# User:Moosey/Googology

## ah

### Part one: Starter notations

define \$_n as any entries (including no entries) in an array. It’s my symbol for we-don’t-care entries. In any one use of any one rule, if n is the same, \$_n is the same

a#b = concatenation of a and b

n@m =
n, m = 1
{n}#(n@(m-1)), m > 1, m not a lim ord
n@(m[n]) if m is a lim ord

g(a,n,B) =
ah_g(a-1,n,B) {B}, a > 1;
n, a = 0

### Basic ah rules

Rule 1.
ah^a_n {\$_0} = g(a,n,\$_0)
Rule 2.
ah_n ({\$_1}#{z}) = ah_n {\$_1}, z = 0
Rule 3.
ah_n{} = n+1
Rule 4.
ah_n{a+1,\$_2} = ah^n_n{a,\$_2}
Rule 5.
ah_n{a,\$_3} = ah_n{a[n],\$_3}, a a lim ord
Rule 6.
ah_n ((0@b)#{a+1,\$_4}) = ah_(n+1) (((a+1)@b)#{a,\$_4}), b > 0
Rule 7.
ah_n ((0@b)#{a,\$_5}) = ah_n ((a@b)#{a[n],\$_5}), a a lim ord & b > 0

### Notating enormous arrays

ah_n{\$_0//(b+1),\$_1} = ah_n(((\$_0)(@^_n)ah_n{\$_0//b})#{//b,\$_1})
ah_n{\$_0//0} = ah_n{\$_0}
ah_n{\$_0//b,\$_1} = ah_n{\$_0//(b[n]),\$_1} for lim ord b
Else: apply ah's 7 rules, starting after legion bar. (e.g. ah_{\$_0//,0,0,w+1} =
ah_{\$_0//,w+1,w+1,w})

(\$_1)(@^_n)a =
(\$_1)@(ah_n((\$_1)(@^_n)(a-1))), a > 1,
ah_n{\$_1}, a=1

ah_n{\$_0@@0} = ah_n{\$_0}
ah_n{\$_0@@a,\$_1} = ah_n{\$_0@@a[n],\$_1},a a lim ord
ah_n{\$_0@@(a+1),\$_1} = ah_n({\$_0//}#{\$_0@@(a),\$_1}
Apply basic ah rules otherwise to everything after the @@

ah_n{\$_0(@@b,\$_2)0} = ah_n{\$_0}
ah_n{\$_0((@@b,\$_2)a,\$_1} = ah_n{\$_0(@@b)a[n],\$_1},a a lim ord
ah_n{\$_0(@@0)(a+1),\$_1} = ah_n({\$_0//}#{\$_0(@@0)(a),\$_1}
ah_n{\$_0(@@b+1,\$_2)(a+1),\$_1} = ah_n({\$_0(@@b,\$_2)}#{\$_0@@(a),\$_1}
ah_n{\$_0(@@b,\$_2)\$_3} = ah_n{\$_0(@@b[n],\$_2)\$_3}, if b is a lim ord
Apply basic ah rules otherwise to everything after the (@@\$) or everything inside the (@@\$)
depending on what is necessary
These array rules can be applied without the ah_n of the array, but any n refers to the n in the ah
subscript.

#### Enormous array update

if unspecified, n is the subscript in the first (closest) ah that is in front of the array.

{\$_0//(b+1),\$_1} = (((\$_0)(@^_n)ah_n{\$_0//b})#{//b,\$_1})
{\$_0//0} = {\$_0}
{\$_0//b,\$_1} = {\$_0//(b[n]),\$_1} for lim ord b
Else: apply ah’s 7 rules, starting after legion bar. This includes nesting and incrementing the subscript (e.g. ah_n{\$_0//,0,0,w+1} = ah_(n+1) {\$_0//,w+1,w+1,w})

(\$_1)(@^_n)a =
(\$_1)@(ah_n((\$_1)(@^_n)(a-1))), a > 1,
ah_n{\$_1}, a=1

{\$_0@@0} = {\$_0}
{\$_0@@a,\$_1} = {\$_0@@a[n],\$_1},a a lim ord
{\$_0@@(a+1),\$_1} = ({\$_0//}#{\$_0@@(a),\$_1}
Apply basic ah rules otherwise to everything after the @@. See legion bar notes for more details

