Yeh, I dont’ wanna touch that yet. I’d rather use yours.Moosey wrote: ↑March 3rd, 2020, 9:36 amActually this use of sup is questionabletestitemqlstudop wrote: ↑March 2nd, 2020, 6:54 amE(0) = 0

E($-0) = $ + 1, if $ is not in parentheses

E((0-a)-0) = sup{a,a-a,a-a-a,a-a-a-a,etc}

E(x) = limit of E if all nodes are <= x-1

E($-x) = E(($-(x-1))-($-(x-1))-(x-1)) + 1, if $ not in parentheses

E((x-$)-x) = sup{((x-1)-$)-(x-1), ((x-1)-$)-((x-1)-$)-(x-1), ((x-1)-$)-((x-1)-$)-((x-1)-$)-(x-1), etc}

## Ordinals in googology

### Re: Ordinals in googology

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- Moosey
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### Re: Ordinals in googology

Thought of the day

Username5243 mentions a possibility for Uncountable FSes in UNOCF. (Note: UNOCF is known for being illdefined/hated).

So that means

A transfinite HH

In a nutshell

f_a(b) is defined (or undefined) thusly:

f_0(b) = b

f_(a+1)(b) = f_a(b+1)

f_a(b) = f_a[{b}](b), when a is a limit ordinal with cof >= b

f_a(b) = b, b > cof(a)

So now

What is [{}] ?

Well it's an extension to FSes obviously

Basically

a[{b}] is defined if cof(a) > b

a[{b}] is like a[ b] (space to avoid bbcode) except you apply all sorts of FS rules to all regulars (Insert unformalized frustration here)

Ex

Ω2[{w+3}] = Ω+Ω[{w+3}] = Ω+w+3

Ω_7[{Ω_6}] = Ω_6

I can't formalize it unfortunately.

But for instance

f_W(w+6) = f_w+6(w+6) = f_w(w+12) = w+12

f_I(M) = M

f_W+6(w) = f_W(w+6) = w+12

etc

Probably could be made more useful, or more interesting at least, if someone found a better version of the b > cof(a) rule

Username5243 mentions a possibility for Uncountable FSes in UNOCF. (Note: UNOCF is known for being illdefined/hated).

So that means

A transfinite HH

In a nutshell

f_a(b) is defined (or undefined) thusly:

f_0(b) = b

f_(a+1)(b) = f_a(b+1)

f_a(b) = f_a[{b}](b), when a is a limit ordinal with cof >= b

f_a(b) = b, b > cof(a)

So now

What is [{}] ?

Well it's an extension to FSes obviously

Basically

a[{b}] is defined if cof(a) > b

a[{b}] is like a[ b] (space to avoid bbcode) except you apply all sorts of FS rules to all regulars (Insert unformalized frustration here)

Ex

Ω2[{w+3}] = Ω+Ω[{w+3}] = Ω+w+3

Ω_7[{Ω_6}] = Ω_6

I can't formalize it unfortunately.

But for instance

f_W(w+6) = f_w+6(w+6) = f_w(w+12) = w+12

f_I(M) = M

f_W+6(w) = f_W(w+6) = w+12

etc

Probably could be made more useful, or more interesting at least, if someone found a better version of the b > cof(a) rule

Last edited by Moosey on March 4th, 2020, 10:11 am, edited 1 time in total.

I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

### Re: Ordinals in googology

I just realized the original definition of E($) uses trees with only zeroes. So maybe we could shorten the notation. Meanwhile, I'll just use the zeroes notation.

∃(0) = 0

∃(ø) = (0-∃(ø[0]))-0

∃(x+1) = ∃(x)-0

I’ll test it.

∃(w) = (0-∃(0))-0 = (0-0)-0

E((0-0)-0) = sup{0,0-0,0-0-0,...} = sup{0,1,2,3,...} = w

∃(w^2+1) = ∃(w^2)-0 = (0-∃(w))-0-0 = (0-(0-0)-0)-0-0

E((0-(0-0)-0)-0-0) = E((0-(0-0)-0)-0) + 1 = sup{(0-0)-0,none of this makes sense}

Janskchdhwhjshsjssahjhsjchcjchchchhhd I do not want to touch the big one

∃(0) = 0

∃(ø) = (0-∃(ø[0]))-0

∃(x+1) = ∃(x)-0

I’ll test it.

