Kenneth Rosen Edition 7 Exercise 1.3 Question 15 (Page No. 34)

Question

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

Answer 1

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

We know that $p \rightarrow q \equiv \sim p\vee q$

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

Convert $\wedge\equiv\cdot ,\vee\equiv +$

Now,   $\overline{q}.(\overline{p}+q)\rightarrow \overline{p}$

        $(\overline{q}.\overline{p}+\overline{q}.q)\rightarrow \overline{p}$

        $(\overline{p}\cdot\overline{q}+0)\rightarrow \overline{p}$

       $\overline{p}\cdot\overline{q}\rightarrow \overline{p}$

       $\overline{\overline{p}\cdot\overline{q}}+\overline{p}$

      $p+q+\overline{p}=p+\overline{p}+q=1+q=1$

Now we can write    $p \vee \sim p\equiv \sim q\vee p\equiv T$

  So, this is Tautology.

Comments

question is whether $(\sim q \;\wedge (p \rightarrow q)) \rightarrow \sim p  $ is a tautology or not. I think, you did small mistake.

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where are the mistakes?

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In question , It is $\sim q$ initially, you have taken $\sim p$

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Ohh i will correct it thank you so much

Answer 2

It is tautology.

Comments

In step 3: q`p` ->p`

In step 4:(q`p`)`+p` instead of  implication symbol