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Determine whether $(\sim p \wedge (p \rightarrow q)) \rightarrow \sim q$ is a tautology.

$(\sim p \wedge (p \rightarrow q)) \rightarrow \sim q$

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.

edited

It is contingency.

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