Conjunction not expressible as conditional If I am to prove that, say, $A \vee B$ can be expressed using only the conditional, I can write out some truth tables until I have found a matching column (in this case, $(A \Rightarrow B) \Rightarrow B$). If, however, I am to prove that a certain wf is not expressible in only the conditional, what can I do? There are infinitely many possibilities to check, and I have finitely much time.. :/
 A: Well, there are infinitely many different formulas you can build, but there are only finitely many different truth tables a formula in your finitely many propositional variables can have.  So, if you start exhaustively listing out all the truth tables you can get by starting from the propositional variables and then repeatedly combining them with $\Rightarrow$, eventually you will have exhausted every possibility.  That is, once you have combined every pair of truth tables on your list and gotten another one that was already on your list, you know you'll never get any new ones again.
In some cases, there may be some shortcut you can use to avoid such exhaustive calculation.  In particular, you can look for some property that will always be true of the truth table of any formula built from $\Rightarrow$ but which is not true for your formula.  For instance, any formula built from $\Rightarrow$ will be true whenever all of the variables are true (think about why!), so any formula that is false when all the variables are true cannot be expressed using only $\Rightarrow$.
A: This follows from Post's theorem on the lattice of clones on a two-element set. The clone $T_1^\infty$ that is generated by implication is not the same as the clone $P_1$ that is generated by implication and conjunction. As discussed in the other answer, there are sometimes shortcuts for identifying clones, but I don't know of one for the clone generated by implication.
