Logical Equivalence for $p \lor q$ I have to prove that
$$p \vee q \equiv (p\wedge q) \vee (\neg p\wedge q) \vee (p\wedge \neg q)$$
Based on the truth table, they are equivalent, but I couldn't figure out how to use logic statements to prove they are equivalent. I have tried many ways but they all go weird.
$(p\wedge q) \vee (\neg p\wedge q) \vee (p\wedge \neg q)$
$\equiv (p\wedge q) \vee ((\neg p\wedge q)\vee p) \wedge ((\neg p\wedge q)\vee \neg q)$
$\equiv (p\wedge q) \vee ((T \wedge (q\vee p)) \wedge (T\wedge \neg(p \wedge q))$
$\equiv (p\wedge q) \vee (q\vee p) \wedge \neg(p \wedge q)$
I couldn't figure out what I'm supposed to from this point. Did I do anything wrong?
Thanks

To prove $p \vee q \equiv (p\wedge q) \vee (\neg p\wedge q) \vee (p\wedge \neg q)$, let's start from RHS as follows.
$(p\wedge q) \vee (\neg p\wedge q) \vee (p\wedge \neg q)$
$\equiv q \wedge (p\vee \neg p) \vee (p\wedge \neg q)$
$\equiv (q \wedge \text{T}) \vee (p\wedge \neg q)$
$\equiv q \vee (p\wedge \neg q)$
$\equiv (q \vee p)\wedge(q\vee \neg q)$
$\equiv (q \vee p)\wedge \text{T}$
$\equiv (q \vee p)$
Q.E.D.
