By checking their truth tables, I understand that these propositions are all the same proposition expressed differently:
$\forall x, \neg P(x) \leftrightarrow C(x) $ (1)
$\forall x, (\neg P(x) \implies C(x)) \wedge (C(x) \implies \neg P(x))$ (2)
$\forall x, (P(x) \vee C(x)) \wedge (\neg C(x) \vee \neg P(x))$(3)
$\forall x, (\neg P(x) \wedge C(x)) \vee (\neg C(x) \wedge P(x))(4)$
I understand how (2) and (3) can be obtained from (1) but I am having trouble obtaining (4) from (1). What would be the laws applied to (1) to obtain (4)?
$$\forall x, \neg P(x) \leftrightarrow C(x) $$ Using definition of iff, $$\forall x, (\neg P(x) \rightarrow C(x)) \land (C(x) \rightarrow \neg P(x))$$ Using definition of if $$\forall x, ( P(x) \lor C(x)) \land (\neg C(x) \lor \neg P(x))$$ Using De Morgan's Law $$\forall x, \neg(( P(x) \lor C(x)) \lor \neg(\neg C(x) \lor \neg P(x)))$$ $$\forall x, (\neg P(x) \land \neg C(x)) \lor (C(x) \land P(x))$$ Using $A\lor(B\land C) \rightarrow (A \lor B) \land (A \lor C)$ $$\forall x, ((\neg P(x) \land \neg C(x)) \lor C(x)) \land ((\neg P(x) \land \neg C(x)) \lor P(x))$$
Using $A\lor(B\land C) \rightarrow (A \lor B) \land (A \lor C)$ $$\forall x, ((\neg P(x) \lor C(x)) \land (\neg C(x) \lor C(x)) \land ((\neg P(x) \lor P(x)) \land (\neg C(x) \lor P(x))$$ $$\forall x, (\neg P(x) \lor C(x)) \land (\neg C(x) \lor P(x))$$ Using Demorgan's Law $$\forall x, \neg(\neg P(x) \lor C(x)) \lor \neg(\neg C(x) \lor P(x))$$ Using Demorgan's Law $$\forall x, (P(x) \land \neg C(x)) \lor ( C(x) \land \neg P(x))$$ $$\forall x, (\neg P(x) \land C(x)) \lor ( \neg C(x) \land P(x))$$