Also related: en.wikipedia.org/wiki/Raven_paradox
Comment on The science is divided
Tiuku@sopuli.xyz 5 weeks ago
Actually a quite interesting article: en.wikipedia.org/…/All_horses_are_the_same_color
Elgenzay@lemmy.ml 5 weeks ago
WildPalmTree@lemmy.world 5 weeks ago
A much more interesting question, if you ask me.
Viking_Hippie@lemmy.dbzer0.com 5 weeks ago
It was stated as a lemma, which in particular allowed the author to “prove” that Alexander the Great did not exist, and he had an infinite number of limbs.[4]
Talk about burying the lede! 😄
sp3tr4l@lemmy.zip 5 weeks ago
From 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.
kogasa@programming.dev 5 weeks 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 5 weeks 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 5 weeks 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.
ThatGuy46475@lemmy.world 5 weeks ago
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.
sp3tr4l@lemmy.zip 5 weeks 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.
cmgvd3lw@discuss.tchncs.de 5 weeks ago
I did do this proof by induction back in the day, but now looking at the article I am clueless.
sp3tr4l@lemmy.zip 5 weeks ago
Do you mean you went through the proof and verified it, or falsified it?
…
As I understand it, it goes something like this:
You have a set of n horses.
Assume a set of n horses are the same color.
Now you also have a set of n+1 horses.
Set 1: (1, 2, 3, … n) Set 2: (2, 3, 4, … n+1)
Referring back to the assumption, both sets have n horses in them, Set 2 is just incremented forward one, therefore, Set 2’s horses are all one color, and Set 1’s horses are all one color.
Finally, Set 1 and Set 2 always overlap, therefore that the color of all Set 1 and Set 2’s horses are the same.
…
So, if you hold the ‘all horses in a set of size n horses are the same color’ assumption as an actually valid assertion, for the sake of argument…
This does logically hold for Set 1 and Set 2 … but only in isolation, not compared to each other.
The problem is that the sets do not actually always overlap.
If n = 1, and n + 1 = 2 then:
Set 1 = ( 1 )
Set 2 = ( 2 )
No overlap.
Thus the attempted induction falls apart.
Set 1’s horse 1 could be brown, Set 2’s horse 2 could be … fucking purple… each set contains only one distinct color, that part is true, but the final assertion that both sets always overlap is false, so when you increment to:
Set 1 = ( 1, 2 )
Set 2 = ( 2, 3 )
We now do not have necessarily have the same colored horse 2 in each set, Set 1’s horse 1 and 2 would be brown, Set 2’s horse 2 and 3 would be purple.
…
I may be getting this wrong in some way, it’s been almost 20 years since I last did set theory / mathematical proof type coursework.
Tiuku@sopuli.xyz 5 weeks ago
Yeah, so the following actually is valid: If all pairs of horses are the same color, all horses are the same color. Just starting the induction one step further.
KingRandomGuy@lemmy.world 5 weeks ago
Yep this is the exact issue. This problem comes up frequently in a first discrete math or formal mathematics course in universities, as an example of how subtle mistakes can arise in induction.
Iceman@lemmy.world 5 weeks ago
existentialcomics.com/comic/2
idunnololz@lemmy.world 5 weeks ago
I took a peak and it is sort of dumb but logically “sound”. Specifically the indictive step.
In the inductive step you assume the statement is true for some number n and use this statement to prove the statement n + 1 is true. If you can do that then you can prove the induction step.
So in this example the statement we assume is true is given n horses, all of them are the same color. To prove the statement for n + 1 horses we look at the n + 1 horses. Then we exclude the first horse. By excluding the last horse we have a set of n horses. By the induction statement this set of horses must all be the same color. So now we’ve proven the first n horses are the same color.
Next we can exclude the first horse. This also gives us a set of n horses. By the induction statement all these horses must also be the same color. Therefore all n + 1 horses must be the same color.
This sounds really dumb but the proof works in the induction step.
The logical issue is that the base case is wrong which is necessary for a complete proof by induction.
sp3tr4l@lemmy.zip 5 weeks ago
I think? I worked through how the induction logic actually fails.
This kind of induction only works if you can actually prove Sets 1 and 2, starting at n and n+1, actually overlap at all stages… and in this case, they don’t.