{\$_0(@@b,\$_2)0} = {\$_0}
{\$_0((@@b,\$_2)a,\$_1} = {\$_0(@@b)a[n],\$_1},a a lim ord
{\$_0(@@0)(a+1),\$_1} = ({\$_0//}#{\$_0(@@0)(a),\$_1}
{\$_0(@@b+1,\$_2)(a+1),\$_1} = ({\$_0(@@b,\$_2)}#{\$_0@@(a),\$_1}
{\$_0(@@b,\$_2)\$_3} = {\$_0(@@b[n],\$_2)\$_3}, if b is a lim ord
Apply basic ah rules otherwise to everything after the (@@\$) or everything inside the (@@\$) depending on what is necessary. See legion bar notes for more details.
(Now things such as ah_n{a(@@(a(@@a)a))a} are well defined.)

{\$_0@@@(a+1),\$_1} = {\$_0(@@(\$_0@@@a,\$_1))\$_0,\$_1}
{\$_0@@@0} = {\$_0}
{\$_0@@@a,\$_1} = {\$_0@@@a[n],\$_1}, a a lim ord.
Apply ah rules otherwise

#### Dimensional arrays: 2D

{\$_0[[1]]0} = {\$_0}
{\$_0[[1]]a+1,\$_1} = {\$_0@@@\$_0[[1]]a,\$_1}
{\$_0[[1]]a,\$_1} = {\$_0,a[[1]]a[n],\$_1}, a a lim ord
{\$_0[[1]]0@m,a+1,\$_1} = {\$_0[[1]](ah_n{\$_0[[1]](a+1)@m,a,\$_1})@m,a,\$_1}
{\$_0[[1]]0@m,a,\$_1} = {\$_0[[1]]a@m,a[n],\$_1}, a a lim ord

## ah v2

a serious revamp of ah. Most rules are preserved.

### Part one: Starter notations

define \$_n as any entries (including no entries) in an array. It’s my symbol for we-don’t-care entries. In any one use of any one rule, if n is the same, \$_n is the same

a#b = concatenation of a and b

n@m =
n, m = 1
{n}#(n@(m-1)), m > 1, m not a lim ord
n@(m[n]) if m is a lim ord

g(a,n,B) =
ah_g(a-1,n,B) {B}, a > 1;
n, a = 0

### Basic ah rules

Rule 1.
ah^a_n {\$_0} = g(a,n,\$_0)
Rule 2.
ah_n ({\$_1}#{z}) = ah_n {\$_1}, z = 0
Rule 3.
ah_n{} = n+1
Rule 4.
ah_n{a+1,\$_2} = ah^n_n{a,\$_2}
Rule 5.
ah_n{a,\$_3} = ah_n{a[n],\$_3}, a a lim ord
Rule 6.
ah_n ((0@b)#{a+1,\$_4}) = ah_(n+1) (((a+1)@b)#{a,\$_4}), b > 0
Rule 7.
ah_n ((0@b)#{a,\$_5}) = ah_n ((a@b)#{a[n],\$_5}), a a lim ord & b > 0

### Enormous arrays

if unspecified, n is the subscript in the first (closest) ah that is in front of the array.

{\$_0(@@b,\$_2)0} = {\$_0}
{\$_0((@@b,\$_2)a,\$_1} = {\$_0(@@b)a[n],\$_1},a a lim ord
{\$_0(@@0)(a+1),\$_1} = ({\$_0@}#{\$_0(@@0)(a),\$_1}
{\$_0(@@b+1,\$_2)(a+1),\$_1} = ({\$_0(@@b,\$_2)}#{\$_0@@(a),\$_1}
{\$_0(@@b,\$_2)\$_3} = {\$_0(@@b[n],\$_2)\$_3}, if b is a lim ord
Apply basic ah rules otherwise to everything after the (@@\$) or everything inside the (@@\$) depending on what is necessary. This includes nesting and incrementing the subscript.

