Difference between revisions of "User:Moosey/Googology"

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(→‎,,,: modified so that the index is clearer)
(→‎,,,: now supporting arrays before :)
Line 426: Line 426:
 
I can do this all in one separator. so let me introduce you to ,,,:
 
I can do this all in one separator. so let me introduce you to ,,,:
  
<nowiki>{$_0[[$_4,,,i:k,$_3]]0} = {$_0}
+
<nowiki>{$_0[[$_4,,,i,$_5:k,$_3]]0} = {$_0}
 
</nowiki><br><nowiki>
 
</nowiki><br><nowiki>
 
{$_0[[0,,,0:$_3]]$_2} = {$_0[[$_3]]$_2}
 
{$_0[[0,,,0:$_3]]$_2} = {$_0[[$_3]]$_2}
 
</nowiki><br><nowiki>
 
</nowiki><br><nowiki>
{$_0[[0,,,i+1:$_3]]$_2} = {$_0[[$_3,,,i:$_3,,,i:$_3,,,...]]$_2} with n copies of ,,,i:$_3 (n+1 total $_3s)
+
{$_0[[0,,,i+1,$_5:$_3]]$_2} = {$_0[[$_3,,,i,$_5:$_3,,,i,$_5:$_3,,,...]]$_2} with n copies of ,,,i:$_3 (n+1 total $_3s)
 
</nowiki><br><nowiki>
 
</nowiki><br><nowiki>
{$_0[[0,,,i:$_3]]$_2} = {$_0[[0,,,i[n]:$_3]]$_2}, i a lim ord
+
{$_0[[0,,,i,$_5:$_3]]$_2} = {$_0[[0,,,i[n],$_5:$_3]]$_2}, i a lim ord
 
</nowiki><br><nowiki>
 
</nowiki><br><nowiki>
{$_0[[$_4,,,i:k+1,$_3]]a+1,$_1} = {$_0[[$_4,,,i:k,$_3]]$_0[[$_4,,,i:k+1,$_3]]a,$_1}
+
 
 +
{$_0[[0,,,0@m,i+1,$_5:$_3]]$_2} = {$_0[[$_3,,,(i+1)@m,i,$_5:$_3,,,(i+1)@m,i,$_5:$_3,,,...]]$_2} with n copies of ,,,i:$_3 (n+1 total $_3s)
 +
</nowiki><br><nowiki>
 +
{$_0[[0,,,0@m,i,$_5:$_3]]$_2} = {$_0[[0,,,i@m,i[n],$_5:$_3]]$_2}, i a lim ord
 +
</nowiki><br><nowiki>
 +
{$_0[[0,,,$_5,0:$_3]]$_2} = {$_0[[0,,,$_5:$_3]]$_2}
 +
</nowiki><br><nowiki>
 +
 
 +
{$_0[[$_4,,,i,$_5:k+1,$_3]]a+1,$_1} = {$_0[[$_4,,,i,$_5:k,$_3]]$_0[[$_4,,,i,$_5:k+1,$_3]]a,$_1}
 
</nowiki><br><nowiki>
 
</nowiki><br><nowiki>
{$_0[[$_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[[$_4,,,i,$_5:k,$_3]]a+1,$_1} = {$_0[[($_4,,,i,$_5:k[n],$_3)]]$_0[[$_4,,,i,$_5:k,$_3]]a,$_1}, k a lim ord
 
</nowiki><br><nowiki>
 
</nowiki><br><nowiki>
{$_0[[b+1,$_4,,,i:0]]a+1,$_1} = {$_0[[b,$_4,,,i:$_0,$_1]]$_0[[b+1,$_4,,,i:0]]a,$_1}
+
{$_0[[b+1,$_4,,,i,$_5:0]]a+1,$_1} = {$_0[[b,$_4,,,i,$_5:$_0,$_1]]$_0[[b+1,$_4,,,i,$_5:0]]a,$_1}
 
</nowiki><br><nowiki>
 
</nowiki><br><nowiki>
{$_0[[b,$_4,,,i:0]]a+1,$_1} = {$_0[[b[n],$_4,,,i:$_0,$_1]]$_0[[b,$_4,,,i:0]]a,$_1}, b a lim ord
+
{$_0[[b,$_4,,,i,$_5:0]]a+1,$_1} = {$_0[[b[n],$_4,,,i,$_5:$_0,$_1]]$_0[[b,$_4,,,i,$_5:0]]a,$_1}, b a lim ord
 
</nowiki><br><nowiki>
 
</nowiki><br><nowiki>
{$_0[[$_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[[$_4,,,i,$_5:0,k+1,$_3]]a+1,$_1} = {$_0[[$_4,,,i,$_5:k+1,k,$_3]]$_0[[$_4,,,i,$_5:k+1,k,$_3]]a+1,$_1}
 
</nowiki><br><nowiki>
 
</nowiki><br><nowiki>
{$_0[[$_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[[$_4,,,i,$_5:0,k,$_3]]a+1,$_1} = {$_0[[$_4,,,i,$_5:k,k[n],$_3]]$_0[[$_4,,,i,$_5:k,k[n],$_3]]a+1,$_1}, k a lim ord
 
