# 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.