# UGC NET NOV 2017 PAPER III Q-57

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57. Let P, Q, R and S be Propositions. Assume that the equivalences P ⇔ (Q ∨ ¬ Q) and Q ⇔ R hold. Then the truth value of the formula (P ∧ Q) ⇒ ((P ∧ R) ∨ S) is always :

(1) True

(2) False

(3) Same as truth table of Q

(4) Same as truth table of S

recategorized

P ⇔ (Q ∨ ¬ Q) "P should be true because RHS will be TRUE always "

Q ⇔ R "when Q is true R is true" and  "when Q is false R is false"

$(P ∧ Q) ⇒ ((P ∧ R) ∨ S)$

there can be only 2 cases (value of S doesn't matter)

1) P = True, Q = True and R = True

$(T ∧ T) ⇒ ((T ∧ T) ∨ S)$

so this case is True

2) P =True, Q = R = False

$(T ∧ F) ⇒ ((P ∧ R) ∨ S)$

In case implication if the premises is false then whole statement is true

this case is also true

The given expression is True in both the cases.

selected
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What if R is false in your ist point ?

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59. Consider the following two well-formed formulas in prepositional logic. F1 : P ⇒ ¬ P F2 : (P ⇒ ¬ P) ∨ (¬ P ⇒ P) Which of the following statements is correct ? (1) F1 is Satisfiable, F2 is valid (2) F1 is unsatisfiable, F2 is Satisfiable (3) F1 is unsatisfiable, F2 is valid (4) F1 and F2 both are Satisfiable