My number is bigger!

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WarDaft
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Re: My number is bigger!

Postby WarDaft » Sun Dec 14, 2014 11:53 am UTC

You're also assuming that the ordinal collapse functions constructed made full use of the recursive structure of the larger ordinals used. That's not necessarily the case. Just because someone may have used אbar doesn't mean an equivalent collapse function could not have been written with אfoo or even smaller.

The Bachmann Howard ordinal, for example, does not require ω1 or even ω1CK. Just using Ω = εψ(ε_{ω_1+1})+1 would be enough to make the collapse work correctly.
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Re: My number is bigger!

Postby Amida » Thu Jan 15, 2015 10:28 pm UTC

It's been nearly two years since I posted my first little thing. I haven't followed this thread much since then either. But what I have done is every few months I'd try to come up with something new on my own that I haven't posted.

In my last post I defined up to chained arrow operations on the function itself to basically be a way of describing plugging the output of a function into itself an obscene number of times. For those who don't want to look it up my last post...

Spoiler:
If we have an arbitrary function f(x), f2(x) = f(f(x)) and fn(x) = f(fn-1(x)). This can be written with up arrows is f^n(x) or chained arrows f→n→1(x).

Moving on, f^^2(x) = ff(x)(x) or x fed into f(x), f(x) times. Extending that brings us to f^^n(x) = f^f^^n-1(x)(x) in chained arrow form this is f→n→2(x).

Extending the pattern based on up arrows and chained arrows, we can have f→m→n(x)


With that out of the way, I can move on to describing newer ideas I've had. Any function I define here can be extended by giving it an additional argument beyond what was originally defined. Mathematically the extended function works like this.

Extending functions:
Spoiler:
If f(x) = whatever then f(x,y) = f→y→y(x) and it can be extended at whim to f(x1, x2,...,xn) = f→xn→xn(x1,x2,...,xn-1).


Subscripts on functions:
Spoiler:
Moving on even further, when extending the function gets too tedious to write it all out, a subscript 2 can be added to the function and where the value x is the number of 4s plugged into the original.
f2(7) = f(4,4,4,4,4,4,4). For those who with to plug in many 4s into a sub 2, you are in luck. You can make it a 3 and extend that sub 2 from here to (something really far away). This works to infinity following the rule fn(x) = fn-1(4,4,...,4) for x 4s.


Multiple subscripts:
Spoiler:
For those who like big numbers going into the subscript, but thinks it looks silly with so much way down there, you can have a 1,2 in subscript where the value of x in the function takes the place of the 1 and the two drops with a 4 plugged into the function(f1,2(9001) = f9001(4))

The power of allowing that doesn't come from just allowing a 1 and a 2, but that you can go on to increase the 1 to a number as high as you like again and following the rule fn, 2(x) = fn-1, 2(4,4,...,4) for x 4s. Same as before, just with a new 2. When you're first spot gets too high again, the 2 can become a 3, a 4, a ... onto infinity. As the second space grows too large, another space can be added. (f1,1,2(9001) = f1,9001(4)= f4,9000(4) = f3,9000(4,4,4,4)= f3,9000→4→4(4,4,4)) eventually that will work down to f3,9000(something obscene) which then becomes f2,9000(4,4,...some obscene number of 4s, 4). At this point, we have a pattern that we can extend these subscripts onto an arbitrarily large number of spaces. Just note that for any slot to work its magic, the one before it must be a 1.


Functions in series:
Spoiler:
When there are too many subscripts, the next function in a series may be given to where the argument of the function is how many subscript slots there are on the function before it in the series with the final slot being a 4. With the alphabetical theme, the most obvious choice to have the next in series be is g(x) and again defaulting to a 4 being plugged in. So g(5) = f1,1,1,1,4(4). After g we can have h(x), i(x) and before f there might be an e(x).

Bracket functions:
Spoiler:
Moving on to brighter pastures, because there are only so many letters, lets switch to numbers. [1](x) (I'll leave undefined for now) is a function. The next in sequence is [2](x)(e.g. [2](7) = [1]1,1,1,1,1,1,4(4)),[3](x) and on to infinity.

Extended Bracket Functions:
Spoiler:
Eventually, that number is going to be big and why not take a page from one of our earlier books. And introduce [2,1](x) = [x](4). The next in sequence after that is [2,2](x) = [2,1]1,1,...,4(4). And after grinding through to get something of the form [2,1](big) we can change to [big](4). When we get far enough into the [2,N](x) sequence, we can switch to [3,1](x) =[2,x](4)= [2,x-1]1,1,1,4(4). This works for [M,N](x) = [M,N-1]...(4)=...[M,1](wow) = [M-1, wow](4). When M gets large, we can keep extending to the left as needed. [x1,x2,...,xn](y).


Matrix Functions:
Spoiler:
I don't like writing out a lot of numbers. So I've included rules that allow you to move on to a second row. With my skill level of forum posting, I'll represent this like so.
[1]
[2](x).
Not pretty, but it works.
When you have a 2 at the bottom, the value of x is how many 4s will fill up the top row.
[1]
[2](5) = [4,4,4,4,4](4) (Just assume a 4 is being plugged into each new function being made. It saves time.)
Next in sequence is
[2]
[2](x) =
[1]
[2]...(4)
We can even have
[2,1]
[2](x) =
[x]
[2](4)
Along with
[3,1]
[2](x) =
[2, x]
[2](4) =
[2, x-1]
[2]...(4)
And when you've filled out the top row too far to the left again, you're allowed to make the 2 on bottom a 3.
[1]
[3](5) =
[4,4,4,4,4]
[2](4)
The 3 can become a 4, 5, ... as needed. When that gets too big, extend the second row to the left as needed following the same rules that every element preceding must be a 1 before that one is allowed to do anything.
Second row growing too large? Add a third.
[1]
[1]
[2](7) =
[1]
[4,4,4,4,4,4,4](4) =
[4,4,4,4]
[4,4,4,4,4,4,3](4) =
[4,4,4,3]
[4,4,4,4,4,4,3]...(4)
After this, as many rows as needed can be added.

