Proof of two statements being equal withe rules of logic How can one prove that $P \to Q$ Is the same thing as $[\text{not } P \text{ or } Q]$. I am guessing I would have to prove that they are the negation of the same thing
 A: The easiest way is to show their truth tables are identical. In more advanced logic, one generally just takes that as the definition.
That is, one generally will just consider $p \to q$ to be short hand for $\neg \, p \lor q$
If you wanted to prove that within a system it would depend on what the rules of inference are for that particular system. Different systems have different rules of inference. With regards to doing it via the truth table, two statements $A$ and $B$ are logically equivalent if the column in the truth table corresponding to $A \gets\! \to B$ is all $T$'s.(this is called a tautology).
So to prove it one would construct the truth table for $ ( P \to Q) \gets \! \to (\neg P \, \lor Q)$ and observe that it's always true. This means both statements will always have the same truth value. They will  always either both be true or both be false. 
A: I'm assuming the framework is classical propositional logic. I further suppose that you want to prove the formulas to be equivalent, i.e. to show that they have the same truth value in any model (function from sentence variables to truth values).
If $v \models p \rightarrow q$  (short for $~p \rightarrow q$ is true in model $v$), then $v \not \models p$ or $v \models q$. In the first case $v \models \neg p$, so $v \models \neg p  \lor q$. Similarly in the second case. 
Conversely, if $v \not \models \neg p  \lor q$, then $v \models p \wedge  \neg q$, so $v \not \models p \rightarrow q$.                    
A: $p\to q$ means that: $q$ will be true if $p$ is true.  
Therefore $p\to q$, entails that: if $q$ is false, then $p$ cannot be true. 
So if $p\to q$ then either $p$ is false or $q$ is true.  $$p\to q\implies \neg p\vee q$$

Conversely.
$\neg p\vee q$ means that either $p$ is false or $q$ is true.
Threfore $\neg p\vee q$ entails: if $p$ is true, then $q$ must be true.
$$\neg p\vee q\implies p\to q$$

$$\therefore \qquad p\to q\iff \neg p\vee q$$
