$(\sim p \wedge (p \rightarrow q)) \rightarrow \sim q$
We know that $p \rightarrow q \equiv \sim p\vee q$
$(\sim p \wedge (\sim p\vee q)) \rightarrow \sim q$
Convert $\wedge\equiv\cdot ,\vee\equiv +$
Now, $\overline{p}.(\overline{p}+q)\rightarrow \overline{q}$
$(\overline{p}.\overline{p}+\overline{p}.q)\rightarrow \overline{q}$
$(\overline{p}+\overline{p}.q)\rightarrow \overline{q}$
$(\overline{p}(1+q))\rightarrow \overline{q}$
$\overline{p}\rightarrow \overline{q}$
$\overline{\overline{p}}+\overline{q}$
$p+\overline{q}$
$\overline{q}+p$
Now we can write $\sim q \vee p\equiv q\rightarrow p$
Now we can make a truth table
$p$ |
$q$ |
$\sim q$ |
$\sim q \vee p$ |
$q\rightarrow q$ |
$T$ |
$T$ |
$F$ |
$T$ |
$T$ |
$T$ |
$F$ |
$T$ |
$T$ |
$T$ |
$F$ |
$T$ |
$F$ |
$F$ |
$F$ |
$F$ |
$F$ |
$T$ |
$T$ |
$T$ |
So,this is not
a tautology.