</nowiki><br><nowiki>
 
</nowiki><br><nowiki>
{$_0[[$_4,,,i:k,$_3]]a,$_1} = {$_0,a[[$_4,,,i:k,$_3]]a[n],$_1}, a a lim ord
+
{$_0[[$_4,,,i,$_5:k,$_3]]a,$_1} = {$_0,a[[$_4,,,i,$_5:k,$_3]]a[n],$_1}, a a lim ord
 
</nowiki><br><nowiki>
 
</nowiki><br><nowiki>
{$_0[[$_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,$_5:k,$_3]]0@m,a+1,$_1} = {$_0[[$_4,,,i,$_5:k,$_3]](ah_n{$_0[[$_4,,,i,$_5:k,$_3]](a+1)@m,a,$_1})@m,a,$_1}
 
</nowiki><br><nowiki>
 
</nowiki><br><nowiki>
{$_0[[$_4,,,i:k,$_3]]0@m,a,$_1} = {$_0[[$_4,,,i:k,$_3]]a@m,a[n],$_1}, a a lim ord</nowiki>
+
{$_0[[$_4,,,i,$_5:k,$_3]]0@m,a,$_1} = {$_0[[$_4,,,i,$_5:k,$_3]]a@m,a[n],$_1}, a a lim ord</nowiki>
 
<br>
 
<br>
 
<br>
 
<br>
Line 456: Line 464:
 
Once again, note that $_4 is still not duplicated in the separators so we can have multiple ,,,i:'s in one separator.
 
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.
+
This gets ah ''much farther''. I would say at least ackermann's ordinal, if you use w as the largest ordinal you can use.

Revision as of 21:33, 27 January 2020

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,$_5:k,$_3]]0} = {$_0}
{$_0[[0,,,0:$_3]]$_2} = {$_0[[$_3]]$_2}
{$_0[[0,,,i+1,$_5:$_3]]$_2} = {$_0[[$_3,,,i,$_5:$_3,,,i,$_5:$_3,,,...]]$_2} with n copies of ,,,i:$_3 (n+1 total $_3s)
{$_0[[0,,,i,$_5:$_3]]$_2} = {$_0[[0,,,i[n],$_5:$_3]]$_2}, i a lim ord
{$_0[[0,,,0@m,i+1,$_5:$_3]]$_2} = {$_0[[$_3,,,(i+1)@m,i,$_5:$_3,,,(i+1)@m,i,$_5:$_3,,,...]]$_2} with n copies of ,,,i:$_3 (n+1 total $_3s)
{$_0[[0,,,0@m,i,$_5:$_3]]$_2} = {$_0[[0,,,i@m,i[n],$_5:$_3]]$_2}, i a lim ord
{$_0[[0,,,$_5,0:$_3]]$_2} = {$_0[[0,,,$_5:$_3]]$_2}
{$_0[[$_4,,,i,$_5:k+1,$_3]]a+1,$_1} = {$_0[[$_4,,,i,$_5:k,$_3]]$_0[[$_4,,,i,$_5:k+1,$_3]]a,$_1}
{$_0[[$_4,,,i,$_5:k,$_3]]a+1,$_1} = {$_0[[($_4,,,i,$_5:k[n],$_3)]]$_0[[$_4,,,i,$_5:k,$_3]]a,$_1}, k a lim ord
{$_0[[b+1,$_4,,,i,$_5:0]]a+1,$_1} = {$_0[[b,$_4,,,i,$_5:$_0,$_1]]$_0[[b+1,$_4,,,i,$_5:0]]a,$_1}
{$_0[[b,$_4,,,i,$_5:0]]a+1,$_1} = {$_0[[b[n],$_4,,,i,$_5:$_0,$_1]]$_0[[b,$_4,,,i,$_5:0]]a,$_1}, b a lim ord
{$_0[[$_4,,,i,$_5:0,k+1,$_3]]a+1,$_1} = {$_0[[$_4,,,i,$_5:k+1,k,$_3]]$_0[[$_4,,,i,$_5:k+1,k,$_3]]a+1,$_1}
{$_0[[$_4,,,i,$_5:0,k,$_3]]a+1,$_1} = {$_0[[$_4,,,i,$_5:k,k[n],$_3]]$_0[[$_4,,,i,$_5:k,k[n],$_3]]a+1,$_1}, k a lim ord
{$_0[[$_4,,,i,$_5:k,$_3]]a,$_1} = {$_0,a[[$_4,,,i,$_5:k,$_3]]a[n],$_1}, a a lim ord
{$_0[[$_4,,,i,$_5:k,$_3]]0@m,a+1,$_1} = {$_0[[$_4,,,i,$_5:k,$_3]](ah_n{$_0[[$_4,,,i,$_5:k,$_3]](a+1)@m,a,$_1})@m,a,$_1}
{$_0[[$_4,,,i,$_5:k,$_3]]0@m,a,$_1} = {$_0[[$_4,,,i,$_5: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, if you use w as the largest ordinal you can use.