Question

Consider the two statements:
$\left(S1\right):(p\to q)\vee (\sim q\to p)$ is a tautology.
$\left(S2\right):(p\wedge \sim q)\wedge (\sim p\vee q)$ is a fallacy.
Then

• A. only $\left(S1\right)$ is true.
• B. only $\left(S2\right)$ is true.
• C. both $\left(S1\right)$ and $\left(S2\right)$ are true.
• D. both $\left(S1\right)$ and $\left(S2\right)$ are false.

Answer 1

The correct option is C both $\left(S1\right)$ and $\left(S2\right)$ are true.

For $S1:$
$\begin{array}{ccccccc}p& q& \sim p& \sim q& p\to q& \sim q\to p& (p\to q)\vee (\sim q\to p)\\ T& T& F& F& T& T& T\\ T& F& F& T& F& T& T\\ F& T& T& F& T& T& T\\ F& F& T& T& T& F& T\end{array}$
Hence, $S1$ is a tautology.

For $S2:$
$\begin{array}{ccccccc}p& q& \sim p& \sim q& p\wedge \sim q& \sim p\vee q& (p\wedge \sim q)\wedge (\sim p\vee q)\\ T& T& F& F& F& T& F\\ T& F& F& T& T& F& F\\ F& T& T& F& F& T& F\\ F& F& T& T& F& T& F\end{array}$
Hence, $S2$ is a fallacy.

Hence, both $S1$ and $S2$ are true.