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Alemagno12

aka Nish

  • I live in no
  • My occupation is you'er mom gayest lmao
  • I am despacito
  • Alemagno12

    Thanks to Ecl1psed276 for the sequence reduction method and for helping me test and fix the rules! (and also for cleaning up the rules)

    Originally copied from the googology Discord, updated and fixed with over time:

    Let x be the number in the square brackets. If the sequence is empty, output x. End. If the first number of the last element is 0, delete the element and replace x with x^2. End. Otherwise:
    Sequence reduction definition, for row k: 1 - Ignore all rows other than row k. 2 - Let the rightmost element be x. 3 - Move to the rightmost non-discarded element to your left. If you are at the leftmost element, stop this subprocess. 4 - Let the current element be y. 5.1 - If y < x, set x to be y. 5.2 - If y >= x, label the current elem…
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  • Alemagno12

    On hiatus. For now, check this instead.

    NOTE: This proof uses this definition for BM2. There is a very good chance that it's right, and in fact ecl1psed276 has tested the rules even on quad sequences and hasn't found any sequences that give different results than the BMS calculator, but it's not proven yet, so we cannot be 100% sure.


    First, let's see how (0,0)(1,1) evaluates. As there are already only two non-discarded elements, the bad root is (0,0) and the bad part is (0,0). Delta is then (1-0,0) = (1,0), C0 is (1,1), S0 is (0,0), and each Sn adds 1 to the first term and keeps the second term the same (it adds 0 to it), giving us Sn = (n,0). The resulting sequence is therefore (0,0)(1,0)(2,0)(3,0)...(x2,0).

    Collorary 1.1: If the last term of ea…


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  • Alemagno12

    Middle Term Theta

    July 6, 2018 by Alemagno12

    This is an OCF I made that has four entries. It has the property Θ(ωx,0,0,0) = ψ(ΩΩx).

    Define:

    • cof(0) = 0
    • cof(ω) = ω
    • cof(Ω) = Ω
    • cof(S(x)) = 0
    • cof(x+y) = cof(y)
    • maxcof(S) = max{x|x=cof(y)∧y∈S}
    • maxcof({a,b,c,d}) = 0: cof(Θ(a,b,c,d)) = ω
    • maxcof({a,b,c,d}) > 0, maxcof({c,d}) < Ω: cof(Θ(a,b,c,d)) = ω
    • maxcof({a,b,c,d}) > 0, maxcof({c,d}) = Ω: cof(Θ(a,b,c,d)) = Ω
    • ω[n] = n
    • Ω[n] = n
    • x+0 = x
    • x+S(y) = S(x+y)
    • cof(y) > 0: (x+y)[n] = x+(y[n])
    • Θ(0,0,0,0) = ω
    • Θ(0,0,0,x+1)[0] = 0
    • Θ(0,0,0,x+1)[n+1] = Θ(0,0,0,x+1)[n]+Θ(0,0,0,x)
    • cof(x) > 0: Θ(0,0,0,x)[n] = Θ(0,0,0,x[n])
    • f = Θ(a,b,c): f(x) = Θ(a,b,c,x)
    • ∃x∀y>x(f(y) > g(y)): f < g
    • lim(F)(0)[0] = 0
    • lim(F)(x+1)[0] = lim(F)(x)+1
    • f ∈ F, ¬∃g∈F(g 0: enum(F,x) = min{f|f∈F∧∀yenum(F,y))}
    • cof(x) = 0, cof(F) = 0, f = max{g|g∈F}: lim(F)(x)[n+1] = g…
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  • Alemagno12

    Take any function F from ordinals to ordinals. Then, we define:

    • ψ0F(0)[0] = 0
    • ψβ+1F(0) = min{x|"x is regular"∧∀y ψβ+1F(0): ψβF(α) = ψβFβ+1F(α))

    The cofinalities of some ordinals are not taken into account and unspecified, but these are either (1) supposed to already be known by the reader or (2) able to be deduced from the ruleset. (NOTE: There are no extended FSes, like ω2[ω+1] or Ω[Ω2] (although something like Ω2[Ω] is fine, as the FS of an ordinal has the same length as its cofinality))

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  • Alemagno12

    First, I prove that (0,0,0)(1,1,1)(2,1,1)(1,1,1) is standard...

    • (0,0,0,0)(1,1,1,1)
    • = (0,0,0)(1,1,1)(2,2,2)(3,3,3)(4,4,4)...
    • > (0,0,0)(1,1,1)(2,2,2)
    • = (0,0,0)(1,1,1)(2,2,1)(3,3,1)(4,4,1)...
    • > (0,0,0)(1,1,1)(2,2,1)
    • = (0,0,0)(1,1,1)(2,2,0)(3,3,1)(4,4,0)(5,5,1)...
    • > (0,0,0)(1,1,1)(2,2,0)
    • = (0,0,0)(1,1,1)(2,1,1)(3,1,1)(4,1,1)...
    • > (0,0,0)(1,1,1)(2,1,1)(3,1,1)
    • = (0,0,0)(1,1,1)(2,1,1)(3,1,0)(4,2,1)(5,1,1)(6,2,0)(7,3,1)(8,1,1)...
    • > (0,0,0)(1,1,1)(2,1,1)(3,1,0)
    • = (0,0,0)(1,1,1)(2,1,1)(3,0,0)(4,1,1)(5,1,1)(6,0,0)(7,1,1)(8,1,1)...
    • > (0,0,0)(1,1,1)(2,1,1)(3,0,0)
    • = (0,0,0)(1,1,1)(2,1,1)(2,1,1)(2,1,1)...
    • > (0,0,0)(1,1,1)(2,1,1)(2,1,1)
    • = (0,0,0)(1,1,1)(2,1,1)(2,1,0)(3,2,1)(4,1,1)(4,2,0)(5,3,1)(6,1,1)...
    • > (0,0,0)(1,1,1)(2,1,1)(2,1,0)
    • = (0,0,0)(1,1,1)(2,1,1)(2,0,0)(3,1,1)(…
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