@steve: I'm not sure I understand your objection, but how about this modification. Let A be defined by $A(n) = \sum_{k=0}^{n-1} A(k)$ for every nonnegative integer n. Then by strong induction we can show that A(n) = 0. That's a trivial example, but if we wanted to, we could make it less trivial: $B(n) = 1 - n + \sum_{k=0}^{n-1} B(k)$, which has solution $B(n) = 1$.