WUCT121 Discrete Mathematics Logic Tutorial Exercises Solutions

(c) ( p ∧ q) ⇒ ( ~ r ∨ ( p ⇒ q))(p ∧ q) ⇒ (~r ∨ (p ⇒ q)Step 1 5* 2 4 3Place F under main connectiveF∧ must be T and ∨ must be F T F∧ p must be T and q must be T∨ ~r must be F and ⇒ must be FT T F F⇒ p must be T and q must be F T Fq cannot be both T and F , thus ( p q) ⇒ ( r ∨ ( p ⇒ q))tautologyQuestion8∧ ~ can only ever be true and is a(a)( )ity(b)p ∧ q ⇒ r ≡~( p ∧ q)∨ rp ⇒≡ (~ p∨~ q)∨ r≡~p∨~ q ∨ r( p ∨ q)≡~p ∨ p ∨ q≡ T ∨ q≡ T≡~p ∨ ( p ∨ q)Implication LawDe Morgan's LawAssociativImplication LawAssociativityNegation LawDominance LawQuestion9(a)LHS = ~(b)LHS =( p ⇒ q)≡~ (~ p ∨ q)≡~~p∧~ q≡ p ∧ ~ q= RHS≡~( p ∧ ~ q)( p ∧ ~ q)≡ (~ p∨~~ q)∨ r≡ (~ p ∨ q)∨ r≡~p ∨ ( q ∨ r)≡ p ⇒= RHS⇒ r( q ∨ r)∨ rImplication LawDe Morgan'sDouble NegationImplication LawDe Morgan'sDouble NegationAssociativityImplicationWUCT121 Logic Tutorial Exercises Solutions 6