Higher Dimensional Matrix Functions:
Spoiler:
When that gets too large, another dimension can be added. I'll represent the next dimension as an extra set side by side.
[1][2](x)
eg
[1][2](3) =
[4]
[4]
[4](4)
This can move on to the form
[m]
[n][o](x) and again you must simply all preceding elements to 1 before using the next 1.
You're third dimensional area can grow as needed, extending to the left or adding rows. When it gets too large, another bit of depth may be added [1][1][2](x) and more on to infinity.

When you have too much depth in the third dimension, you can try adding a 4th dimension where a two in the 4d spot represents how deep to go into three with 4s. (I don't know a good way to represent adding the 4th dimension without risking confusion, so take my word for a 1 and then a 2 off in 4d land of 5) = [4][4][4][4][4](4).
Then we can have as many dimensions as we like beyond this.


Too Many Dimensions?:
Spoiler:
For convenience, I'm introducing a(x,y) where there are x dimensions all of y length filled with 4s of 4.
examples:
a(1,2) = [4,4](4)
a(1,5) = [4,4,4,4,4](4)
a(2,3) =
[4,4,4]
[4,4,4]
[4,4,4] (4)
a(3,2) =
[4,4] [4,4]
[4,4] [4,4](4)
a(4,2) = (If you have a suggestion for how to show 4d I'll take it)(4)

Hopefully helpful so far.
Spoiler:
At this point there is a lot to take in. So I'll run through a bit of an example and hope it helps someone out.
Take a(3,2), not the worst that a could be.
That is equal to
[4,4][4,4]
[4,4][4,4](4)
I think of the first set as the first depth (if anyone knows the 3d equivalent of columns and rows, please share)
So we have to take the leading 4 and reduce it to a 3, resulting in
[4,3][4,4]
[4,4][4,4]1,1,1,4(4)
Carrying on with what we have, and reducing a bit
[4,3][4,4]
[4,4][4,4]3,3,3,3(4,4,4,4)
Still moving on...
[4,3][4,4]
[4,4][4,4]3,3,3,3→4→4(4,4,4)
Simplifying that following our chained arrow rules, eventually we will get
[4,3][4,4]
[4,4][4,4]3,3,3,3large power tower(4,4,4)
And still following them
[4,3][4,4]
[4,4][4,4]3,3,3,3(big, big, big) (we are plugging the function back into itself many times after all)
At this point we use the extending function rules again.
[4,3][4,4]
[4,4][4,4]3,3,3,3→big→big(big,big)
Simplify
[4,3][4,4]
[4,4][4,4]3,3,3,3(huge, huge)
Extending
[4,3][4,4]
[4,4][4,4]3,3,3,3→huge→huge(huge)
Simplify
[4,3][4,4]
[4,4][4,4]3,3,3,3(giant)
We are now back at subscript level
[4,3][4,4]
[4,4][4,4]2,3,3,3(4,4,...giant number of 4s...,4)
Simplify using the extending function rules as far as it will go.
[4,3][4,4]
[4,4][4,4]2,3,3,3(monstrous)
Again at the subscript level
[4,3][4,4]
[4,4][4,4]1,3,3,3(4,4,...monstrous 4s...,4)
Simplify
[4,3][4,4]
[4,4][4,4]1,3,3,3(wtf)
Multiple subscript rules
[4,3][4,4]
[4,4][4,4]wtf,2,3,3(4)
Simplify
[4,3][4,4]
[4,4][4,4]1,2,3,3(seriously?)
Mult Sub rules
[4,3][4,4]
[4,4][4,4]seriously,1,3,3(4)
...
[4,3][4,4]
[4,4][4,4]1,1,3,3(godly?)
Mult Sub rules
[4,3][4,4]
[4,4][4,4]1,godly?,2,3(4)
Ditto
[4,3][4,4]
[4,4][4,4]4,godly?-1,2,3(4)
...
[4,3][4,4]
[4,4][4,4]1,1,1,3(Zeus Weeps)
...
[4,3][4,4]
[4,4][4,4]1,1,Zeus Weeps,2(4)
...
[4,3][4,4]
[4,4][4,4]3,3,Zeus Weeps-1,2(4,4,4,4)
...
[4,3][4,4]
[4,4][4,4]1,1,1,2(Colorful Metaphor)
...
[4,3][4,4]
[4,4][4,4]1,1,Colorful Metaphor(4)
...
[4,3][4,4]
[4,4][4,4](Out of Metaphors)
Matrix Rules
[4,2][4,4]
[4,4][4,4]1,1,...Out of Metaphors..., 4(4)
Simplify...
[4,2][4,4]
[4,4][4,4](Bigger)
Matrix
[4,1][4,4]
[4,4][4,4](Even bigger)
Matrix
[3,Even Bigger][4,4]
[4,4][4,4](4)
...
[3,Even Bigger-1][4,4]
[4,4][4,4][sub]1,1,1,4[/sub](4)
...
[3,1][4,4]
[4,4][4,4](Still Bigger)
...
[2, Still Bigger][4,4]
[4,4][4,4](4)
...
[2,1][4,4]
[4,4][4,4](Yet Bigger)
...
[1,Yet Bigger][4,4]
[4,4][4,4](4)
...
[1][4,4]
[4,3][4,4](wow...)
...
[4,4,...wow...,4][4,4]
[4,3][4,4](4)
...
[1][4,4]
[4,3][4,4](ouch)
...
[4,4,...ouch...,4][4,4]
[4,2][4,4](pain)
...
[1][4,4]
[4,2][4,4](still bothering?)
...
[4,4...still bothering?...,4][4,4]
[4,1][4,4](really?)
...
[1][4,4]
[4,1][4,4](okay...)
...
[1][4,4]
[3,okay...][4,4](skip a few)
...
[1][4,4]
[1][4,4](wasn't a problem?)

...

[4]
[4]
...
wasn't a problem?
...[4,3]
[4][4,4](hope not)
...

[1][4,3]
[1][4,4](too much stuff)
...

[4]
...[4,2]
[4][4,4](still more)

[1][4,1]
[1][4,4](just in case)
...
[1][3,just in case]
[1][4,4](got it?)
Hopefully by now, it makes sense how to simplify the rest. If not, speak up.


My number.
My number is a(g64, g64) where [1](x)=3→3→x.

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Yakk
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Re: My number is bigger!

