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(\mathrm{S} 1)(p \Rightarrow q) \vee(p \wedge(\sim q))  is a tautology

(\mathrm{S} 2)((\sim p) \Rightarrow(\sim q)) \wedge((\sim p) \vee q)   is a contradiction.
Then

Option: 1

both (S1) and (S2) are correct


Option: 2

only ( S1) is correct
 


Option: 3

only (S2) is correct


Option: 4

both (S1) and (S2) are wrong


Answers (1)

best_answer

S_1 :(P\Rightarrow q)V(P\Lambda (\sim Q))

P q P\Rightarrowq \sim q P\Lambda \simq (P \Rightarrow q) V (P\wedge  \simq)
T     T     T F F T
T F F T T T
F T T F F T
F F T T F T

 S1 is a tautology

\mathrm{S}_2:((\sim \mathrm{P}) \Rightarrow(\sim \mathrm{q})) \Lambda((\sim \mathrm{P}) \mathrm{Vq})

  \simP          \simq       \simP\Rightarrow\simq \simP v q     ((\simP) \Rightarrow (\simq)) \Lambda (\simP) vq)
F F T T T
F T T F F
T F F T F
T T T T T

S2 is not a contradiction

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manish

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