WUCT121 Discrete Mathematics Logic Tutorial Exercises Solutions

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(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
Question10(a) If x is a positive integer and x2 ≤ 3 then x = 1 .The proposition is True.2If x is a positive integer, then x ≤ 3 ⇒ x ≤ 3 .Now 3 ≈ 1. 7 and so x = 1.(b) ( ~ ( x > 1) ∨ ~ ( y ≤ 0)) ⇔ ~ (( x ≤ 1) ∧ ( y > 0)).The proposition is false. (You should have tried proving it using De Morgan’s Laws andfailed.)Now find values of x and y that make the statement false.Let x = 0 and y = 1.~ x > 1 ∨ ~ y ≤ 0 is True( ) ( )( x ≤ 1) ∧ ( y > 0)Thus, ~ (( x ≤ 1) ∧ ( y > 0))is also Trueand the proposition is False.is FalseQuestion11(a)~ ( x > 1) ⇒ ~ ( y ≤ 0)≡~ (~ ( x > 1)) ∨ ~ ( y ≤ 0)≡ ( x > 1) ∨ ~ ( y ≤ 0)≡ ( x > 1) ∨ ( y > 0)Implication LawDouble NegationNegation of ≤(b)( y ≤ 0 ) ⇒ ( x > 1 )≡~( y ≤ 0) ∨ ( x > 1)Implication≡ ( y > 0) ∨ ( x > 1) Negation of ≤Law .Question12~ ( ~ ( p ∨ q ) ∧ ~ q )≡~~( p ∨ q)∨ ~~ q≡ ( p ∨ q)∨ q≡ p ∨ q ∨ q≡ p ∨ qDe Morgan'sDouble NegationAssociativityIdempotent LawWUCT121 Logic Tutorial Exercises Solutions 7
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WUCT121 Discrete Mathematics Logic Tutorial Exercises Solutions

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