It’s not circular reasoning, it’s a step of mathematical induction. First you show that something is true for a set of 1, then you show that if it’s true for a set of n it is also true for a set of n+1.
Comment on The science is divided
sp3tr4l@lemmy.zip 19 hours agoFrom that link:
Assume that n horses always are the same color.
… I mean… yes, the logic follows… if you… make and hold that assumption… which is ostensibly what you are trying to prove.
This is otherwise known as circular reasoning.
Apparently this arose basically as a joke, a way of illustrating that you actually have to prove the induction is valid every step of the way, instead of just asserting it.
ThatGuy46475@lemmy.world 17 hours ago
sp3tr4l@lemmy.zip 17 hours ago
As with Kogasa, you’re right that this is not circular reasoning, it is induction.
I judged it a bit too quickly.
However, it isn’t a valid proof of induction.
I tried to work through exactly where and how it fails in another comment.
So… it is still fallacious reasoning of some kind, but yes, not the circular reasoning fallacy.
kogasa@programming.dev 17 hours ago
No, that’s what induction is. You prove the base case (e.g. n=1) and then prove that the (n+1) case follows from the (n) case. You may then conclude the result holds for all n, since we proved it holds for 1, which means it holds for 2, which means it holds for 3, and so on.
KingRandomGuy@lemmy.world 17 hours ago
Exactly, the assumption (known as the inductive hypothesis) is completely fine by itself and doesn’t represent circular reasoning. The issue in the “proof” actually arises from the logic coming after this, in which they assume that they can form two differeny overlapping sets by removing a different horse from the total set of horses, which fails if n=2 (as then they each have a distinct element).
sp3tr4l@lemmy.zip 17 hours ago
You are correct that in the mathematical sense, this is not circular reasoning, it is induction.
The problem is that this is an example of a failed, invalid proof of induction.
I investigated it a bit further and tried to work through the actual point at which the proof fails in another comment.