∃(w) = (0-∃(0))-0 = (0-0)-0

E((0-0)-0) = sup{0,0-0,0-0-0,...} = sup{0,1,2,3,...} = w

∃(w^2+1) = ∃(w^2)-0 = (0-∃(w))-0-0 = (0-(0-0)-0)-0-0

E((0-(0-0)-0)-0-0) = E((0-(0-0)-0)-0) + 1 = sup{(0-0)-0,none of this makes sense}

Janskchdhwhjshsjssahjhsjchcjchchchhhd I do not want to touch the big one

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### Re: Ordinals in googology

This can be simplified a bittestitemqlstudop wrote: ↑March 2nd, 2020, 6:54 amE(0) = 0

E($-0) = $ + 1, if $ is not in parentheses

E((0-a)-0) = sup{a,a-a,a-a-a,a-a-a-a,etc}

E(x) = limit of E if all nodes are <= x-1

E($-x) = E(($-(x-1))-($-(x-1))-(x-1)) + 1, if $ not in parentheses

E((x-$)-x) = sup{((x-1)-$)-(x-1), ((x-1)-$)-((x-1)-$)-(x-1), ((x-1)-$)-((x-1)-$)-((x-1)-$)-(x-1), etc}

E(0) = 0

E($-0) = $ + 1

E((0-a)) = sup{a,a-a,a-a-a,a-a-a-a,etc}

E(x) = limit of E if all nodes are < x

E($-(x+1)) = E(($-x)-($-x)-x) + 1, if $ not in parentheses

E(((x+1)-$)-(x+1)) = sup{(x-$)-x, (x-$)-(x-$)-x, (x-$)-(x-$)-(x-$)-x, etc}

This makes it considerably more legible, of course at the price of trees- (0-0) is the same tree as 0-0, but of course it's still a notation

I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- Moosey
**Posts:**3285**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
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### Re: Ordinals in googology

As I'm incredibly bored, I think I'll describe an ordinal-inspired array system like TAN.

So

a is a number

a{} = a

a{$,0} = a{$}

a{b} = a+b

a{$,b,c+1} = (a+1){$,(a+1){$,(a+1){$,(a+1){...(a+1){$,b,c}...},c},c},c}) with a copies of a+1

a{$,w} = a{$,a}

So for instance

2{2,1} = 3{3{2}} = 3^9

Not much, okay

2{2,2} = 3{3{2,1},1} = 3{4{4{4{2}}},1} = 3{4^4^16,1} = 4^4^4^4^4^16

Still not much

What about

2{w,w}?

Well...

3{3{w,1},1} = 3{4{4{4{4}}},1}

Still not very good, okay

But that's only the first part!

Now for this part

a{$,b,,c} = a{$,b,b,b,...b} w/ c bs

3{w,,w} = 3{w,w,w} = 3{w,w,3} = a whole darn lot

But what's much bigger

Is 3{w,,w,,w}

Which is ultimately: 3{w,w,w,w,w,w,w.........with lots of ws...w,w,w,w}

Of course

That means we can generalize like with ExE (come to think of it this is like ExE)

meaning

a{$,b,^c!d} = a{$,b,^(c-1)!b,^(c-1)!b,...} with d bs

w of course becomes a, but only when it clogs the system

In fact

Let's call this separator notation, SeN

In SeN, let % be a generic separator

%^c = %%%%%% with c %

! is a separator needed for generic separation when , is not appropriate

Brackets are used if necessary to show what is being applied to by separator ops

So for instance

3{w,^w!w} is 3{w,,,w} which is 3{w,,w,,w} which is, as stated, VERY large

now

3{w,^w!w,^w!w} = 3{w,,,,,....,,,,,,,,w} with about 3{w,,w,,w} ,s

So now let's have more fun

Since w "clogs the system" only when left alone, and we can subtract one from w+1, which is technically not disallowed, ...

3{w,^(w+1)!w} = that

3{w,^(w+2)!w} = about that many w,^w!s

Next

%^^ is shorthand for [%^]^

So

a,[^]^m+1!n+1!b = a,[^]^m!b{a,[^]^m+1!^n!b}!a

a,[^]^1!b!c = a,^b!c

3{w,[^]^w!w!w} is big

So

a is a number

a{} = a

a{$,0} = a{$}

a{b} = a+b

a{$,b,c+1} = (a+1){$,(a+1){$,(a+1){$,(a+1){...(a+1){$,b,c}...},c},c},c}) with a copies of a+1

a{$,w} = a{$,a}

So for instance

2{2,1} = 3{3{2}} = 3^9

Not much, okay

2{2,2} = 3{3{2,1},1} = 3{4{4{4{2}}},1} = 3{4^4^16,1} = 4^4^4^4^4^16

Still not much

What about

2{w,w}?

