The statement pattern $[(p \vee q) \wedge \sim p] \wedge(\sim q)$ is

\begin{array}{|c|c|c|c|c|c|c|} \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mathrm{p} & \mathrm{q} & \sim \mathrm{p} & \sim \mathrm{q} & \mathrm{p} \vee \mathrm{q} & (\mathrm{p} \vee \mathrm{q}) \wedge \sim \mathrm{p} & {[(\mathrm{p} \vee \mathrm{q}) \wedge \sim \mathrm{p}] \wedge \sim \mathrm{q}} \\ \hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\ \hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\ \hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\ \hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\ \hline \end{array} All entries in last column are F. $\therefore$ It is contradiction.