Question

Haskell implementation of the familiar Fibonacci function

fibSlow n

| n == 0 = 1 --fib.1

| n == 1 = 1 --fib.2

| otherwise = fibSlow(n-1) + fibSlow(n-2) --fib.3


What is the induction proof of correctness for fibSlow?

Answer 1

To prove correctness of a function on the natural numbers by induction, you would show that it’s correct for certain base cases, and then that it’s correct for higher values of the parameter given the assumption that it’s correct for lower ones. So you’d verify first that fibSlow 0 = 1, and then that fibSlow 1 = 1, and then that for n > 1, fibSlow n is equal to the (n-1)th fibonacci number plus the (n-2)th fibonacci number. Here you get to assume that those numbers are fibSlow (n-1) and fibSlow (n-2), since fibSlow is correct for all inputs less than n by the inductive hypothesis.

This might seem all rather trivial… because it is! The whole point of such an example in Haskell is that you can write code that’s obviously correct. When you go to prove it correct, the proof just writes itself and amounts to looking at the code and noting that it clearly says exactly what you’re trying to prove. This is one of the nice properties of a declarative language like Haskell.