q ∨ q) … (Distributive Law)

≡ ~ p ∧ T … (Complement Law)

≡ ~ p … (Identity Law)

∴ ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.