• sp3ctr4l@lemmy.zip
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        4 months 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
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          4 months 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.