E# vs. BEAF
ABSTRACT
In this article we will prove that E# can keep up with up with Bowers' trientrical arrays, a subsystem of BEAF. To do this formally we will first establish many of the basic properties we take for granted about BEAF, E#, and large number sequences in general, and then use these properties to prove the E# keeps up with trientricals at every stage of development.
TRIENTRICAL ARRAYS
First we will define trientrical arrays, and then quickly establish their key properties. We will use the following definition:
Let {a,b,c} be defined when a,b,c are elements of the positive integers as follows:
when c=1; {a,b,1} = a^b
otherwise when b=1; {a,1,c} = a
otherwise; {a,b,c} = {a,{a,b-1,c},c-1}
Let us define a family of functions:
{10,n,1}
{10,n,2}
{10,n,3}
{10,n,4}
{10,n,5}
etc.
We want to establish that all of these functions are Strictly Increasing (S.IN) and everywhere abundant (E.AB). To do so we begin with the base function.
{10,n,1} = 10^n
We know that:
{10,n+1,1} = 10^(n+1) = 10*10^n = 10*{10,n,1}
By Axiom 16 (See Appendix D), the Axiom of Product and Factor [AxP&F], we have:
10*{10,n,1} > {10,n,1}
So we know that {10,n,1} is S.IN. We also know that:
{10,1,1} = 10 > 1
and that:
{10,2,1} > {10,1,1} --> {10,2,1} >= {10,1,1}+1 [Axiom 14] > 1+1 = 2
{10,3,1} > {10,2,2} --> {10,3,1} >= {10,2,1}+1 [Axiom 14] > 2+1 = 3
and in general...
{10,k+1,1} > {10,k,1} --> {10,k+1,1} >= {10,k,1}+1 [Axiom 14] > k+1
So {10,n,1} is E.AB
So {10,n,1} is S.IN/E.AB
Assume {10,n,k} is S.IN/E.AB
{10,n+1,k+1} = {10,{10,n,k+1},k} > {10,n,k+1} [bec. {10,n,k} is E.AB]
Therefore {10,n,k+1} is S.IN
{10,1,k+1} = 10 > 1
{10,2,k+1} > {10,1,k+1} --> {10,2,k+1} >= {10,1,k+1}+1 > 1+1 = 2
{10,3,k+1} > {10,2,k+1} --> {10,3,k+1} >= {10,2,k+1}+1 > 2+1 = 3
...
{10,n+1,k+1} > {10,n,k+1} --> {10,n+1,k+1} >= {10,n,k+1}+1 > n+1
Therefore {10,n,k+1} is E.AB
Therefore for all "k" {10,n,k} is a S.IN/E.AB function of "n".
Next we prove that:
10n < {10,n,1} : n>1
and
{10,n,a} < {10,n,b} : a<b & n>1
10(1) = 10 = 10^1 = {10,1,1}
10(2) = 20 < 100 = 10^2 = {10,2,1}
Assume 10k < {10,k,1} : k>1
10(k+1) = 10k+10 [Axiom 20] < 10k+10k = 20k < 100k = 10*10k < 10*{10,k,1}
= {10,k+1,1}
Therefore 10n < {10,n,1} : n>1
Now we show that each new function is also faster growing.
we want to prove that:
{10,n,k} < {10,n,k+1}
Firstly we have:
{10,1,k} = 10 = {10,1,k+1}
For n=2 we have:
{10,2,k+1} = {10,{10,1,k+1},k} = {10,10,k} > {10,2,k} [S.IN]
Assume:
{10,j,k} < {10,j,k+1}
{10,j+1,k+1} = {10,{10,j,k+1},k} > {10,{10,j,k},k} [S.IN]
{10,n,k} is E.AB, therefore:
{10,j,k} > j --> {10,j,k} >= j+1 [Axiom 14 - AxMS]
Therefore:
{10,{10,j,k},k} >= {10,j+1,k}
Therefore we have:
{10,n,1} < {10,n,2} < {10,n,3} < {10,n,4} < ... : n>1
E#
Briefly we define E#:
if there is only one argument; En = 10^n
otherwise, if last argument = 1; @m#1 = @m
otherwise; @m#n = @(@m#(n-1))
In the previous article where we compared E# to FGH, it was shown that all functions of the form E100#*(k)n are S.IN/E.AB functions of "n", and that:
E100#*(a)n < E100#*(b)n : a<b
E# is at least as strong as {10,n,k}
Firstly we note that:
{10,n,1} = 10^n = En
So here they are exactly the same thing. Next we have:
E100#n = EEE...EEE100 w/n Es
= 10^10^10^...^10^10^10^100 w/n 10s
We want to demonstrate that this is greater than {10,n,2}.
