If any one can help me with this problem i would be grateful, since i have no past experience with propositional calculus . I tried using the following laws:
$¬ (¬ P) \equiv P\\ (P ∨ Q) \equiv (¬ P → Q)$
Contrapositive law: $(P → Q) \equiv (¬ Q→¬P)$
De Morgan’s law: $¬ (P ∨ Q) \equiv (¬ P ∧ ¬ Q) \text{ and } ¬ (P ∧ Q) \equiv (¬ P ∨¬ Q)$
Commutative laws: $(P ∧ Q) \equiv (Q ∧ P) \text{ and } (P ∨ Q) ≡ (Q ∨ P)$
Associative law: $((P ∧ Q) ∧ R) \equiv (P ∧ (Q ∧ R))$
Associative law: $((P ∨ Q) ∨ R) \equiv (P ∨ (Q ∨ R))$
Distributive law: $P ∨ (Q ∧ R) \equiv (P ∨ Q) ∧ (P ∨ R)$
Distributive law: $P ∧ (Q ∨ R) \equiv (P ∧ Q) ∨ (P ∧ R)$
HINT
Since almost all your laws involve $\land$'s, $\lor$'s and $\neg$'s, but few involve $\rightarrow$, it makes sense to rewrite any $\rightarrow$ in terms of $\lor$'s and $\neg$'s, so that then you can just focus on the $\land$'s, $\lor$'s and $\neg$'s.
So, for example, you can take $P \rightarrow Q$ and say:
$$P \rightarrow Q \equiv \neg \neg P \rightarrow Q \equiv \neg P \lor Q$$
Notice that in the first step I used $P \equiv \neg \neg P$, but I only applied that to the component statement $P$, rather than the whole statement $P \rightarrow Q$ ... and that's perfectly ok! If you're new to this, that is something you may not have yet realized (in fact, maybe that's why you go stuck!).