Postby Yakk » Fri Jan 16, 2015 3:35 pm UTC

Amida, I think you are working on the wrong scale.
Myself wrote:In short, in order to generate a decent value, you first need to describe not just recursion, but a method to generate recursion and a way to name extremely complex recursive structures. The fast growing hierarchy is a famous one that is hard to naively beat. In this thread, if you aren't doing something at least that powerful, your submission could be described as small. If you have at least used the fast growing hierarchy, I would denote your answer as at least medium.

The previous post, at a quick glance, looks like a small submission.

Each level you describe seems to define a new way to lay on recursive structure. Which makes it small. You lack the notation to climb the fast growing hierarchy fast enough with that kind of plan.

To generate medium numbers, you have to define a way to define new orders of recursion rather than defining new orders of recursion.

Imagine if someone went and described a way to take each of your fancy recursive systems in each spoiler, and said "that is one pebble of extra recursion", instead of having to custom write a system in your spoiler. Each of them can be measured as a spoiler. They did this formally, not with hand waves, and their pebbles are (roughly) as powerful as your spoilers.

Then, they defined a rock as "given input N, we use N pebbles of recursion". They define a rock plus a pebble as "we take the input, pass it to the rock. We evaluate rock recursively that many times on the original input". With two pebbles, we take the input, pass it to the pebble plus rock, then recursively apply rock+pebble on the input that many times. Etc.

Two rocks means we take the input, pass it to the first rock, get a bunch of pebbles, then evaluate rock + bunch of pebbles. Note that we are now building pebble-based recursive structures whose structure is described by other recursive functions.

A boulder does the same with rocks as rocks do with pebbles.

Now, we generalize the pebble->rock->boulder->etc path. We call a mountain a function that takes an input n, and looks up the nth entry in that path. Then evaluates that.

A mountain plus a pebble first takes an input, applies mountain to it, then applies mountain that many times recursively on the input. Then we reach mountain+rock, mountain+many rocks, mountain+boulder, mountain+next level boulder, mountain+blah.

Then we reach two mountains.

Now, we have the pebble->mountain->??? generalization (which can be formalized). We call the function that uses this the grain of sand. The grain of sand takes an input n, and looks up the nth entry in the pebble->mountain->?? hierarchy, and evaluates it with n as an input.

A grain of sand plus a pebble first takes an input, calls grain of sand on it, then recursively calls grain of sand that many times on the original input.

Then we get two grains of sand. etc.

etc.

Note that the -> bits: the meaning of the "kind" of step that I am referring to with -> changes between pebble->rock->boulder->??? and pebble->mountain->???.

Rock is the "end" of the infinite pile of pebbles. Boulder is the "end" of the infinite pile of rocks. There is another thing that is the "end" of the infinite pile of boulders.

Mountain is the "end" of the infinite game of making infinite piles of things, and finding their end, and making infinite piles of the end, etc. We label each of the infinite piles with a number, and mountain uses those indexes.

The thing after mountain in the pebble->mountain->??? sequence isn't an infinite pile of mountains, because at this point -> is referring to the game that takes pebble to mountain, not the game that takes pebble to rock.

After we build such a sequence, the next layer of sequence appears.

We can then talk about the nth entry in the nth such sequence of ->s as a sequence.

And it keeps on going.

Now, this is hand wavey. But fortunately, ordinal notation has already well defined how to build these ridiculously complex recursive structures and the resulting generalizations. And with a bit of work, the fast growing hierarchy maps said recursive structures in a well defined way to increasingly complex recursive function calls (very, very roughly like what I typed above).

Which means once we have defined the mapping from ordinal -> function reasonably well, we just have to name an reasonably well known ordinal, and we get insanely complex recursive structures out of it that grow at mind bogglingly fast speeds.

This is what is required to have a medium sized submission in this thread. Your submission, lacking that level of power in defining recursive structure, is delegated to small.

The reason why these beat yours is that they leverage someone else's research and knowledge (the ordinal heirarchy), abstract it, map it over to a system that makes fast growing functions, and poof.

The large submissions are harder to understand. One of them is proof theoretic Turing machines doing a particular trick (plus recursion), and another involves extremely large ordinals. Extremely large ordinals could be more powerful than the proof theoretic machine, but if they are the proof that they are indeed well defined is intractable in whatever proof theory we made our proof theoretic Turing machine work with.

As the Turing Machine recursion ends up recursing on the strength of proof systems instead of on recursive functions, I think it grows fundamentally faster than the ordinal system. But I am not 100% convinced. It takes a bit of care to make the system well defined and fast growing, admittedly. (The hard part is proving that it does indeed halt).

Now, we can beat the Turing machine recursion by using a more powerful proof theory. You can see this with something like Tree3 or hydras, where weak proof theories simply cannot prove that the system in question ever halts (in general) -- they may be able to prove that it halts on a given input, but not on all inputs, and the proof grows extremely fast (it may not be fundamentally shorter than the output of the function!) -- but stronger ones can prove the halting in general in an extremely short proof.

In any case, using large ordinals or careful Turing machine proof machines is the category of answer that I would call large.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

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Vytron
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Re: My number is bigger!

Postby Vytron » Fri Jan 16, 2015 4:18 pm UTC


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Re: My number is bigger!

Postby Daggoth » Fri Jan 16, 2015 4:22 pm UTC

Yeah. I think your entry would land a nice 4th or maybe even third place.

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Re: My number is bigger!

Postby Amida » Fri Jan 16, 2015 4:22 pm UTC

Aren't ordinals the different levels infinity? And wasn't the infinite banned?
The Turing Machine stuff you described being unsure if it even halts, sounds like the Busy Beavers, or at least non-computable. Which again, wasn't that banned at the beginning?

Am I misunderstanding something or did I miss some change?

Also, is there a number that most people think is the biggest on this thread? 37 pages of posts of multiple page proofs is too much for me to really catch up on.
Last edited by Amida on Fri Jan 16, 2015 4:46 pm UTC, edited 1 time in total.

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Re: My number is bigger!

Postby Vytron » Fri Jan 16, 2015 4:35 pm UTC

The Fast Growing Hierarchy defines methods to collapse infinite ordinals into finite numbers. It starts with a function that returns f(n)=n+1, and then defines methods of recursion that have equivalences with ordinals.

