TITLE: Show that (p ∧ q) → (p ∨ q) is a tautology?
QUESTION [14 upvotes]: I am having a little trouble understanding proofs without truth tables particularly when it comes to → 
Here is a problem I am confused with: 
Show that (p ∧ q) → (p ∨ q) is a tautology
The first step shows: (p ∧ q) → (p ∨ q) ≡ ¬(p ∧ q) ∨ (p ∨ q)
I've been reading my text book and looking at Equivalence Laws. I know the answer to this but I don't understand the first step.
How is (p ∧ q)→ ≡ ¬(p ∧ q)?
If someone could explain this I would be extremely grateful. I'm sure its something simple and I am overlooking it.
The first thing I want to do when seeing this is
(p ∧ q) → (p ∨ q) ≡ ¬(p → ¬q)→(p ∨ q)
but the answer shows:
¬ (p ∧ q) ∨ (p ∨ q)     (by logical equivalence)
I don't see a equivalence law that explains this.

REPLY [2 votes]: The following is an inference rule approach to showing that $P \to Q \equiv \neg P \lor Q$, using the Constructive Dillema inference rule: 
$$ \large \frac{P \to Q,~ R \to S, ~P \lor R}{ Q \lor S}$$
It can be shown that $\neg P \lor  P$ and $\neg P \to \neg P$ are tautologies, and given that we know $ P \to Q $ , we can substitute into the above inference rule.   
$$  \large \frac{ \neg P \to \neg P,~P \to Q,~ ~\neg P \lor P}{  \neg P \lor Q }$$   
So far we have shown that $ (P \to Q) ~\vdash  (\neg P \lor Q)$. To finish proving the equivalency $ P \to Q \equiv \neg P \lor Q ~$ we also need to show $  (\neg P \lor Q)  \vdash  (\neg P \lor Q)     $. I don't see an obvious inference rule at this point, but we could show it by contradiction.