$$(\mathrm{S} 1)~(p \Rightarrow q) \vee(p \wedge(\sim q))$$ is a tautology

$$(\mathrm{S} 2)~((\sim p) \Rightarrow(\sim q)) \wedge((\sim p) \vee q)$$ is a contradiction.

Then

Consider the following statements:

P : I have fever

Q: I will not take medicine

$\mathrm{R}$ : I will take rest.

The statement "If I have fever, then I will take medicine and I will take rest" is equivalent to :

Among the statements :

$$(\mathrm{S} 1)~((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow(\mathrm{p} \Rightarrow \mathrm{r})$$

$$(\mathrm{S} 2)~((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow((\mathrm{p} \Rightarrow \mathrm{r}) \vee(\mathrm{q} \Rightarrow \mathrm{r}))$$

If $$p,q$$ and $$r$$ are three propositions, then which of the following combination of truth values of $$p,q$$ and $$r$$ makes the logical expression $$\left\{ {(p \vee q) \wedge \left( {( \sim p) \vee r} \right)} \right\} \to \left( {( \sim q) \vee r} \right)$$ false?