I believe your number is between epsilon_0 and zeta_0. Mainly, I believe your number matches a recursion equivalent to epsilon_X, where X is a value less than epsilon, so that every extra recursion that you add just adds ordinal exponentiation of omegas to epsilon (say, your number would be between epsilon_omega and epsilon_omega^omega...omega^omega, for an arbitrary number of omegas) so the limit of your notation would be epsilon_epsilon_0.

(Add an f_x, where x is the ordinal, to the above sentence so it stands for a function of the Fast Growing Hierarchy with that growth)

The biggest number of the thread managed to implement Turing Machines and Busy Beavers in the recursions while remaining computable.

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Re: My number is bigger!

Postby Vytron » Fri Jan 16, 2015 4:37 pm UTC

Amida wrote:Also, is there a number that most people think is the biggest on this thread? 37 pages of posts of multiple page proofs is too much for me to really catch up on.


It seems you edited that in while I was writing, please double post to add new content on the future (forum games doesn't increase postcount) so that people that already read your message don't miss newly edited content.

Here's the top numbers of the thread:

Vytron wrote:This is the biggest number of the thread:

EliezerYudkowsky's

It's so big that it could use a much smaller number as input, but it does use 10 as input.

Second biggest is here. tomtom2357 claims it's even bigger than Yudkowsky's, but doesn't prove it.

(tomtom2357's was challenged and changed his number to "F(n) as the largest number such that there exists a statement that uniquely defines it, where that is provable in a proof of at most f(n) symbols (where f(n) is your favorite function - I guess tomtom2357 would be using Deedlit's function below here.).)

And here's the third biggest one:

Deedlit's

The one to beat before Yudkowsky arrived, and, it may actually be the current champion since Yudkowsky never proved that his was bigger (people just went like, "oh, yeah, Yudkowsky's should be bigger", because you can't invent a notation that represents Yudkowsky number by definition, or something.)

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Re: My number is bigger!

Postby Yakk » Fri Jan 16, 2015 5:11 pm UTC

The medium numbers use ordinals to define how the recursion is structured. The value returned from the function is still a finite (if ridiculous) value.

Define f_0(n) = n+1.
Define f_{x+1}(n) = f_x(n) invoked on n f_x(n) times.

This defines a finite family of functions.

We then take the limit ordinal of this family (0, 1, 2, 3) and call it omega.

Define f_omega(n) = f_n(n)

"omega" is an "infinite ordinal" (it comes after 0, 1, 2, 3, ... -- every natural number), but the function f_omega is well defined and runs a finite number of steps. Now it doesn't look that strong -- until you look at f_{omega+1}(n). ("omega" is not the number infinity, and +1 is well defined here). "omega" is not a number, t is an ordinal. So we don't follow the same rules as for numbers. When we talk about "omega^2" we are talking about symbols arranged in a particular way, not "squaring" like the natural numbers. Same for "omega^omega". These are all well defined: see ordinal theory.

Now if we feed in 3 we get:
f_{omega}(3) = f_3(3) = (f_2)^(f_2(3))(3)
f_2(3) = (f_1)^(f_1(3))(3)
f_1(3) = (f_0)^(f_0(3))(3)
f_0(3) = 4
f_0^4(3) = 7
so f_1(3) = 7
In fact, f_1(k) = 2k+1
f_1^7(3) is thus at least 3*2^7
and in general f_2(k) > k * 4^k
so f_3(3) = f_2^(3*2^7)(3)
which is greater than 4^...^4^3 with a tower length of 3*2^7

The f_n(3) keep on generalizing, each one massively more powerful than the last.

f_{omega+1}(3) we first run f_3(3) and get a result (that huge tower).

Then we run f_omega that many times on 3. So the first iteration is the f_{huge power tower}(3). The second iteration is f_{ f_{huge power tower}(3) }(3). And we do this huge power tower times.

That is f_omega+1. f_{omega+2} is more ridiculous. f_{2omega} is also defined, as is f_{omega^2}, and f_{omega^3} and f_{omega^omega} and f_{omega^omega^2} and f_{omega^omega^3} and f_{omega^omega^omega} and f_{ omega^omega^...^omega^omega } (aka f_epsilon_0).

As noted, the hard part is doing the work to show that we can map these structures to extremely fast growing, yet finite, functions. And people have done that. Then we steal the ordinal structure and map that to fast growing functions.


The large numbers are carefully constructed so we can show that the resulting Turing Machine does halt. Usually after obscene numbers of steps, where the number of steps it takes to halt is recursively generated by prior Turing Machine runs. It will literally check every proof up to a certain length (finite number of such proofs) for every function definition (defined so there are a finite number) up to a certain size, and then runs the Turing Machine a bounded number of steps (as proof checking is "easy", usually polynomial in the length of the proof) to see if the function is well defined. It stores finds the largest one. It then uses that value to bound later executions of itself to search even deeper among all functions that can be defined certain ways.
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Re: My number is bigger!

Postby Daggoth » Fri Jan 16, 2015 5:31 pm UTC

Omegas are "increased" by add multiply and exponentiate.
You may use natural numbers or omegas
The limit to that is a tower of omegas of omega height
that is equivalent to epsilon zero
Epsilon numbers introduce a new "operation" called subscript. Basically an exponentiation tower of epsilon_n is equivalent to epsilon_{n+1}.
You may use omegas and even epsilons such as epsilon zero in the subscript so the limit for that is a downward line of epsilon subscripts. That is equivalent to zeta zero

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Re: My number is bigger!

Postby Vytron » Fri Jan 16, 2015 6:25 pm UTC

And, doing all the job again from zeta_0 gets you to zeta_0*2, so you can specify a function that recurses on that, up to zeta_0*omega, and begin the process again at this level until reaching zeta_0*epsilon_0, and so on.

Once you increase the recursions levels up to zeta_0*zeta_0 you reach zeta_0^2, and you begin again on your way to zeta_0^omega.

That is, the higher the ordinal that you reach, the more recursions necessary to reach the next one.

zeta_1 is the limit of zeta_0^zeta_0^zeta_0^zeta_0...

eta_0 is the limit of ...zeta_zeta_zeta_0

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Re: My number is bigger!

Postby Yakk » Fri Jan 16, 2015 6:58 pm UTC

And while these ordinals are more than "infinitely far" down a given path in the ordinals, they can also be expressed finitely.

