The statement $(\mathrm{p} \Rightarrow \mathrm{q}) \Leftrightarrow(\sim p \wedge q)$ is a

\begin{array}{c|c|c|c|c|c} \hline \mathrm{p} & \mathbf{q} & \mathbf{p} & \mathbf{p} \Rightarrow \mathbf{q} & \sim \mathbf{p}^{\wedge} \mathbf{q} & (\mathbf{p} \Rightarrow \mathbf{q}) \Leftrightarrow\left(\sim \mathbf{p}^{\wedge} \mathbf{q}\right) \\ \hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\ \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\ \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\ \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\ \hline \end{array} Hence, given statement is neither tautology nor contradiction.