User:Vel!/pu/ABHAN

ABHAN claims to be very very powerful, but is it? How does it compare to other things, including the FGH, BEAF, and my extended arrow notation?

So we have the rules for the basic notation.
 * b(a,b)=a*b=ab
 * 1) b(# 0)=b(#)
 * b(a,b,c #)=b(a,b(a,b-1,c #),c-1 #)
 * b(a,1,#)=a
 * b(a,b,0,.....,0,c,#)=b(a,a,0,.....,b,c-1,#)

b(3,3) = 3*3 = 9

b(10,100) = 1000

b(10100,10100) = 10200 ("gargoogol")

Not that googological... but hey $$f_{1}(n)$$ isn't very googological either, but it can be recursed over and over again to create very fast-growing functions. So maybe, more of this notation will give us larger numbers. Of course it will! Not all ABHAN arrays are limited to two entries.

b(3,3,0) = b(3,3) = 3*3 = 9

b(3,3,1) = b(3,b(3,2,1),0) = b(3,b(3,b(3,1,1),0),0) = b(3,b(3,3,0),0) = b(3,b(3,3),0) = b(3,9,0) = b(3,9) = 27

b(3,3,2) = b(3,b(3,b(3,1,2),1),1) = b(3,b(3,3,1),1) = b(3,b(3,b(3,b(3,1,1),0),0),1) = ... = 7,625,597,484,987 = 33 3

b(3,3,3) = 3^^^3 ("tritri")

b(3,3,4) = 3^^^^3 ("grahal")

b(3,3,5) = 3^^^^^3

...

b(3,3,1,1) = b(3,b(3,b(3,1,0,1),0,1),0,1) = b(3,b(3,3,0,1),0,1) = b(3,b(3,3,3),0,1) = b(3,3,b(3,3,3)) = 3{3{3}3}3 = 33 - 3 expanded to 3. So b(a,b,1,1) is equivalent to expansion.

b(a,b,2,1) = multiexpansion

b(a,b,3,1) = powerexpansion

b(a,b,4,1) = expandotetration

b(a,b,1,2) = explosion

b(a,b,1,3) = detonation

b(a,b,1,4) = pentonation

b(a,b,1,0,1) ~ megotion

b(a,b,1,0,2) ~ gigotion

b(a,b,1,0,3) ~ terotion

b(a,b,1,0,0,1) ~ {a,b,1,1,1,2}

b(a,b,1,0,0,0,1) ~ {a,b,1,1,1,1,2}

b(a,b,1,0,0,...,0,1) with n 0's ~ {a,b,1,1,...,1,2} with n+1 1's

So this has a limit ordinal of $$\omega^{\omega}$$, making it on par with BEAF's linear arrays, and a diagonalizing function is not provable complete in PRA.

Add some new rules:


 * b(a,b{k+1}m+1} = b(a,a{k}0{k}0{k}.......0{k}a,b{k+1}m) with b {k}'s

He also unclearly defines another rule called "M2" with the previous one being "M1":

b(A{n}A2)=b(/A/{n}A2/) - this means that the array before the seperator is solved normally until it's worked down into 2 characters and the array after is then solved but also depends on the array before it. - Example:b(3,2,1{5}0,1)=b(3,b(3,1,1{5}0,1){5}0,1)=b(3,3{5}0,1)=b(3,3{5}3). For now we let b(n,m@#@#2) be shorthand for b(n,m@n,m,#@#2). The limit of this extension is w^w^w

Whatever that rule means...

So then nested arrays are created, with no formal definition. I don't know exactly what happens but whatever it is it has $$\varepsilon_0$$ limit (unless he flunks something). Now we have the diagonalizer \ which expands to:

M3:b(a,b{#,0\(k+1)}1)=b(a,a{#,a,a{#,a,a{.....}1\k}1\k}1) with b nests from the center out.

But what's in the center??

So now we have \, which appears to be a hyperseparator-diagonalizer not unlike one I thought of for my Z function (page coming). But this should reach $$\zeta_0$$ with {0\0\...\0\1}. Then a series of \-separators are defined. I haven't analyzed, but it should get you to $$\eta_0$$. Then a colon separator extends all this, possibly reaching $$\varphi(\omega,0)$$, but all this is just guesswork. also has that level.

Really all this is guesswork.