And the function that we build from them is a recursive function that only recurses a finite number of times, and it traditionally terminates with a function that says "add one to the input". (the point being that any increase, no matter how small, becomes rather large once recursed enough times).

The reason this is powerful is that we describe new ways to describe recursion. Then we steal the entire theory of ordinals, which describe a rather large and complex hierarchy, and use those to describe increasingly deep recursions. Once we have done so, we don't have to custom craft our ridiculous recursive systems: instead, we just have to name an ordinal that is large, and then use its structure to build a recursive structure for our f, then feed it a modest number like 3.

If you custom craft ridiculous recursive systems, you will end up with a number in the category of small.

If you use the above, your number can reach medium. You can usually describe an upper bound on a small number by examining its recursive structure -- the number of layers used to define how it recurses -- and name an ordinal that generates at least that much recursive structure. Quite often someone will end up defining a stage of recursion that doesn't fundamentally increase their value all that much (they do something like that omega^omega, and add 1 to it, in the ordinal-mapping-to-recursion space). Or worse, they do something like 1+omega^omega (which shows no significant difference from omega^omega). But you can hand-wave around that by looking for an upper bound on how big their value is.

And, as I have said before, the large numbers are trickier, but there is reason to believe they are larger than any "reasonable" medium number.
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Re: My number is bigger!

Postby Vytron » Fri Jan 16, 2015 7:31 pm UTC

With the result that, once you get recursive enough, it doesn't matter what you're doing or where do you start (e.g. no difference from starting with a function that does n+1, n*2, n^2, n^^2, or n^^n... or with an input of 3, a googol plex, g_64, xkcd... given the number of recursions the resulting number will have about the same magnitude.)

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Re: My number is bigger!

Postby Daggoth » Fri Jan 16, 2015 9:09 pm UTC

The only thing wrong with his entry is the thread its in

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Re: My number is bigger!

Postby Amida » Fri Jan 16, 2015 10:28 pm UTC

I really like how everyone here is posting and helping me grow my understanding of mathematics.
So now, I'm gonna try to define something even better.

Vytron wrote:With the result that, once you get recursive enough, it doesn't matter what you're doing or where do you start (e.g. no difference from starting with a function that does n+1, n*2, n^2, n^^2, or n^^n... or with an input of 3, a googol plex, g_64, xkcd... given the number of recursions the resulting number will have about the same magnitude.)


With that in mind and because I like the graham's number series I'll start with the function G0(x) = gx so G0(64) is Graham's Number itself. Those with a close eye, might have noticed my last number was also on the chain (assuming I set it up right).

Since Yakk mentioned using the ordinals to define function recursion I'll define how I will interpret them for my functions.
Omega and epsilons will be the same as they have been. I haven't seen an article on zeta numbers, so I'll define them to be of the form Zeta = EpsilonZeta for my purposes. This is just like epsilon = omegaepsilon. The smallest Zeta number is Zeta0. Zeta1 = epsilonepsilon_...epsilon_(zeta_0+1).After that, in my readings of Wikipedia, I learned that the Greek alphabet has a finite number of characters[Citation Needed] so I'll take a different road here and define something new. an is the nth ordinal class. n=0 is omega, n = 1 is epsilon, n = 2 is zeta and higher numbered classes follow the same pattern. So the an class is an infinite string of subscripts of the previous (an-1) class.

Again using Yakk's comment.
Gn+1(x) = Gn^(Gn(x))(x)
Onward and upward to Gomega(x) = Gx(x)
Gomega +1(x) and what not.
Gomega*n(x) = Gomega*(n-1) + omega(x) = Gomega*(n-1)+x(x)
Gomega^n(x) = Gomega^(n-1)*omega(x) = Gomega^(n-1)*x(x)
Yada Yada
Gepsilon_0(x) = Gomega^^x(x)
Gepsilon_n(x) = Gepsilon_n-1^^x(x)
Gzeta_0(x) = Gepsilon_epsilon_x epsilons total_epsilon_0(x)
Ga_n(x)=Gnth ordinal null(x)

For my new number, I choose
Ga_G(64)(64)
How did I do?

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Re: My number is bigger!

Postby Vytron » Fri Jan 16, 2015 10:44 pm UTC

Seems to max out at phi(omega,0), so it's below orangedragonfire's 9[_9 9]_9 on page 11 of the other thread we've kept mentioning, because he recurses what you're doing (but just 9 times).

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Re: My number is bigger!

Postby Daggoth » Sat Jan 17, 2015 4:04 am UTC

Not the first person to notice the alphabet size as a limitation to work around..


But there is in fact a more canonical continuation from zeta_zeta_...0 and on its called the phi function it starts with two variables looks like phi_a(b) one of 'em is the subindex and the other is the ath member of that greek letter sequence ( epsilon zeta eta etc...) so you could have something like f_phi_999(999)(n) even though there arent 999 greek letters.
phi_1(0) is epsilon zero. phi_2(0) zeta zero and so on
Again, you may use lesser ordinals like omega or epsilon to replace a 999. Even phi() itself. I.e. you can use f_phi_{phi_999(999)}(n)
Obviously there is a limit to that idea too. A downward line of phi_phi_....0 it is called gamma zero.

but that is still not top shelf (were still between small/medium in the yakk scale)

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Re: My number is bigger!

Postby Vytron » Sat Jan 17, 2015 1:23 pm UTC

(and you can't reach medium unless, instead of trying new things one by one, you find a way to iterate "doing new things", and recurse on that)

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Re: My number is bigger!

Postby WarDaft » Sat Jan 17, 2015 3:26 pm UTC

Vytron wrote:And, doing all the job again from zeta_0 gets you to zeta_0*2, so you can specify a function that recurses on that, up to zeta_0*omega, and begin the process again at this level until reaching zeta_0*epsilon_0, and so on.

Once you increase the recursions levels up to zeta_0*zeta_0 you reach zeta_0^2, and you begin again on your way to zeta_0^omega.

That is, the higher the ordinal that you reach, the more recursions necessary to reach the next one.

zeta_1 is the limit of zeta_0^zeta_0^zeta_0^zeta_0...

eta_0 is the limit of ...zeta_zeta_zeta_0

Not quite.

epsilon_(zeta_0+1) is the limit of 0, zeta_0, zeta_0^zeta_0, zeta_0^zeta_0^zeta_0^zeta_0...

