JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year DISCRETE STRUCTURES Previous Question Paper (March - 2022 - R18)

To determine if the propositional function ((p→q)∧(r→s)∧(p∨r))→(q∨s) is a tautology, we construct a truth table or use logical equivalence. Using truth table analysis, we evaluate the expression for all 16 combinations of p, q, r, and s.

This expression is a form of the constructive dilemma rule in propositional logic. If we assume the antecedent ((p→q)∧(r→s)∧(p∨r)) is true, then (p→q) is true, (r→s) is true, and (p∨r) is true. If p is true, then q must be true. If r is true, then s must be true. Since (p∨r) is true, at least one of p or r must be true, which forces at least one of q or s to be true, making (q∨s) true. Since the implication holds true for all possible truth values of the variables, the expression is a tautology.

To demonstrate that S is a valid inference, we apply rules of inference to the given premises:

1. P → ~Q (Premise 1)
2. Q ∨ R (Premise 2)
3. ~S → P (Premise 3)
4. ~R (Premise 4)
5. Q (From 2 and 4 using Disjunctive Syllogism)
6. ~P (From 1 and 5 using Modus Tollens)
7. S (From 3 and 6 using Modus Tollens)

Since we have derived S from the given premises, the inference is valid.