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# Alemagno12

## aka Nish

My favorite wikis
• I live in no
• My occupation is you'er mom gayest lmao
• I am despacito
• ## A program-free definition for BM2?

July 9, 2018 by 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…``` Read more >
• ## Proof of termination of Pair Sequence System (BM2)

July 9, 2018 by 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…

• ## 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…
• ## Creating an OCF based on any function from ordinals to ordinals

July 1, 2018 by 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))

• ## BM3 has an infinite loop

June 29, 2018 by 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)(…