Then you can have epsilon_(zeta_0+1)^epsilon_(zeta_0+1), epsilon_(zeta_0+2), epsilon_(zeta_0*2), epsilon_(zeta_0*zeta_0), epsilon_(zeta_0^zeta_0), epsilon_(zeta_0^zeta_0^zeta_0)... up to the limit of epsilon_(epsilon_(zeta_0+1))
Then you can have things like epsilon_(epsilon_(zeta_0+1)+1), epsilon_(epsilon_(zeta_0+1)*2), epsilon_(epsilon_(zeta_0+1)*epsilon_(zeta_0+1)), epsilon_(epsilon_(zeta_0+1)^epsilon_(zeta_0+1)), epsilon_(epsilon_(zeta_0+1)^epsilon_(zeta_0+1)^epsilon_(zeta_0+1)), up to the limit of epsilon_(epsilon_(zeta_0+2)).
Likewise for epsilon_(epsilon_(zeta_0+3)), and epsilon_(epsilon_(zeta_0+4)). Each subscript here acts like going up an exponent did when originally counting to epsilon_0, in that it is an entirely new kind of growth compared to what came before.

It is only once you have the sequence zeta_0+1, epsilon_(zeta_0+1), epsilon_(epsilon_(zeta+1)), epsilon_(epsilon_(epsilon_(zeta+1)))... that you reach zeta_1.
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Re: My number is bigger!

Postby Vytron » Sat Jan 17, 2015 3:50 pm UTC

Oh, I see, thanks.

Then Amida's number is much lower than estimated (should be between zeta_0 and zeta_omega, since their defined functions don't behave like the FGH.)

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Re: My number is bigger!

Postby Daggoth » Sat Jan 17, 2015 6:48 pm UTC

I estimate F_gamma_0(4) to be a upper bound for his entry

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Re: My number is bigger!

Postby Vytron » Sat Jan 17, 2015 7:04 pm UTC

How come? In his list zeta_0 behaves like epsilon_0, not stronger, and so his nth ordinal null never seems to reach eta_0 or zeta_zeta_0 or zeta_epsilon_0.

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Re: My number is bigger!

Postby Daggoth » Sat Jan 17, 2015 7:19 pm UTC

Well i said upper bound, but i didnt say very tight

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Re: My number is bigger!

Postby Amida » Sat Jan 17, 2015 10:43 pm UTC

In my post, I described Zeta1 as the limit of the series Zeta0+1, epsilonzeta_0+1.
Amida wrote:Zeta1 = epsilonepsilon_...epsilon_(zeta_0+1).

And all my following ordinals follow the same pattern. Either I'm missing the point, or people assumed I copied exactly.

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Re: My number is bigger!

Postby Daggoth » Sat Jan 17, 2015 11:10 pm UTC

yeah i totally get what you've been saying. on the other hand, you arent paying attention to what me and vytron are implying. That there's another thread where you stand a much better chance competing while here, well you're looking at top 10 or 11 material.

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Re: My number is bigger!

Postby Amida » Sun Jan 18, 2015 12:09 am UTC

I've seen that, I guess I just didn't want to pull a comic 1254 for the parts that have been talked about here. My next submission will be there.

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Re: My number is bigger!

Postby SuperJedi224 » Fri Apr 17, 2015 12:59 am UTC

So, what are we talking here... gamma_0 order?

How many of you know who Sbiis Saibian is?

Alright, so his Extended Cascading-E Notation (I'd include a link, but this darn forum won't let me), one of the upper levels of a series of increasingly complex large number notations, is an incredibly insanely powerful notation. On the order of f_phi(w,0,0) in the fast-growing hierarchy.

It starts with relatively simple expressions (Hyper-E level) such as:
E100 = A googol
E100#2 = A googolplex
E1#100 = 10^^100 (Hectalogue)
E100#100 = Grangol
E100#100#2 = E100#(E100#100) = Grangoldex
E100##3=E100#100#100=Greagol
E100##4=Gigangol
...
E100##8=Ginorgol
E100##9=Gargantuul
...
E100##100=Gugold
E100##100#2 = E100##(E100##100)= Gugolda-suplex
...
E100###100=Throogol
E100####100=Teroogol

Eventually building up to expressions like these:
E100#^#100 (Godgahlah)
E100#^#^#100 (Godgathor)
E100#^#^#^#100 (Godtothol (?))
...
E100#^^#100 (Tethrathoth, on the order of f_e(0))
E100#^^^#100 (Pentacthulhum, on the order of f_gamma(0))
E100#^^^^#100 (Hexacthulhum, on the order of f_phi(2,0,0))
E100#^^^^^#100 (Heptacthulhum (?), on the order of f_phi(3,0,0))

With, of course, absurd numbers of other possible expressions in between.

So, using the aforementioned notation, the number I'm going to submit is the Colossigog (which is actually Saibian's official name for it): E100#^^...^^#100, where there are 50,000,000,000,000,000 carats.
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Re: My number is bigger!

Postby Vytron » Fri Apr 17, 2015 1:56 am UTC

SuperJedi224 wrote:So, what are we talking here... gamma_0 order?


Um, no, that was defeated on the first 10 pages of the thread, or something.

You probably want to play in this game which is currently at ψ_0(ψ_9(Ω_9)) order.

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Re: My number is bigger!

Postby SuperJedi224 » Fri Apr 17, 2015 10:17 am UTC

Vytron wrote:
SuperJedi224 wrote:So, what are we talking here... gamma_0 order?


Um, no, that was defeated on the first 10 pages of the thread, or something.

[link] which is currently at [link] order.

Oh. Okay then. Sorry.
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Re: My number is bigger!

Postby Vytron » Fri Apr 17, 2015 3:33 pm UTC

No problem!

What happened was that in page 32 of the thread, EliezerYudkowsky basically killed the game by sending a number so big that nobody has been able to send a number provably bigger than his.

He managed to do it by using Turing Machines and the halting problem (with Zermelo-Fraenkel set theory and a couple axioms in there) while keeping his number computable, so since then, several big number games alternatives have been created, some more popular than others.

