Prove that the set {→, ¬} is functionally complete
Solution 1:
The implication $\to$ is defined by $$ a\to b \equiv \neg a \vee b. $$
This means, that $$ \neg a \to b \equiv a \vee b$$ and thus, you can express logical or using $\to$ and $\neg$. Furthermore you know de Morgans rules and you have $$ \neg(\neg a \vee \neg b) \equiv \neg\neg a\wedge \neg\neg b = a\wedge b.$$
Thus you can express all logical operators using $\to$ and $\neg$. The set of these is known to be functionally complete.
Solution 2:
No, $\to$ alone is not complete because you cannot get to negation, for example.