Well...

3{3{w,1},1} = 3{4{4{4{4}}},1}

Still not very good, okay

But that's only the first part!

Now for this part

a{$,b,,c} = a{$,b,b,b,...b} w/ c bs

3{w,,w} = 3{w,w,w} = 3{w,w,3} = a whole darn lot

But what's much bigger

Is 3{w,,w,,w}

Which is ultimately: 3{w,w,w,w,w,w,w.........with lots of ws...w,w,w,w}

Of course

That means we can generalize like with ExE (come to think of it this is like ExE)

meaning

a{$,b,^c!d} = a{$,b,^(c-1)!b,^(c-1)!b,...} with d bs

w of course becomes a, but only when it clogs the system

In fact

Let's call this separator notation, SeN

In SeN, let % be a generic separator

%^c = %%%%%% with c %

! is a separator needed for generic separation when , is not appropriate

Brackets are used if necessary to show what is being applied to by separator ops

So for instance

3{w,^w!w} is 3{w,,,w} which is 3{w,,w,,w} which is, as stated, VERY large

now

3{w,^w!w,^w!w} = 3{w,,,,,....,,,,,,,,w} with about 3{w,,w,,w} ,s

So now let's have more fun

Since w "clogs the system" only when left alone, and we can subtract one from w+1, which is technically not disallowed, ...

3{w,^(w+1)!w} = that

3{w,^(w+2)!w} = about that many w,^w!s

Next

%^^ is shorthand for [%^]^

So

a,[^]^m+1!n+1!b = a,[^]^m!b{a,[^]^m+1!^n!b}!a

a,[^]^1!b!c = a,^b!c

3{w,[^]^w!w!w} is big

I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- Moosey
**Posts:**3285**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
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### Re: Ordinals in googology

I challenge someone to analyse my notation (ASEN, or Applied SEparator Notation) above

a{} = a

a{$,0} = a{$}

a{b} = a+b

a{$,b,c+1} = (a+1){$,(a+1){$,(a+1){$,(a+1){...(a+1){$,b,c}...},c},c},c}) with a copies of a+1

a{$,w} = a{$,a}

a{$,b,,c} = a{$,b,b,b,...b} w/ c bs

a{$,b,^c!d} = a{$,b,^(c-1)!b,^(c-1)!b,...} with d bs

a,[^]^m+1!n+1!b = a,[^]^m!b{a,[^]^m+1!^n!b}!a

a,[^]^1!b!c = a,^b!c

In SeN, let % be a generic separator

%^c = %%%%%% with c %

! is a separator needed for generic separation when , is not appropriate

Brackets are used if necessary to show what is being applied to by separator ops

a{} = a

a{$,0} = a{$}

a{b} = a+b

a{$,b,c+1} = (a+1){$,(a+1){$,(a+1){$,(a+1){...(a+1){$,b,c}...},c},c},c}) with a copies of a+1

a{$,w} = a{$,a}

a{$,b,,c} = a{$,b,b,b,...b} w/ c bs

a{$,b,^c!d} = a{$,b,^(c-1)!b,^(c-1)!b,...} with d bs

a,[^]^m+1!n+1!b = a,[^]^m!b{a,[^]^m+1!^n!b}!a

a,[^]^1!b!c = a,^b!c

In SeN, let % be a generic separator

%^c = %%%%%% with c %

! is a separator needed for generic separation when , is not appropriate

Brackets are used if necessary to show what is being applied to by separator ops

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

### Re: Ordinals in googology

Without w

3{3,3} = 363
Without ,,

3{w,w} = 366
Without !

3{w,,w} = a super-absurdly large number
Ok that got large quick

3{3,3} = 363

Code: Select all

```
3{3,3} = 4{4{4{3,2},2},2} = 4{4{123,2},2} = 4{243,2} = 363
4{3,2} = 5{5{5{5{3,1},1},1},1} = 5{5{5{33,1},1},1} = 5{5{63,1},1} = 5{93,1} = 123
5{3,1} = 6{6{6{6{6{3,0},0},0},0},0} = 6{6{6{6{6{3}}}}} = 33
5{33,1} = 6{6{6{6{6{33,0},0},0},0},0} = 6{6{6{6{6{33}}}}} = 63
4{123,2} = 5{5{5{5{123,1},1},1},1} = 5{5{5{153,1},1},1} = 5{5{183,1},1} = 5{213,1} = 243
```

