$(S1) (p \Rightarrow q) \vee (p \wedge (\sim q))$ is a tautology. $(S2) ((\sim p) \Rightarrow (\sim q)) \wedge ((\sim p) \vee q)$ is a contradiction. Then
- Aonly $(S2)$ is correct
- Bboth $(S1)$ and $(S2)$ are correct
- Cboth $(S1)$ and $(S2)$ are wrong
- Donly $(S1)$ is correct
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Similar Questions
If $(\sim p \wedge q) \rightarrow r$ is false,then the truth values of $p, q, r$ are respectively:
Consider the following three statements:
$(A)$ If $3+3=7$ then $4+3=8$.
$(B)$ If $5+3=8$ then earth is flat.
$(C)$ If both $(A)$ and $(B)$ are true then $5+6=17$.
Which of the following statements is correct?
$(A)$ If $3+3=7$ then $4+3=8$.
$(B)$ If $5+3=8$ then earth is flat.
$(C)$ If both $(A)$ and $(B)$ are true then $5+6=17$.
Which of the following statements is correct?
The statement pattern $[(p \land q) \to (\sim p \lor r)] \lor [(\sim p \lor r) \to (p \land q)]$ is
$p$: If $7$ is an odd number,then $7$ is divisible by $2$.
$q$: If $7$ is a prime number,then $7$ is an odd number.
If $V_1$ and $V_2$ are the respective truth values of the contrapositive of $p$ and $q$,then $(V_1, V_2) \equiv$
$q$: If $7$ is a prime number,then $7$ is an odd number.
If $V_1$ and $V_2$ are the respective truth values of the contrapositive of $p$ and $q$,then $(V_1, V_2) \equiv$
Which of the following is not logically equivalent to the proposition: "$A$ real number is either rational or irrational"?
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