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

For better readability I use $+$ for "or", $\cdot$ for "and" and $()'$ for negation.
So, you have
\begin{eqnarray*} pq + p'q+pq'
& = & (p+p')q+pq' \\
& = & q + pq' \\
& \stackrel{DeMorgan}{=} & (q'(p'+q))' \\
& \stackrel{qq' = F}{=} & (q'p')' \\
& = & p+q
\end{eqnarray*}
