• Steve Dice@sh.itjust.works
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    2 days ago

    Let M and f be as in the hypotheses of the meme.

    Since f is a meme, then f∈M. That means f can be applied to itself. It follows that (f, f(f))∈f. We have then:

    f ∈ {f, {f, f(f)}} = (f, f(f)) ∈ f

    Thus, f∈f, violating the Axiom of Regularity. We conclude the meme is mathematical bullshit and I will not have it.

    • wolframhydroxide@sh.itjust.works
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      2 days ago

      One of your precepts is flawed. f is not a meme any more than the word “all” is a meme in “all your base are belong to us”. f is defined by text within the overall meme, but while it is part of the meme, it is not the meme itself, as it lacks the context of the remainder of the meme. Your precept is like saying “9 is prime, because it is the prime number ‘19’”.

      9 is not prime. It is part of the representation of the number 19.

      f is not the meme. It is part of the context which defines the meme.

        • wolframhydroxide@sh.itjust.works
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          2 days ago

          It does not.

          I could be wrong, but I’m fairly sure that, while ‘f’ is a function, ‘f(x)’ is the function’s output, not the function itself. So f(x) is a meme (as.long as x is a meme), because function f’s output is a meme. The function itself is a mathematical operation, not its output.

          f and F are not memes, but functions which output memes.