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GO Classes DPP | GATE CS | Discrete Mathematics| Propositional Logic

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If $p, q$ and $r$ are three propositions, then which of the following combination of truth values of $p, q$ and $r$ makes the logical expression $\{(p \vee q) \wedge((\sim p) \vee r)\} \rightarrow((\sim q) \vee r)$ false?

  1. $p=F, q=T, r=F$
  2. $p=T, q=T, r=F$
  3. $p=T, q=F, r=T$
  4. $p=T, q=F, r=F$

7 Answers

0 0 votes

{ (p ∨ q) ∧ (¬p ∨ r) → (¬q ∨ r)}

a. p = F, q = T, r = F

= { (F ∨ T) ∧ (T ∨ F) → (F ∨ F)}

= (T ∧ T) → F

= T → F 

= False

b.  p = T, q = T, r = F

= { (T ∨ T) ∧ (F ∨ F) → (F ∨ F)}

= (T ∧ F) → F

= F → F 

= True

c. p = T, q = F, r = T

= { (T ∨ F) ∧ (F ∨ T) → (T ∨ T)}

= (T ∧ T) → T

= T → T

= True

d. p = T, q = F, r = F

= { (T ∨ F) ∧ (F ∨ F) → (T ∨ F)}

= (T ∧ F) → T

= F → T

= True

Answer: A

0 0 votes

In propositional logic, the statement P-->Q  is false only when P is True and Q is False. Because only Option A follows the True -->False pattern.

Answer : A

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