Question

Given the following two statements

$\left(S_1\right):(q\vee p)\to (p\leftrightarrow ~q)$ is a tautology.

$\left(S_2\right):~q\wedge (~p\leftrightarrow q)$ is a fallacy. Then:

• A. only $\left(S_1\right)$ is correct.
• B. both $\left(S_1\right)$ and $\left(S_2\right)$ are correct.
• C. only $\left(S_2\right)$ is correct
• D. both $\left(S_1\right)$ and $\left(S_2\right)$ are not correct.

Answer 1

The correct option is D

both $\left(S_1\right)$ and $\left(S_2\right)$ are not correct.

Explanation for the correct option:

$\left(S_1\right):(q\vee p)\to (p\leftrightarrow ~q)$

$p$	$q$	$~q$	$q\vee p$	$p\leftrightarrow ~q$	$\left(q\vee p\right)\to \left(q\leftrightarrow ~q\right)$
$T$	$T$	$F$	$T$	$F$	$F$
$T$	$F$	$T$	$T$	$T$	$T$
$F$	$T$	$F$	$T$	$T$	$T$
$F$	$F$	$T$	$F$	$F$	$T$

$S_1$ is not correct

$\left(S_2\right):~q\wedge (~p\leftrightarrow q)$

$p$	$q$	$~q$	$~p$	$~p\leftrightarrow q$	$~q\wedge \left(~p\leftrightarrow q\right)$
$T$	$T$	$F$	$F$	$F$	$F$
$T$	$F$	$T$	$F$	$T$	$T$
$F$	$T$	$F$	$T$	$T$	$F$
$F$	$F$	$T$	$T$	$F$	$F$

$S_2$ is not correct.

Hence, option(D) is correct.