Conditionals involving disjunctions How would i prove a conditional in this form:
$ A \rightarrow (B \vee C)$
through 'algebraic' manipulations i come up with the equivalence
$ A \rightarrow (B \vee C)$ $\Leftrightarrow$ $( A \wedge ¬B \rightarrow C)\wedge (A \wedge ¬C \rightarrow B)$
So, to prove $ A \rightarrow (B \vee C)$  one must prove $ A \wedge ¬B \rightarrow C$ and $A \wedge ¬C \rightarrow B$ just like proving biconditionals? or there is an easier way to do so? 
Example: 
Proposition:
Suppose $ a, b \in N$. If $gcd(a, b) > 1$, then $b|a$ OR $b$ is not prime.
Proof:
Suppose $gcd(a, b) > 1$ and b $\not|$ $a$
...
Therefore, $b$ is not prime.
Now, suppose that $gcd(a, b) > 1$ and $b$ is prime.
...  
therefore, $ b|a$
Q.E.D.
thanks in advance and sorry for the bad english.
 A: $B \lor C$ is equivalent - by Material Implication - to : $\lnot B \to C$.
Thus, $A → (B ∨ C)$ is equivalent to : $A \to (\lnot B \to C)$ that, in turn - by Exportation - is equivalent to :

$(A \land \lnot B) \to C$.

In conclusion, your first proof is enough, due to the fact that it amounts to:

assume $\text {gcd}(a,b) > 1$: if $\lnot (b|a)$, then $b$ is not prime.

A: As a side note: A good way to think about what's going on instead of resorting to using manipulations of the syntax is the following: You want to show $A$ implies $B\vee{C}$. Now assume that $A$ is true. Then you want to show that $B\vee{C}$ is true. If $B$ is true then $B\vee{C}$ is true so you are done. So assume not: then you just need to show that $C$ holds.
For your example: Assume that $gcd(a,b)>1$. If $b|a$ then you are done. So suppose this isn't the case.....
The syntax and the rules formalize this intutive understanding; but sometimes its better to take a step back and work with the intuitive understanding than deal directly with the syntax.
