Induction on a is unflawed: every occurrence of a in the conjecture
(equal (app (app a b) c)
(app a (app b c)))
is in a position being recursively decomposed!
Now look at the occurrences of b. The first (shown in bold below)
is in a position that is held constant in the recursion of (app a b).
It would be ``bad'' to induct on b here.
(equal (app (app a b) c)
(app a (app b c)))
