Construct a truth table for the statement.
-$w \vee ( w \wedge \sim w )$
A)
$\begin{array}{ll}
\mathrm{W} & \mathrm{W} \vee(\mathrm{w} \wedge \sim \mathrm{w}) \
\hline \mathrm{T} & \mathrm{F} \
\mathrm{F} & \mathrm{F}
\end{array}$

B)
$\begin{array}{lc}
\mathrm{W} & \mathrm{W} \vee(\mathrm{w} \wedge \sim \mathrm{W}) \
\hline \mathrm{T} & \mathrm{T} \
\mathrm{F} & \mathrm{F}
\end{array}$

C)
$\begin{array}{lc}
\mathrm{W} & \mathrm{W} \vee(\mathrm{W} \wedge \sim \mathrm{W}) \
\hline \mathrm{T} & \mathrm{T} \
\mathrm{F} & \mathrm{T}
\end{array}$

D)
$\begin{array}{lc}
\mathrm{W} & \mathrm{W} \vee(\mathrm{W} \wedge \sim \mathrm{W}) \
\mathrm{T} & \mathrm{F} \
\mathrm{F} & \mathrm{T}
\end{array}$

Question

Construct a truth table for the statement.
-$w \vee ( w \wedge \sim w )$
A)
$\begin{array}{ll}
\mathrm{W} & \mathrm{W} \vee(\mathrm{w} \wedge \sim \mathrm{w}) \
\hline \mathrm{T} & \mathrm{F} \
\mathrm{F} & \mathrm{F}
\end{array}$

B)
$\begin{array}{lc}
\mathrm{W} & \mathrm{W} \vee(\mathrm{w} \wedge \sim \mathrm{W}) \
\hline \mathrm{T} & \mathrm{T} \
\mathrm{F} & \mathrm{F}
\end{array}$

C)
$\begin{array}{lc}
\mathrm{W} & \mathrm{W} \vee(\mathrm{W} \wedge \sim \mathrm{W}) \
\hline \mathrm{T} & \mathrm{T} \
\mathrm{F} & \mathrm{T}
\end{array}$

D)
$\begin{array}{lc}
\mathrm{W} & \mathrm{W} \vee(\mathrm{W} \wedge \sim \mathrm{W}) \
\mathrm{T} & \mathrm{F} \
\mathrm{F} & \mathrm{T}
\end{array}$

Quizplus · Accepted Answer

The expression $w \wedge \sim w$ is always false because $w$ and $\sim w$ (not $w$) are opposites and cannot both be true at the same time. Therefore, $w \vee (w \wedge \sim w)$ simplifies to $w \vee F$, which is just $w$. So, if $w$ is true, the whole expression is true; if $w$ is false, the whole expression is false.

Construct a Truth Table for the Statement w∨(w∧∼w)w \vee ( w \wedge \sim w )w∨(w∧∼w)

Construct a Truth Table for the Statement $w \vee ( w \wedge \sim w )$