(S1) (p Rightarrow q) vee (p wedge (sim q)) i | vedclass

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(D) To check if $(S1)$ is a tautology,we construct the truth table for $(p \Rightarrow q) \vee (p \wedge (\sim q))$.Since all values in the final colu

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only $(S1)$ is correct

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(D) To check if $(S1)$ is a tautology,we construct the truth table for $(p \Rightarrow q) \vee (p \wedge (\sim q))$.Since all values in the final column are $T$,$(S1)$ is a tautology.To check if $(S2)$ is a contradiction,we construct the truth table for $((\sim p) \Rightarrow (\sim q)) \wedge ((\sim p) \vee q)$.Since the final column contains both $T$ and $F$,it is a contingency,not a contradiction.Therefore,only $(S1)$ is correct.

$(S1) (p \Rightarrow q) \vee (p \wedge (\sim q))$ is a tautology. $(S2) ((\sim p) \Rightarrow (\sim q)) \wedge ((\sim p) \vee q)$ is a contradiction. Then

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