A riddle in logic and propositional logic The professor gave the class a riddle
Suppose the following two statements are true:


*

*I love A or I love B

*If I love A, then I love B


Does it necessarily follow that I love A? Does it necessarily follow that I love B?
I'm not quite sure what to do here but here goes...
Let a: I love A
Let b: I love B
So $1$ becomes $ a \lor b $ 
and $2$ is $a \to b $
We are looking to find if 
$ ((a \lor b) \land (a \to b)) \to  a$ 
or if
$ ((a \lor b) \land (a \to  b)) \to b$
From the truth table, we see that the latter is a tautology. Therefore, it follows that I love B. Is that right?
 A: The conclution can't be only $A$ cause in that case you are dening $B$ and that implies $\lnot A$ because of (2), so it yelds a contradiction. On the other hand it could, in principle, be just $B$ because if you deny $A$ (2) would be always true, but it could also be $A \land B$, so you can conclude $B \lor (A \land B)$ which finally is equivalent to $B$.
A: One way to see that "$I \, love \, B \, $" follows from the premises, is to note that the 2nd premise material conditional is defined as $\lnot A \lor B$, so in conjunction with the 1st premise and conjunction introduction we have $(A \lor B) \land (\lnot A \lor B)$, which is equivalent to $(A \land \lnot A) \lor B$ by distributivity of $\land$. Finally, conclude $B$ by Disjunctive Syllogism, since it's never the case that $A \land \lnot A$.
A: Since the other answers haven't mentioned it, your method is correct. Every tautology is necessarily true, and every contradiction is necessarily false, whereas every other statement is true in some models but false in others. So indeed you look for which option corresponds to a tautology when asked for which option is necessarily true.
$\def\imp{\Rightarrow}$
Another approach is via the tableaux method. If you start with the given premises and the conclusion, and every branch closes (impossible), then you know that the conclusion is necessarily false given the premises. Likewise, if you start with the given premises and the negation of the conclusion, and every branch closes, then you know that the conclusion is necessarily true given the premises. If in both cases some branch is left open, then that branch shows you how to construct a situation which satisfies everything that you start with. In your example, starting with "$a \lor b$" and "$a \imp b$" and "$\neg a$" gives you a tree with an open branch containing "$\neg a$" and "$b$", and you see that indeed the situation where these are true satisfies all the premises but not "$a$". Thus we conclude that $a$ does not necessarily follow.
If you just want to prove the tautology, yet another way is via some deductive system. I strongly recommend natural deduction (Fitch-style), for which the proof of the second option goes like this:
  $a \lor b$. [premise]
  $a \imp b$. [premise]
  If $a$:
    $b$. [implication elimination]
  If $b$:
    $b$.
  Therefore $b$. [disjunction elimination]