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Re: My number is bigger!

Postby zabing12 » Fri Apr 17, 2015 4:14 pm UTC

Delete this pls mods
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Re: My number is bigger!

Postby zabing12 » Fri Apr 17, 2015 4:14 pm UTC

0.
0 can serve some important roles in equations, including those that involve addition, multiplication, and subtraction. Numbers can also be raised by the power 0, which will always produce the value of 1. And if you raise 0 to power of anything, you still get 0. But, if try to do 0^0, math goes all squirrel again and the answer becomes basically anything (an “indeterminate form”).

The sum of 0 numbers is 0, but the product of 0 numbers is 1. And 0 is neither positive, nor negative. It’s not a prime number, and it's not a unit.
Also, going back to the answer becomes basically anything (an “indeterminate form”), all numbers, including negatives, equal 0. (1+-1)+(2+-2)... and so on would equal 0. A line, subtracted form a line, plus a square, minus a square, Plus a cube and so on would mean that ALL dimensions equal the 0th dimension. So, mathematically, the 0th dimension IS ALL DIMENSIONS. 0 is everything.
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Re: My number is bigger!

Postby firesoul31 » Fri Apr 17, 2015 7:40 pm UTC

what
Pronouns: she/her/hers or they/them please.

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Re: My number is bigger!

Postby SuperJedi224 » Fri Apr 17, 2015 9:45 pm UTC

The largest entry so far, from what I can understand of it, seems to be at most on the order of (about) w1CK+2 (correct me if I'm wrong).
What about (using Adam Goucher's Xi function) Ξ(Ξ(100))?
There is some debate about the growth rate of this function, but the consensus seems to be that the limit ordinal is at least wwCK and at most the first fixed point of a->waCK, so, assuming my first conclusion is correct, than this number is, in fact, bigger.
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Re: My number is bigger!

Postby Vytron » Fri Apr 17, 2015 11:13 pm UTC

SuperJedi224 wrote:The largest entry so far, from what I can understand of it, seems to be at most on the order of (about) w1CK+2 (correct me if I'm wrong).


From what I gather (and correct me if I'm wrong :P ) you can't use symbols from the FGH to describe EliezerYudkowsky's number, because by definition, it would surpass it.

So EliezerYudkowsky would be over...

Well, it should be over all the notations described on this page.

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Re: My number is bigger!

Postby Nethertaker » Sat Feb 06, 2016 2:25 am UTC

w1CK+3

...

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Re: My number is bigger!

Postby emlightened » Sun Feb 07, 2016 7:07 pm UTC

Nethertaker wrote:ω1CK+3

Y'know, as far as countable nonrecursive ordinals go, that one's miniscule.

Oh, and it's not finite, hence doesn't count.
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Re: My number is bigger!

Postby tcamps » Thu Apr 14, 2016 6:32 pm UTC

There's an issue with Yudkowsky's construction. Yudkowsky's construction starts with a theory T (he uses T = ZFC + I0), and looks through all proofs in T that a given Turing machine M will halt. Then he runs that M, and waits for it to halt. Assuming that M actually does halt, he can look at its running time, and use that in the construction of his function.

But we need to know that M actually _does_ halt. In order to be sure of this, we have to assume that the theory T actually does describe "actual reality" correctly. The trouble is, mathematicians and philosophers have found no knock-down arguments that can tell us whether T actually describes "actual reality". This goes much deeper than just not knowing whether T is consistent, which is probably a mathematical problem. It's really a philosophical problem.

In fact, there are other theories which contradict T which might also plausibly describe the actual world. For example we could take ZFC + V=L (an axiom [introduced by Gödel][https://en.wikipedia.org/wiki/Axiom_of_constructibility]), or we could take ZF + AD (by ZF, I mean set theory without the axiom of choice, and by AD, I mean the [axiom of determinacy][https://en.wikipedia.org/wiki/Axiom_of_determinacy]) or ZF + the existence of a [Berkeley cardinal][http://cantorsattic.info/Berkeley].

So it's not clear that Yudkowsky's algorithm will actually halt.

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Re: My number is bigger!

Postby emlightened » Thu Apr 14, 2016 11:00 pm UTC

Okay, this seems to be a misunderstanding on how mathematics works.

I find it interesting as to how mathematics can model the real world, and have wondered in the past if there is a 'truer' theory: a theory that better describes the world we live in than another. This does not exist. This is the problem with confusing mathematics and physics. Physics is what happens. Mathematics is not what happens. It is an extremely complex use of logic, and most of mathematics is used to model what happens in the real world, such as the maths used in physics or game theory or statistics.

The misunderstanding here, as I see it, is that mathematics is but a model for what happens, no mater how complex. It may appear that the equations are exact, and we have nothing to show they aren't. But we also can't prove they are. In mathematics, we choose what is real and what isn't, because it isn't related to physics. Mathematics is, formally, built on logic in set theory, and uses symbols like ∈, ∃ and ∧, to say something. We can decide what's true or false, with statements like ∀a∀b∃S∀e(e∈S⇔(e=a∨e=b)) called axioms. The thing about maths, though, is that it also makes sense to say ¬∀a∀b∃S∀e(e∈S⇔(e=a∨e=b)) as well; either could be true.

We get to decide what's true or false through axioms. There is no 'right' group of axioms — no correct theory — because it depends on how we use the mathematics to model real life. Even if there was an exact way of measuring and verifying these things, and we knew them to be true everywhere, multiple theories could model the same thing perfectly. ZFC+I0 could model the universe just as well as ZFC+(V=L) — yet ZFC+I0+(V=L) leads to a contradiction, where anything can be both true and false.

If we take a step back, and simply look at the logical symbols, at the axioms, behind it, the only way we can say that something is 'false' is if it isn't consistent: if we can prove anything to be either true or false based on what we have. There are no inherent assumptions, except those of the theory you choose.

Although Yudowski's number is well defined, he did not actually define it, he simply described its definition. It's not very hard to make a definition, though, with the right sources. Assuming the consistency of ZFC+I0 (the inconsistency of the theory in question is the only way the number would not be well-defined), all you need to do (well, most of it) is turn the text 'ZFC+I0' into the conjunction of several axioms, encode a Turing machine and basic arithmetic into a set theory (it doesn't come with it, at least not in ZFC), and model this appropriately. This means that the only question as to whether a number is defined, and whether ZFC+I0 can tell if one halts, correctly, is whether that theory is consistent. The question of what is 'real' doesn't matter, as we're purely dealing with logic.


