Show that (p ∧ q) → (p ∨ q) is a tautology? 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.

This is a classic equivalence
$$p\to q \equiv \neg p\lor q.\tag 1$$
We can examine equivalence using a truth table
\begin{array}{c|c|c|c|c}
p&q&\neg p&p\to q&\neg p\lor q\\\hline
T&T&F&T&T\\\hline
T&F&F&F&F\\\hline
F&T&T&T&T\\\hline
F&F&T&T&T
\end{array}
Hence, $p\to q\equiv \neg p\lor q$.
Further, if $a \equiv p\land q$ and $b\equiv p\lor q$, and so
$$a\to b\equiv \neg a\lor b\equiv \neg(p\land q)\lor(p\lor q)\equiv (\neg p\lor \neg q)\lor (p\lor q),$$
by using $(1)$.
