32 views
Determine whether $(\sim q \wedge (p \rightarrow q)) \rightarrow \sim p$ is a tautology.
edited | 32 views

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

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.
edited
0
question is whether $(\sim q \;\wedge (p \rightarrow q)) \rightarrow \sim p$ is a tautology or not. I think, you did small mistake.
0
where are the mistakes?
0
In question , It is $\sim q$ initially, you have taken $\sim p$
+1
Ohh i will correct it thank you so much

It is tautology.

0
In step 3: qp ->p

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

1
2