Construct a truth table for the statement.
-$\sim p \wedge \sim s$
A)
$\begin{array}{ccc}
\mathrm{p} & \mathrm{s}(\sim \mathrm{p} \wedge \sim \mathrm{s}) \
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} \
\mathrm{T} & \mathrm{F} & \mathrm{F} \
\mathrm{F} & \mathrm{T} & \mathrm{F} \
\mathrm{F} & \mathrm{F} & \mathrm{F}
\end{array}$
B)
$\begin{array}{ccc}
\mathrm{P} & \mathrm{s}(\sim \mathrm{P} \wedge \sim \mathrm{s}) \
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} \
\mathrm{T} & \mathrm{F} & \mathrm{T} \
\mathrm{F} & \mathrm{T} & \mathrm{T} \
\mathrm{F} & \mathrm{F} & \mathrm{T}
\end{array}$
C)
$\begin{array}{ccc}
\mathrm{p} & \mathrm{s} & (\sim \mathrm{p} \wedge \sim \mathrm{s}) \
\hline \mathrm{T} & \mathrm{T} & \mathrm{T} \
\mathrm{T} & \mathrm{F} & \mathrm{F} \
\mathrm{F} & \mathrm{T} & \mathrm{F} \
\mathrm{F} & \mathrm{F} & \mathrm{T}
\end{array}$
D)
$\begin{array}{ccc}
\mathrm{p} & \mathrm{s} & (\sim \mathrm{p} \wedge \sim \mathrm{s}) \
\hline \mathrm{T} & \mathrm{T} & \mathrm{F} \
\mathrm{T} & \mathrm{F} & \mathrm{F} \
\mathrm{F} & \mathrm{T} & \mathrm{F} \
\mathrm{F} & \mathrm{F} & \mathrm{T}
\end{array}$

Question

Quizplus · Accepted Answer

The correct truth table for $\sim p \wedge \sim s$ shows that the statement is true only when both $p$ and $s$ are false. This is because the negation ($\sim$) flips the truth value of each variable, and the conjunction ($\wedge$) requires both conditions to be true for the overall statement to be true.

Construct a Truth Table for the Statement ∼p∧∼s\sim p \wedge \sim s∼p∧∼s

Construct a Truth Table for the Statement $\sim p \wedge \sim s$