{10,1,2} = 10
E100#1 = E100 = 10^100 = {10,100,1} > 100 [{10,n,1}>n] > 10
Assume {10,k,2} < E100#k
{10,k+1,2} = {10,{10,k,2},1} = E{10,k,2} < E(E100#k) = E100#(k+1)
Therefore:
{10,n,2} < E100#n
Now we want to prove the general case:
{10,n,k} < E100#100#...#100#n w/k-1 100s = E100#*(k-1)n
{10,1,k+1} = 10
E100#*(k)1 = E100#*(k-1)100 > {10,100,k} > 100 > 10
Assume {10,j,k+1} < E100#*(k)j
{10,j+1,k+1} = {10,{10,j,k+1},k} < {10,E100#*(k)j,k} < E100#*(k-1)(E100#*(k)j)
= E100#*(k-1)(E100#*(k-1)100#j)
let @ = 'E100#*(k-1)'
thus we have:
@(@100#j) = @100#(j+1) [def. of E#]
@100#(j+1) = E100#*(k-1)100#(j+1) = E100#*(k)(j+1)
Thus we have proven our general conjecture:
{10,n,k} < E100#*(k-1)n
This proves that every hyper-operator up to trientrical arrays is dominates by a function of E#H. Thus E# is at least as strong as trientrical arrays.
E# is dominated by {10,n,1,2}
On the other hand we can show that:
E100#*(k)n < {10,n,k+2} : n>1
We begin with:
En < {10,n,2}
E1 = 10 & {10,1,2} = 10
but we have:
{10,2,2} = {10,{10,1,2},1} = {10,10,1} > {10,2,1} = E2
Assume Ek < {10,k,2}
{10,k+1,2} = {10,{10,k,2},1} = 10^{10,k,2} > 10^(Ek) > 10*Ek = E(k+1)
Therefore:
En < {10,n,2}
E100#1 = E100 & {10,1,3} = 10
but:
E100#2 = 10^10^100
&
{10,2,3} = 10^^^2 = 10^^10 = 10^10^10^10^10^10^10^10^10^10
so...
E100#2 < {10,2,3}
Assume E100#k < {10,k,3}
{10,k+1,3} = {10,{10,k,3},2} > {10,E100#k,2} > {10,E100#k,1} = E100#(k+1)
Therefore:
E100#n < {10,n,3}
Assume E100#*(k)n < {10,n,k+2}
E100#*(k+1)1 = E100#*(k)100 > 100 [Every function of E# is E.AB]
& {10,1,k+3} = 10
however
E100#*(k+1)2 = E100#*(k)(E100#*(k)100) < E100#*(k){10,100,k+2}
< 10{k+2}10{k+2}100
E100#n < E(10^10)#n = E1#(n+2) = {10,n+2,2}
E100#100#1 < {10,102,2}
E100#100#2 = E100#(E100#100) < {10,{10,102,2}+2,2}
< {10,{10,103,2},2}
YET TO PROVE (CONJECTURE)
E100#*(k)n < {10,n+2,k+1} < {10,n,k+2}
Once this is obtained:
{10,n,1,2}
= {10,10,{10,n-1,1,2}}
> E100#*({10,n-1,1,2}-2)10
now prove that this eventually dominates:
E100#*(k)n for any k.