Use truth tables to test the validity of the argument.
-$$\begin{array} { l }
\mathrm { p } \rightarrow \mathrm { q } \
\frac { \sim \mathrm { p } } { \therefore \sim \mathrm { q } }
\end{array}$$
A) Valid
B) Invalid

Question

Use truth tables to test the validity of the argument.
-$$\begin{array} { l } 
\mathrm { p } \rightarrow \mathrm { q } \
\frac { \sim \mathrm { p } } { \therefore \sim \mathrm { q } }
\end{array}$$
A) Valid
B) Invalid

Quizplus · Accepted Answer

The argument is invalid because the premise $p \rightarrow q$ does not logically lead to the conclusion $\sim q$ from the premise $\sim p$. In logical terms, just because $p$ implies $q$, it does not follow that the negation of $p$ ($\sim p$) implies the negation of $q$ ($\sim q$). This is a misunderstanding of the implication relationship in logic.

Use Truth Tables to Test the Validity of the Argument p→q∼p∴∼q\begin{array} { l } \mathrm { p } \rightarrow \mathrm { q } \\\frac { \sim \mathrm { p } } { \therefore \sim \mathrm { q } }\end{array}p→q∴∼q∼p​​

Use Truth Tables to Test the Validity of the Argument $\begin{array} { l } \mathrm { p } \rightarrow \mathrm { q } \\\frac { \sim \mathrm { p } } { \therefore \sim \mathrm { q } }\end{array}$