3{w,w} = 366

Code: Select all

```
3{w,w} = 3{w,3} = 4{4{4{w,2},2},2} = 4{4{126,2},2} = 4{246,2} = 366
4{w,2} = 5{5{5{5{w,1},1},1},1} = 5{5{5{36,1},1},1} = 5{5{66,1},1} = 5{96,1} = 126
5{w,1} = 6{6{6{6{6{w,0},0},0},0},0} = 6{6{6{6{6{w}}}}} = 6{6{6{6{6{6}}}}} = 36
```

3{w,,w} = a super-absurdly large number

Code: Select all

```
3{w,,w} = 3{w,,3} = 3{w,w,w} = 3{w,w,3} = 4{w,4{w,4{w,w,2},2},2}
4{w,w,2} = 5{w,5{w,5{w,5{w,w,1},1},1},1}
5{w,w,1} = 6{w,6{w,6{w,6{w,6{w,w,0},0},0},0},0} = 6{w,6{w,6{w,6{w,6{w,w}}}}} = 6{w,6{w,6{w,6{w,6{w,6}}}}} = 6{w,6{w,6{w,6{w,3991692}}}} = an absurdly large number
6{w,6} = 7{7{7{7{7{7{w,5},5},5},5},5},5} = 3991692
7{w,5} = 8{8{8{8{8{8{8{w,4},4},4},4},4},4},4} = 665292
8{w,4} = 9{9{9{9{9{9{9{9{w,3},3},3},3},3},3},3},3} = 9{9{9{9{9{9{9{11892,3},3},3},3},3},3},3} = 11880*7+11892
9{w,3} = 10{10{10{10{10{10{10{10{10{w,2},2},2},2},2},2},2},2},2} = 95052
10{10{10{10{10{10{10{10{1332,2},2},2},2},2},2},2},2} = 10{10{10{10{10{10{10{2652,2},2},2},2},2},2},2} = 11892
10{w,2} = 11{11{11{11{11{11{11{11{11{11{w,1},1},1},1},1},1},1},1},1},1} = 11{11{11{11{11{11{11{11{11{144,1},1},1},1},1},1},1},1},1} = 11{11{11{11{11{11{11{11{276,1},1},1},1},1},1},1},1} = 11{11{11{11{11{11{11{408,1},1},1},1},1},1},1} = 1332
11{w,1} = 12{12{12{12{12{12{12{12{12{12{12{w}}}}}}}}}}} = 12{12{12{12{12{12{12{12{12{12{12{12}}}}}}}}}}} = 144
11{144,1} = 12{12{12{12{12{12{12{12{12{12{12{144}}}}}}}}}}} = 276
```

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### Re: Ordinals in googology

Do 3{w[^]^w!w!w}

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

### Re: Ordinals in googology

If this is accidentally posted, do not reply until it is deleted.

Without [^]^

3{w,^w!w} =
Without w+n (as requested by Moosey)

3{w,[^]^w!w!w} = add some curly brackets please i cannot comprehend it

Without [^]^

3{w,^w!w} =

Code: Select all

```
3{w,^w!w} = 3{w,^3!3} = 3{w,^2!w,^2!w,^2 does this end with } or !w}
```

3{w,[^]^w!w!w} = add some curly brackets please i cannot comprehend it

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`Function(‘a’+’lert(1)’)()`

- Moosey
**Posts:**3285**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
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### Re: Ordinals in googology

screw the above post

I want to share this REAL bad

So, googology discord user Despacit2.0 and I conceived "ARE notation"
And we've made an analysis sheet

but we've gotten stuck at zw = sup{(A)[(R)(E)]E, (A)[([(R)(E)]E)(E)]E, (A)[([([(R)(E)]E)(E)]E)(E)]E...}

does anyone have an idea for an extension?

I want to share this REAL bad

So, googology discord user Despacit2.0 and I conceived "ARE notation"

Code: Select all

```
(x)A = x + 1
(x)R = ((...(x)x)x)...
(x)(y)(z)(.....)(a)(b)R = ((x)(y)(z)(.....)(((...(a)...b)b)b)
(...)[#]E = (...)(#########... (the last parenthesis pair is deleted)
```

but we've gotten stuck at zw = sup{(A)[(R)(E)]E, (A)[([(R)(E)]E)(E)]E, (A)[([([(R)(E)]E)(E)]E)(E)]E...}

does anyone have an idea for an extension?

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- Moosey
**Posts:**3285**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

### Re: Ordinals in googology

bump

update: we're at phi(4,0)

hopefully we'll get to phi(w,0) and then g0 soon

update: we're at phi(4,0)

hopefully we'll get to phi(w,0) and then g0 soon

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"