Add predicates on predicates in FOOT to get second order oodle theory. Continue to get nth order oodle theory. Define FOOT+(a, b) to be the largest natural number uniquely definable in ath order oodle theory with at most b symbols. Consider FOOT+^{10}(1, 10^{100}). FOOT+ recurs into a. I call this number BIGFOOTPLEX. Is this the biggest defined number so far?
BTD6 maker (talk) 20:08, January 30, 2015 (UTC)

- What is "predicate on predicate"? LittlePeng9 (talk) 20:25, January 30, 2015 (UTC)

- By a predicate on predicates, I mean that in ordinary FOOT a predicate refers to oodles. Now consider a predicate that refers to predicates of FOOT rather than oodles.BTD6 maker (talk) 21:05, January 30, 2015 (UTC)
- Language of FOOT doesn't include any predicates. LittlePeng9 (talk) 21:22, January 30, 2015 (UTC)
- I can still define nth order oodle theory to diagonalise over n-1th order oodle theory. An example of a diagonalisation is to add expressions for FOOT+(n-1, b). Using this diagonalisation, BIGFOOTPLEX is defined. I am working on an even more powerful function. In the meantime, it seems that BIGFOOTPLEX is the biggest known number.BTD6 maker (talk) 20:40, January 31, 2015 (UTC)
- If you want to have your number recognized, try to publish it on an external source. If you are really trying to beat my number, you have to make sure that you formally define the new language and how it's interpreted. LittlePeng9 (talk) 21:07, January 31, 2015 (UTC)
- "add expressions for FOOT+(n-1, b)"... i'm not sure precisely what this means, but it sounds pretty salady. at the very least a naive extension. just augmenting FOOT with a new predicate isnt really a very strong extension (kinda like what kyodaisuu did with FOST), rebuilding the system from scratch is the better way to go -- vel! 09:57, February 1, 2015 (UTC)

- By a predicate on predicates, I mean that in ordinary FOOT a predicate refers to oodles. Now consider a predicate that refers to predicates of FOOT rather than oodles.BTD6 maker (talk) 21:05, January 30, 2015 (UTC)

## New idea!!!!

Hey everyone I have a new idea for how to extend BIG FOOT. Add a new predicate D(n) where "D(n) = max number of symbols expressible in less than n statements in the language of FOUT (First Order Ultra Theory)." FOUT is defined using the Tarskian definition of truth with a new symbol, <X>. If you encounter <X> in a formula, you tally up all the googologisms ever invented in the world up until year X. Then I define BIG FOUT to be FOUT^10(10^10^100). Whaddy'all think? -- xXx~*~ NiNjAdAvE1234 ~*~xXx 08:31, February 9, 2016 (UTC)

- ... "largest googologism ever + 1", right?... why don't you make your own googologism from scratch rather? anyway, what would be <3017>? Fluoroantimonic Acid (talk) 22:47, February 9, 2016 (UTC)
- Ninjadave is vel Billicusp (talk) 23:52, February 9, 2016 (UTC)

## Ordinals to describe FOST & FOOT?

I've been studying the constructions of famous ordinals (such as the LVO and the TFBO) while simultaneously observing functions (such as Extensible E and HAN) with ordinals of similar magnitude to approximate their growth rates. That all got me thinking, "what about uncomputable functions like Rayo(n) or FOOT(n)?" Are there not ordinals describing their growth rates because such functions diagonalize over all standard set theory in the first place? I just want to know about what's going on with them a little better so that I could truly understand their immensities compared to the average 'computable' function. QuasarBooster (talk) 00:14, May 9, 2016 (UTC)

you can define such ordinals on the fly ie. "i define Q to be the least ordinal such that f_Q(n) > Rayo(n) . " Of course, this is useless for the intents and purposes of your question but i thought that it was worth commenting anyway**(Chronolegends (talk) 18:20, October 9, 2016 (UTC)).**

- You have to define all the fundamental sequences blah blah blah <typical ramble about people thinking that you can just plug an ordinal into FGH and everything will be fine>. LittlePeng9 (talk) 18:41, October 9, 2016 (UTC)

- Depending on the meaning of "have to". When FS-es are not fixed, there are just infinitely many ordinals Q. If we're pretending to "get" a single ordinal, it's okay to say that "there exists at least one ordinal Q with some system of FS-es so that f_Q(n) > Rayo(n)". 185.137.18.186 13:28, November 13, 2016 (UTC)

- That's indeed true, but very boring - with proper choice of fundamental sequence, \(f_\omega(n)>\text{Rayo}(n)\). LittlePeng9 (talk) 15:02, November 13, 2016 (UTC)

- If you consider that boring == trivial, then it's not boring at all (at least for me) to find some other ordinals in that set. For example, which set of FS-es could be for the inequalities \(f_{\omega+1}(n) > Rayo(n)\), \(f_{\omega^2}(n) > Rayo(n)\), ..., \(f_{\omega_1^\text{CK}}(n) > Rayo(n)\)? And is the number of choices of set of FS-es always infinite for each transfinite ordinal? 185.137.18.186 15:24, November 13, 2016 (UTC)

- It's easy to find fundamental sequences so that, for each infinite, countable ordinal we have \(f_\alpha(n)>\text{Rayo}(n)\). However, if we require \(\alpha\) to be the
*smallest*such ordinal, then this turns into a somewhat interesting and rather nontrivial question. It's not hard to believe that this is the case for any countable ordinal \(\alpha\geq\omega\). I suspect this question to be more tedious than difficult, but feel free to investigate it. LittlePeng9 (talk) 15:38, November 13, 2016 (UTC)

- It's easy to find fundamental sequences so that, for each infinite, countable ordinal we have \(f_\alpha(n)>\text{Rayo}(n)\). However, if we require \(\alpha\) to be the

- What about equality? Can we really pick fundamental sequence for \(\alpha\) and below so that \(f_\alpha(n) = \text{Rayo}(n)\)? It's not very clear if we can do it even for \(\alpha = \omega\). 185.137.18.186 19:13, November 13, 2016 (UTC)

## Bounds

Did you know bounds on FOOT()? AarexWikia04 - 23:01, August 1, 2016 (UTC)

## Just an Idea

Make BIG FOOT illion.

That's only slightly larger than Big FOOT itself... Username5243 (talk) 09:59, October 7, 2016 (UTC)

- It is salad number so no. AarexWikia04 - 10:49, October 7, 2016 (UTC)

## Fun fact

The article states that "BIG FOOT is a large number that is a derivative of the famously extremely large Rayo's number."

Derivative of Rayo's number is 0. Thus BIG FOOT = 0. Also it implies that 0 is a large number. Now isMindExploded = true.

185.137.18.186 13:11, November 13, 2016 (UTC)

LOL I get it. but that's not what it means here. Username5243 (talk) 19:19, November 13, 2016 (UTC)

- BIG FOOT = 0? It is just a number larger than Rayo's Number, meaning 1. AarexWikia04 - 23:28, November 13, 2016 (UTC)

## LEET FOOT

Can we add this as a FOOT^{133713371337}(10^{10101337})? ^{—Preceding unsigned comment added by Wikillion (talk • contribs) 07:39, November 23, 2016 (UTC)}

- No, it's a naive extension. 80.98.179.160 09:14, November 24, 2017 (UTC)

## FOOT^FOOT^FOOT^FOOT...

What about numbers such as FOOT^FOOT^10(10^100) or FOOT^^10(10^100)? Are they considered as naive extensions? PlantStar/Alpineer 00:23, April 2, 2018 (UTC)