Notes:
a) ZFC is not set theory. It's the group of axioms that most mathematicians study. NF (New Foundations) is also studied, as well as simpler theories like PA (Peano Arithmetic) and Z2 (Second-Order logic).
b) The description of 'actual reality' may or may not be the whim of some known or unknown omnipotent being(s). I don't think it's necessary that there exists a theory that can fully model it.
c) I recently found this video, which also covers this topic (probably better than me). It also has a nice introduction to ordinals and cardinals, for anyone more interested in that. (The video is incorrect in regards to the (generalised) continuum hypothesis - it has in fact been proven equiconsistent with and independent of ZFC for ℵf(α) = ℶα for any normal function f with f(0) = 0, and in particular that 2α = ℵβ for any β>α (that includes both 20=ℵ1 and 20≠ℵ1) are consistent with ZFC).
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Re: My number is bigger!

Postby tcamps » Fri Apr 15, 2016 3:02 am UTC

Let me clarify. Yudkowsky's algorithm goes like this. We let T be the theory ZFC+I0. We define the following function f. For input N, the output f(N) is defined as follows:

Code: Select all

Initialize f(N) = 0.
For each Turing machines M with at most N states:
      For each proof P in T of length at most 2^^N:
            If P is a proof in T that M, (*)
            Then run M until it halts. (***)
                 add the running time of M to f(N).

Then Yudkowsky's number is f^10(10).

The issue is with the step I've labeled (***). We need M to halt. Why should it halt? Because T proved that it halts in (*)! My point is that this is not enough.

If you object to the idea that there is a "real world" where we are "actually running M" then very well, I'm sympathetic -- in the relevant cases M will take longer than the age of the universe to halt if it does halt. So let us analyze this further.

If we want to prove that f is a well-defined function, then we need some sort of formal metatheory T* in which we define the notions we've used here, such as what the formal theory T is, what a proof in T is, what a Turing machine is, and what it means for T to prove that a Turing machine halts (after all, "Turing machine" is not a notion that appears in the language of T, which is just built from the elementhood relation between sets). In order to prove that f is a well-defined function, it looks like we need to prove in our metatheory T* that the condition in (*) implies that (***) will work, i.e. we need that

Phi(M): T* proves that [ (T proves M halts) => M halts ]

Now what is a reasonable choice of metatheory T*? The default would be to take T* = ZFC. But ZFC does not prove Phi(M) (unless ZFC is inconsistent). This is intuitively obvious: since T* is a stronger theory than ZFC, its consequences should not automatically become consequences of ZFC. I'll provide a proof at the end.[footnote] So f is not well-defined if we take our metatheory T* to be ZFC.

I could stop here: the fact that f is not well-defined for a reasonable choice of metatheory is already fatal, I think. But maybe we can salvage things by making some stronger assumptions about our metatheory, though? The only way I can see to get something like Phi(M) is to choose T* to be T itself. And now we come to the problem I mentioned in my previous post. Namely, it is ultimately a philosophical question whether a given metatheory is "appropriate". It's philosophically reasonable to deny that T is an appropriate metatheory. For example, one might object that the evidence that T is even consistent is not particularly compelling. Even if one grants that T is likely to be consistent, one might object to it on other grounds. For instance, one might simply have a favorite metatheory which happens not to be compatible with T. For instance, there are other "reasonable" choices of metatheory T* which clash with T on the truth values of certain statements, such as the existence of a measurable cardinal, or the axiom of choice.

So to convince somebody that f is well-defined, you have to convince them to use a metatheory satifying Phi(M) for all Turing machines M. Suppose you do this. Then f will be well-defined for them, but this is still note enough. For one thing, the values of f that they calculate may still be different from the values of f that you calculate, because your metatheories may get different answers on the question of the running time of M for various M.

So in order to convince somebody that f is well-defined _and_ agree on what its values are, it looks like you need to convince them to use exactly the same metatheory that you happen to be using, and that metatheory has to be closely related to T itself. At this point we're basically saying that everybody has to agree to use T as their metatheory for mathematics in order to get agreement on what f is, which is a very, very tall philosophical order.

Actually, it's even worse than this. There could be a Turing machine M such that T proves that M halts, and (assuming Phi(M)), then T* proves that M halts, but there is no explicit number X such that T* proves that M halts in X steps. So although T* proves that f is well-defined, T* does not provide an actual explicit value for f(N) for every N. Actually, I'm pretty sure that this will be the case so long as T* has a recursively enumerable set of axioms, by some version of the incompleteness theorems, but I think I need to learn more logic to verify this. This issue is why mathematicians typically adopt the fiction of assuming that we are working in a _complete_ metatheory, where such things can't happen -- this is pretty much like assuming there's a "real world" that decides the answers to all questions. Of course, by Godel's incompleteness theorems, no complete metatheory exists which is sufficiently powerful to formalize mathematics and also has an explicit way to enumerate all of its axioms. In these terms, the issues I've raised can be phrased as "we need to assume that T is actually true in the real world / in our complete metatheory".

[footnote] Construction of a Turing machine M such that Phi(M) does not hold (with T*=ZFC). Define M in the following (stupid, but totally legal) way:

If ZFC is consistent, then M is a one-state Turing machine that halts after one step. Otherwise, M is a 1-state Turing machine that immediately enters an infinite loop and stays there.

  1. Obviously, T and ZFC both prove that [M halts if and only if ZFC is consistent].
  2. Now, T proves that ZFC is consistent (this is because T has higher consistency strength than ZFC). Moreover, ZFC proves [T proves that ZFC is consistent].
  3. Hence, by (1) and (2), T proves that M halts, and ZFC proves [T proves that M halts].
  4. But by Godel's 2nd incompleteness theorem, (assuming that ZFC is consistent) ZFC does not prove that ZFC is consistent. So by (1), ZFC does not prove that M halts.
  5. By (3) and (4), ZFC does not prove that [T proves that M halts => M halts]. That is, Phi(M) does not hold when we take T* to be ZFC.


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