        o---o
|       |                   |
|       @          =        @
|
|   ( A B ( C ))   =   ( A ( ( B ( C )) ))

It would be fudging a bit, however, to call this equivalence "one-step".
That is, the "one-step-ness" of the above proof is qualified by the fact
that one first has to prove the prior theorem of "Double Negation" (DN).

Strictly speaking, DN is a theorem, and one that requires quite a bit of
ingenuity to prove, because it says something more than the simple axiom:

|     o
|     |
|     o
