• KthulhuHimself

    Well, I'm back, and I have a new idea for a function to show everyone. I am aware of how similar it may be to previous functions, but it's interesting nonetheless.

    If you see any immediate problems with it, feel free to comment below. I can already imagine that you may have something to say.

    Hell, this is pretty much a function identical to what Norminals was meant to be.

    Now, consider the language M0. It is, in essence, first order logic.

    Now, M1 is in essence first order logic, but with an added symbol that allows it to predicate over the order of logic it is using. In essence, think of it as ath order logic, where a can be any variable defined within a previously established bth order logical expression. Let's use the symbol | for that. To …

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  • KthulhuHimself

    Norminals Theory

    March 19, 2017 by KthulhuHimself

    Well, might as well start writing...

    I've lately seen a few interesting things on the wiki, the Little Biggedon being one of them. Ultimately, I admire that we finally have yet another wiki record for the highest coined number officially present on the site; yet a matter comes to mind when reading through it.

    That matter, my friends and colleagues, is something many of you may probably remember; the norminals function. The norminals, by their very design, are variables built in a way such that they will always be able to ascend beyond their old selves, and diagonalise their entire structure in a single step, every time. This does, of course raise the question "well, are 'norminals' well defined?" and the answer is, to a degree, no.

    Norminals …

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  • KthulhuHimself

    Well, I'll try to put this in the simplest terms (even though it's pretty simple either way).

    (I know this is basically the Godel's-incompleteness-theorem thing, but we'll go on with it anyways.)

    Let P be some formal system in which there are no contradictions between the axioms, and A be some axiom.

    \(\forall P\exists A:A\not\in P\)

    It almost seems trivial; but I'd stil like some nice formal proof.

    We should also say that

    \(\forall P\exists P':P'\not\subseteq P\)

    But, well; I guess it's up for debate whether or not this is even necessary.

    (We'll get to this once the previous are proven.)

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  • KthulhuHimself

    The root of all evil.

    September 20, 2016 by KthulhuHimself


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  • KthulhuHimself

    I haven't

    July 18, 2016 by KthulhuHimself


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