I can write this proposition in sequent notation:
$$(P\rightarrow Q)\rightarrow (\neg P \lor Q)$$
as this one in rule form (see here):
$$\frac{(P\rightarrow Q)}{(\neg P \lor Q)}$$
But how can I transform, for example, these into rule form?
$$(P\rightarrow Q)\leftrightarrow(\neg(P\land\neg Q))$$ $$((P\land Q)\rightarrow R)\rightarrow((P\land\neg R)\rightarrow\neg Q)$$ $$((P\lor Q)\rightarrow (P\land S))\rightarrow(\neg P\lor\neg Q)$$
Are there any specific rules to follow? I saw a similar example on wikipedia but I cannot follow it.
First, in sequent notation it would be:
$$(P \to Q) \vdash (\neg P \lor Q)$$
rather than
$$(P \to Q) \to (\neg P \lor Q)$$
The latter is a tautology and as such can be used as an axiom in an axiom system, but the latter is not a sequent.
Second, for
$$(P\rightarrow Q)\leftrightarrow(\neg(P\land\neg Q))$$
you would need two rules:
$$\frac{P\rightarrow Q}{\neg (P \land \neg Q)}$$
and
$$\frac{\neg (P \land \neg Q)}{P\rightarrow Q}$$
The
$$((P\land Q)\rightarrow R)\rightarrow((P\land\neg R)\rightarrow\neg Q)$$
is easy, as that just becomes:
$$\frac{(P\land Q)\rightarrow R}{(P\land\neg R)\rightarrow\neg Q}$$
The
$$((P\lor Q)\rightarrow (P\land S))\rightarrow(\neg P\lor\neg Q)$$
can be written as a rule
$$\frac{(P\lor Q)\rightarrow (P\land S)}{\neg P\lor\neg Q}$$
but please note that this is not a valid inference, and indeed
$$((P\lor Q)\rightarrow (P\land S))\rightarrow(\neg P\lor\neg Q)$$
is not a tautology!