Kenneth Rosen Edition 7 Exercise 1.3 Question 14 (Page No. 35)

Question

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

Answer 1

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

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. 

Answer 2

It is contingency.

 

Answer 3

Try to make hypothesis True and conclusion False. If it's possible then compound proposition is not a tautology. 

If p is false and q is true then we can make hypothesis true and conclusion false. Hence, given compound proposition is not a tautology.