Prove or refute contingent: If A implies B is contingent, then B is too The question is:

If $A, A \to B$ are contingent, then so is $B$

$A, A \to B$ (implies) is a contingent, but how exactly to show «so is $B$»?
If I'm using a truth table, how should I show that $B$ is also contingent?
 A: This is not necessarily true.
We explore each of the alternatives to contingency: Suppose $B$ is a tautology. Suppose $B$ is a contradiction. (If both of these lead to contradicting our premises, then $B$ would necessarily be contingent. Otherwise, it is not.


*

*Suppose $B$ is a tautology (hence always evaluating to true), then
$A\rightarrow B$ would necessarily be true always...hence a
tautology. This contradicts one of our premises.
$\therefore$ $B$ cannot be a tautology.

*Suppose $B$ is a contradiction (always false). This is entirely
possible, given the contingency of $A$ and of $A\rightarrow B$. That is,
$B$ could always be false, and then the truth value of $A\rightarrow
   B$ is entirely determined by (contingent upon) the truth-value of  $A$, which is, in
turn, contingent.
$\therefore\; B$ could be a logical contradiction.
Hence, by definition then, $B$ is not necessarily contingent given only the constraints of the premises. $B$ could very well be a logical contradiction, and in that event, it is not contingent.
A: Contradiction example:  
\begin{matrix}
A & B & A \to B \\
F & F & T \\
T & F & F \\
\end{matrix}
$A$ is contingent, $A \to B$ is contingent, but $B$ is contradiction.