{\$_0@@@(a+1),\$_1} = {\$_0(@@(\$_0@@@a,\$_1))\$_0,\$_1}
{\$_0@@@0} = {\$_0}
{\$_0@@@a,\$_1} = {\$_0@@@a[n],\$_1}, a a lim ord.
Apply ah rules otherwise

### Dimensional arrays

{\$_0[[k,\$_3]]0} = {\$_0}
{\$_0[[k+1,\$_3]]a+1,\$_1} = {\$_0[[k,\$_3]]\$_0[[k+1,\$_3]]a,\$_1}
{\$_0[[k,\$_3]]a+1,\$_1} = {\$_0[[(k[n],\$_3)]]\$_0[[k,\$_3]]a,\$_1}, k a lim ord
{\$_0[[0]]a+1,\$_1} = {\$_0@@@\$_0[[0]]a,\$_1}
{\$_0[[0,k+1,\$_3]]a+1,\$_1} = {\$_0@@@\$_0[[k+1,k,\$_3]]a+1,\$_1}
{\$_0[[0,k,\$_3]]a+1,\$_1} = {\$_0@@@\$_0[[k,k[n],\$_3]]a+1,\$_1}, k a lim ord
{\$_0[[k,\$_3]]a,\$_1} = {\$_0,a[[k,\$_3]]a[n],\$_1}, a a lim ord
{\$_0[[k,\$_3]]0@m,a+1,\$_1} = {\$_0[[k,\$_3]](ah_n{\$_0[[k,\$_3]](a+1)@m,a,\$_1})@m,a,\$_1}
{\$_0[[k,\$_3]]0@m,a,\$_1} = {\$_0[[k,\$_3]]a@m,a[n],\$_1}, a a lim ord

The leftmost of the highest-valued (by k) of the longest [[\$]] is evaluated first.

### New round of separators

Generalization from dimensions and (@@\$)

{\$_0[[\$_4,,k,\$_3]]0} = {\$_0}
{\$_0[[0,,\$_3]]\$_2} = {\$_0[[\$_3]]\$_2}
{\$_0[[\$_4,,k+1,\$_3]]a+1,\$_1} = {\$_0[[\$_4,,k,\$_3]]\$_0[[\$_4,,k+1,\$_3]]a,\$_1}
{\$_0[[\$_4,,k,\$_3]]a+1,\$_1} = {\$_0[[(\$_4,,k[n],\$_3)]]\$_0[[\$_4,,k,\$_3]]a,\$_1}, k a lim ord
{\$_0[[b+1,\$_4,,0]]a+1,\$_1} = {\$_0[[b,\$_4,,\$_0,\$_1]]\$_0[[b+1,\$_4,,0]]a,\$_1}
{\$_0[[b,\$_4,,0]]a+1,\$_1} = {\$_0[[b[n],\$_4,,\$_0,\$_1]]\$_0[[b,\$_4,,0]]a,\$_1}, b a lim ord
{\$_0[[\$_4,,0,k+1,\$_3]]a+1,\$_1} = {\$_0[[\$_4,,k+1,k,\$_3]]\$_0[[\$_4,,k+1,k,\$_3]]a+1,\$_1}
{\$_0[[\$_4,,0,k,\$_3]]a+1,\$_1} = {\$_0[[\$_4,,k,k[n],\$_3]]\$_0[[\$_4,,k,k[n],\$_3]]a+1,\$_1}, k a lim ord
{\$_0[[\$_4,,k,\$_3]]a,\$_1} = {\$_0,a[[\$_4,,k,\$_3]]a[n],\$_1}, a a lim ord
{\$_0[[\$_4,,k,\$_3]]0@m,a+1,\$_1} = {\$_0[[\$_4,,k,\$_3]](ah_n{\$_0[[\$_4,,k,\$_3]](a+1)@m,a,\$_1})@m,a,\$_1}
{\$_0[[\$_4,,k,\$_3]]0@m,a,\$_1} = {\$_0[[\$_4,,k,\$_3]]a@m,a[n],\$_1}, a a lim ord

Note: Since \$_4 is never duplicated, \$_4 can include ,, and similar separator-y things

Total power using nothing any larger than w in the arrays is currently at least Gamma_w, but that's very very very conservative

## ah v3

a more serious revamp of ah, beyond ,,

### Part one: Starter notations

define \$_n as any entries (including no entries) in an array. It’s my symbol for we-don’t-care entries. In any one use of any one rule, if n is the same, \$_n is the same

a#b = concatenation of a and b

n@m =
n, m = 1
{n}#(n@(m-1)), m > 1, m not a lim ord
n@(m[n]) if m is a lim ord

g(a,n,B) =
ah_g(a-1,n,B) {B}, a > 1;
n, a = 0

### Basic ah rules

Rule 1.
ah^a_n {\$_0} = g(a,n,\$_0)
Rule 2.
ah_n ({\$_1}#{z}) = ah_n {\$_1}, z = 0
Rule 3.
ah_n{} = n+1
Rule 4.
ah_n{a+1,\$_2} = ah^n_n{a,\$_2}
Rule 5.
ah_n{a,\$_3} = ah_n{a[n],\$_3}, a a lim ord
Rule 6.
ah_n ((0@b)#{a+1,\$_4}) = ah_(n+1) (((a+1)@b)#{a,\$_4}), b > 0
Rule 7.
ah_n ((0@b)#{a,\$_5}) = ah_n ((a@b)#{a[n],\$_5}), a a lim ord & b > 0

### Enormous arrays

if unspecified, n is the subscript in the first (closest) ah that is in front of the array.

