Answer
$(\neg q\wedge(p\rightarrow q))\rightarrow\neg p $ is a tautology.
Work Step by Step
$\left[\begin{array}{ll}
p & q\\
\hline & \\
T & T\\
T & F\\
F & T\\
F & F
\end{array}\right. \left.\begin{array}{lll}
\neg q & p\rightarrow q & \neg q\wedge(p\rightarrow q)\\
\hline & & \\
F & T & F\\
T & F & F\\
F & T & F\\
T & T & T
\end{array}\right]\left[\begin{array}{cc}
\neg p & (\neg q\wedge(p\rightarrow q))\rightarrow\neg p\\
\hline & \\
F & T\\
F & T\\
T & T\\
T & T
\end{array}\right]$
$(\neg q\wedge(p\rightarrow q))\rightarrow\neg p $ has all T values in its column,
so it is a tautology.
