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.
 A: To show
 (p ∧ q) → (p ∨ q).
If (p ∧ q) is true,
then both p and q
are true,
so (p ∨ q) is true,
and $T \to T$
is true.
If (p ∧ q) is false,
then
(p ∧ q) → (p ∨ q)
is true,
because false implies anything.
Q.E.D.
A: I know you asked specifically about a given proof, but here is another way:
(1) Assume p ∧ q
(2) By ∧-elimination, p
(3) By ∨-introduction, p ∨ q
(4) By →-introduction and marking the assumption (1), (p ∧ q) → (p ∨ q).
In less formal language: if P and Q is true, then you can look at either P or Q separately and it must be true. Now from any true statement you can create a longer true statement by creating a disjunction with any statement: if P is true, then "P or R" is also true for any statement R (e.g. if you are sure "it is raining", then it is also the case that "it is raining or you are a dragon"). In particular taking R = Q in this case allows you to reason that P or Q is true. 
A: Combine your first step with De Morgan's Law:
$$ \begin{array}{rcll}
(p ∧ q) → (p ∨ q) & ≡ & ¬(p ∧ q) ∨ (p ∨ q) & \text{Definition of $→$} \\
& ≡ & (¬p ∨ ¬q) ∨ (p ∨ q) & \text{De Morgan's Law}\\
& ≡ & (¬p ∨ p) ∨ (¬q ∨ q) & \text{Associativity and commutativity of $∨$} \\
& ≡ & T ∨ T  \\
& ≡ & T\\
\end{array}
$$
A: Alternately:
 p ∧ q        Assumed           <--------\
 p            base Extract P ∧ Q ⊢ P     |
 p ∨ q        base Widening P ⊢ P ∨ Q    |
 (p∧q)→(p∨q)  pop assumption            -/

This isn't quite rendering right -- need a fixed width font.
A: It is because of the following equivalence law, which you can prove from a truth table:
$$r\rightarrow s\equiv \lnot r\lor s.$$
If you let $r = p\land q$ and $s = p\lor q$, you get what you are looking for, namely that
$$(p\land q)\rightarrow (p\lor q)\equiv \lnot(p\land q)\lor(p\lor q).$$
A: As $\lnot p\lor p$ is trivial and $p→q$ means $p$ necessitates $q$. This gives $\lnot p\lor q$.
A: 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.
A: 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 $(1)$.