{\$_0(@@b,\$_2)0} = {\$_0}
{\$_0((@@b,\$_2)a,\$_1} = {\$_0(@@b)a[n],\$_1},a a lim ord
{\$_0(@@0)(a+1),\$_1} = ({\$_0@}#{\$_0(@@0)(a),\$_1}
{\$_0(@@b+1,\$_2)(a+1),\$_1} = ({\$_0(@@b,\$_2)}#{\$_0@@(a),\$_1}
{\$_0(@@b,\$_2)\$_3} = {\$_0(@@b[n],\$_2)\$_3}, if b is a lim ord
Apply basic ah rules otherwise to everything after the (@@\$) or everything inside the (@@\$) depending on what is necessary. This includes nesting and incrementing the subscript.

{\$_0@@@(a+1),\$_1} = {\$_0(@@(\$_0@@@a,\$_1))\$_0,\$_1}
{\$_0@@@0} = {\$_0}
{\$_0@@@a,\$_1} = {\$_0@@@a[n],\$_1}, a a lim ord.
Apply ah rules otherwise

### Dimensional arrays

{\$_0[[k,\$_3]]0} = {\$_0}
{\$_0[[k+1,\$_3]]a+1,\$_1} = {\$_0[[k,\$_3]]\$_0[[k+1,\$_3]]a,\$_1}
{\$_0[[k,\$_3]]a+1,\$_1} = {\$_0[[(k[n],\$_3)]]\$_0[[k,\$_3]]a,\$_1}, k a lim ord
{\$_0[[0]]a+1,\$_1} = {\$_0@@@\$_0[[0]]a,\$_1}
{\$_0[[0,k+1,\$_3]]a+1,\$_1} = {\$_0@@@\$_0[[k+1,k,\$_3]]a+1,\$_1}
{\$_0[[0,k,\$_3]]a+1,\$_1} = {\$_0@@@\$_0[[k,k[n],\$_3]]a+1,\$_1}, k a lim ord
{\$_0[[k,\$_3]]a,\$_1} = {\$_0,a[[k,\$_3]]a[n],\$_1}, a a lim ord
{\$_0[[k,\$_3]]0@m,a+1,\$_1} = {\$_0[[k,\$_3]](ah_n{\$_0[[k,\$_3]](a+1)@m,a,\$_1})@m,a,\$_1}
{\$_0[[k,\$_3]]0@m,a,\$_1} = {\$_0[[k,\$_3]]a@m,a[n],\$_1}, a a lim ord

The leftmost of the highest-valued (by k) of the longest [[\$]] is evaluated first.

### New round of separators

Generalization from dimensions and (@@\$)

{\$_0[[\$_4,,k,\$_3]]0} = {\$_0}
{\$_0[[0,,\$_3]]\$_2} = {\$_0[[\$_3]]\$_2}
{\$_0[[\$_4,,k+1,\$_3]]a+1,\$_1} = {\$_0[[\$_4,,k,\$_3]]\$_0[[\$_4,,k+1,\$_3]]a,\$_1}
{\$_0[[\$_4,,k,\$_3]]a+1,\$_1} = {\$_0[[(\$_4,,k[n],\$_3)]]\$_0[[\$_4,,k,\$_3]]a,\$_1}, k a lim ord
{\$_0[[b+1,\$_4,,0]]a+1,\$_1} = {\$_0[[b,\$_4,,\$_0,\$_1]]\$_0[[b+1,\$_4,,0]]a,\$_1}
{\$_0[[b,\$_4,,0]]a+1,\$_1} = {\$_0[[b[n],\$_4,,\$_0,\$_1]]\$_0[[b,\$_4,,0]]a,\$_1}, b a lim ord
{\$_0[[\$_4,,0,k+1,\$_3]]a+1,\$_1} = {\$_0[[\$_4,,k+1,k,\$_3]]\$_0[[\$_4,,k+1,k,\$_3]]a+1,\$_1}
{\$_0[[\$_4,,0,k,\$_3]]a+1,\$_1} = {\$_0[[\$_4,,k,k[n],\$_3]]\$_0[[\$_4,,k,k[n],\$_3]]a+1,\$_1}, k a lim ord
{\$_0[[\$_4,,k,\$_3]]a,\$_1} = {\$_0,a[[\$_4,,k,\$_3]]a[n],\$_1}, a a lim ord
{\$_0[[\$_4,,k,\$_3]]0@m,a+1,\$_1} = {\$_0[[\$_4,,k,\$_3]](ah_n{\$_0[[\$_4,,k,\$_3]](a+1)@m,a,\$_1})@m,a,\$_1}
{\$_0[[\$_4,,k,\$_3]]0@m,a,\$_1} = {\$_0[[\$_4,,k,\$_3]]a@m,a[n],\$_1}, a a lim ord

