[phkahler]

>> The inductive step assumes the hypothesis is true for n

You don't get to assume the thing you're trying to prove is true as a part of the proof that it's true. The only time we assume something like that is in a proof by contradiction, where the contradiction implies either an error in our logic, or an error in the assumptions.

[mannerheim]

An inductive proof has two steps:

1) Prove H(0)

2) Prove that if H(n), then H(n+1)

Then, by the axiom of induction, this is proven for all positive integer values of n. Because if those two conditions hold, we would have

H(0) is true

H(1) is true because H(0) -> H(1) ((2) with n=0) and H(0)

H(2) is true because H(1) -> H(2) ((2) with n=1) and H(1)

and so on.

The assumption isn't what's being proven, it's proving the inductive hypothesis for the next integer.

[phkahler]

Yes, I was being dumb.

As the article says there is a problem talking about equality among members of a set with size 1. So saying "all the horses in a set of size 1 are the same color" begs the question "same as what?" and that's where the real problem is, and the difficulty in pinning down that kind of thing is why I'm not a mathematician ;-)

[mannerheim]

There are several 'mathematical' ways you could phrase it (glossing over how to define the colour of a horse):

1) In a set of horses with size n, if horses A and B being both members of this set implies horse A has the same colour as horse B, then all horses in the set have the same colour.

2) In a set of horses with size n, and let f be a function on this set that produces each horse's colour. Then all horses in this set have the same colour if and only if the codomain of f on this set has size 1.

You could probably prove these are equivalent definitions.