Question about using disjunction rule of inference to obtain material implication? I'm seeing math logic and I have a question.
Let $p$ be a proposition. Let's suppose I have $\lnot p$. 
By disjunction rule, this implies 
                                  $\lnot p \vee q$, where $q$ is any proposition.
This is equivalent (looking at the truth tables) to
                                 $p \implies q$
Does this mean that we only have to prove $\lnot p$ in order to prove $p \implies q\;$?
 A: This is correct.   A false statement implies anything.  However, in practice, we do not always prove an implication by proving the premise is never true; hence, this technique has limited scope.
A: As has already been said, yes, it is true.  
In mathematical work, the usual net effect of this fact is that if we wish to prove $p \longrightarrow q$, we do not need to even think about the situations in which $p$ is false, so we don't.  
So for example if we want to prove that every finite integral domain is a field (crudely, $p$ is then "is finite integral domain" and $q$ is "is field") there is no reason to bother examining things that are not finite integral domains.   
Although "false implies everything" is in the background of the way we use "implies," it is seldom if ever explicitly invoked by most working mathematicians.
A: If the logic involved come as a two-valued logic, then yes.  However, if we have some other system of logic involved, perhaps, perhaps not.  To see what's going on here, consider how from Np (the negation of p) we can legitimately infer Cpq (that's (p->q) in Polish notation).  Now, "from Np, we may infer Cpq", though it could get used as such, doesn't usually get taken as a basic rule of inference for two-valued propositional logic.  Instead,  CNpCpq exists as a theorem of classical two-valued logic.  Consequently if "Np" gets proven, then by application of modus ponendo ponens (the rule of detachment, conditional elimination), we can obtain "Cpq".  So, we can simply obtain a proof for Cpq also by adding one "Cpq" to the end of the proof of "Np".  So, basically if a system of logic has the rule of modus ponendo ponens, has CNpCpq as a logical theorem, then in order to prove Cpq, you just have to prove Np.
A: As previous answers state $\lnot p$ implies $p \to q$ (the converse however is not true).
I'll just provide some motivation for such choice of truth values for material implication. Consider for example the the statement "for each natural number $x$, if $x$ is equally divisible by 4, then it is equally divisible by 2." You obviously want this statement to be true. Now if you take $x = 2$ then you'll have $false \to true$, and if you take $x = 5$ you'll have $false \to false$. Thus a $false$ statement implies everything.
