Logical proof using equivalencies

Question

I am trying to do a question which goes like this:

Show that (P ∨((P∧( P ∨~Q)) ∧~Q)) ∧(Q →(Q∧~P)) ≡ (P ∧~Q), using the logical equivalence.

So this is how i went about doing it:

(Q →(Q∧~P)) = (~Q ∨ (Q ^ ~P)) using implication equivalence

(P∧( P ∨~Q) ∧~Q) = (P^P) ∨ (P^~Q) ^ ~Q using distributive law

(P^P) ∨ (P^~Q) ^ ~Q = p ∨ p^(~Q^~Q) using idempotent law and associative law

p^(~Q^~Q) = P^~Q using idempotent law

(~Q ∨ (Q ^ ~P)) = (~Q ∨ Q) ^ (~Q ∨~P) using distributive law

(~Q ∨ Q) ^ (~Q ∨~P) = t ^ (~Q ∨ ~P) using negation property

t ^ (~Q ∨ ~P) = identity property

After this i have problem solving it. My equation stacks up like this:

(P ∨ p ∨ (P^~Q) ^ (~Q ∨ ~P)

Now i can combine P ∨ P tom make it P using idempotent law

So now it becomes like this:

P ∨ (P^~Q) ^ (~Q ∨ ~P) // i can still apply distributive law but it leads me to nowhere. I believe i have messed up something using the laws.

SE community please point me in the right direction.

Answer 1

The first conjunct of the LHS is :

$P ∨ [(P ∧ ( P ∨ \sim Q)) ∧ \sim Q]$

i.e.

$P ∨ [(P ∧ \sim Q) ∧ ( P ∨ \sim Q)]$.

By Distributivity:

$$(P ∨ (P ∧ \sim Q)) ∧ (P ∨ ( P ∨ \sim Q)) \equiv (P \lor P) \land (P \lor \sim Q) \land (P \lor \sim Q) \equiv$$

$$P \land (P \lor \sim Q).$$

Now, considering also the second conjunct of the LHS, that amounts to : $(\sim P ∨ \sim Q)$ we have :

$$P \land (P \lor \sim Q) \land (\sim P ∨ \sim Q) \equiv P \land [(P \land \sim P) \lor \sim Q] \equiv$$

$$ (P \land \sim Q).$$