Proving A $\implies$ B or A $\implies$ C To prove A $\implies$ B or A $\implies$ C with a direct proof, I'm confused on what we can assume. Normally for $A \implies B$ we assume A then prove B is true. 
For proving A $\implies$ B or A $\implies$ C, do we just assume A is true then show the following:


*

*B is true and C is false

*B is false and C is true

*B is true and C is true


Or are there other ways using a direct proof?
 A: Is your $\Rightarrow $ used to mean logical implication, or is it used as the material implication?  It is typically used to mean the former ... if you mean the latter then please use $\rightarrow$
This distinction is crucial!
It is true that $(A \rightarrow B) \lor (A \rightarrow C)$ is logically equivalent to $A \rightarrow (B \lor C)$, i.e. that $(A \rightarrow B) \lor (A \rightarrow C) \Leftrightarrow (A \rightarrow B) \lor (A \rightarrow C)$, and hence it is true that $A \rightarrow (B \lor C) \Rightarrow (A \rightarrow B) \lor (A \rightarrow C)$, and therefore proving that $A \rightarrow (B \lor C)$ will indeed prove that $(A \rightarrow B) \lor (A \rightarrow C)$
However, it is not true that $A \Rightarrow B \lor C$ is the same as $A \Rightarrow B$ or $A \Rightarrow C$: if it is true that $A \Rightarrow B$ or $A \rightarrow C$, then it will also be true that $A \Rightarrow B \lor C$, but the other way around does not hold.
Consider:
$B=P$, $C = \neg P$, and $A=Q$
Now, clearly you have $Q \Rightarrow P \lor \neg P$, and yet you have neither $Q \Rightarrow P$ nor $Q \Rightarrow \neg P$
So, showing that $A$ logically implies $B \lor C$ by showing that the assumption that $A$ is true means that at least one of $B$ and $C$ has to be true shows that $A \Rightarrow B \lor C$, but that does not show that either $A$ logically implies $B$ or that $A$ logically implies $C$, i.e. your method would not show that $A \Rightarrow B$ or $A \Rightarrow C$
A: The following truth table may help show what needs to be done:

To derive the consequent, we need the antecedent, $B \lor C$, as a premise or a derivable proposition from other propositions. Notice that we don't need to know anything about $A$.
Assuming we have $B \lor C$ as a premise we can prove the consequent as follows using a Fitch-style natural deduction proof:


Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