Note: Since \$_4 is never duplicated, \$_4 can include ,, and similar separator-y things

Total power using nothing any larger than w in the arrays up to and through ,, at least Gamma_w, but that's very very very conservative

### ,,,

so honestly I was going to use something like the ,, system but the base of ,,, was like iterated ,, and the base of ,,,, was iterated ,,, ; but then I was like: hold it: I can do this all in one separator. so let me introduce you to ,,,:

{\$_0[[\$_4,,,i,k,\$_3]]0} = {\$_0}
{\$_0[[0,,,0,\$_3]]\$_2} = {\$_0[[\$_3]]\$_2}
{\$_0[[0,,,i+1,\$_3]]\$_2} = {\$_0[[\$_3,,,i,\$_3,,,i,\$_3,,,...]]\$_2} with n copies of ,,,i,\$_3 (n+1 total \$_3s)
{\$_0[[0,,,i,\$_3]]\$_2} = {\$_0[[0,,,i[n],\$_3]]\$_2}, i a lim ord
{\$_0[[\$_4,,,i,k+1,\$_3]]a+1,\$_1} = {\$_0[[\$_4,,,i,k,\$_3]]\$_0[[\$_4,,,i,k+1,\$_3]]a,\$_1}
{\$_0[[\$_4,,,i,k,\$_3]]a+1,\$_1} = {\$_0[[(\$_4,,,i,k[n],\$_3)]]\$_0[[\$_4,,,i,k,\$_3]]a,\$_1}, k a lim ord
{\$_0[[b+1,\$_4,,,i,0]]a+1,\$_1} = {\$_0[[b,\$_4,,,i,\$_0,\$_1]]\$_0[[b+1,\$_4,,,i,0]]a,\$_1}
{\$_0[[b,\$_4,,,i,0]]a+1,\$_1} = {\$_0[[b[n],\$_4,,,i,\$_0,\$_1]]\$_0[[b,\$_4,,,i,0]]a,\$_1}, b a lim ord
{\$_0[[\$_4,,,i,0,k+1,\$_3]]a+1,\$_1} = {\$_0[[\$_4,,,i,k+1,k,\$_3]]\$_0[[\$_4,,,i,k+1,k,\$_3]]a+1,\$_1}
{\$_0[[\$_4,,,i,0,k,\$_3]]a+1,\$_1} = {\$_0[[\$_4,,,i,k,k[n],\$_3]]\$_0[[\$_4,,,i,k,k[n],\$_3]]a+1,\$_1}, k a lim ord
{\$_0[[\$_4,,,i,k,\$_3]]a,\$_1} = {\$_0,a[[\$_4,,,i,k,\$_3]]a[n],\$_1}, a a lim ord
{\$_0[[\$_4,,,i,k,\$_3]]0@m,a+1,\$_1} = {\$_0[[\$_4,,,i,k,\$_3]](ah_n{\$_0[[\$_4,,,i,k,\$_3]](a+1)@m,a,\$_1})@m,a,\$_1}
{\$_0[[\$_4,,,i,k,\$_3]]0@m,a,\$_1} = {\$_0[[\$_4,,,i,k,\$_3]]a@m,a[n],\$_1}, a a lim ord

Once again, note that \$_4 is still not duplicated in the separators so we can have multiple ,,,i,'s in one separator.

This gets ah much farther. I would say at least ackermann's ordinal but probably nearing if not at the SVO, if you use w as the largest ordinal you can use.