User talk:Deedlit11

Welcome
Hi, welcome to Googology Wiki! Thanks for your edit to the Talk:Goodstein sequence page.

Please leave a message on my talk page if I can help with anything! -- FB100Z (Talk) 03:39, September 20, 2012

Strength of & operator
Hey, Deedlit. You probably thought that \(f_{\Gamma_0}(n) \approx \{n,n / 2\}\) from the reason that:

\(f_\omega(n) \approx \{n,n,n\}\) and \(f_{\varphi(\omega,0)}(n) \approx \{X,X,X\} \&\ n\).

\(f_{\omega+1}(n) \approx \{n,n,1,2\}\) and \(f_{\varphi(\omega+1,0)}(n)\) therefore should be \(\{X,X,1,2\} \&\ n\).

That seems reasonable, but we can look at that by the other way: \(f_\varphi(\alpha,0)(n)\) is the:

\(\alpha+1\)-th hyper-operator applied to X's. (When \(\alpha\) is a finite number)

\(\alpha\)-th hyper-operator applied to X's. (When \(\alpha\) is a transfinite ordinal)

So f_{\(f_\varphi(n,0)\)}(n) corresponds to \(\{X,X,n+1\} \&\ n\) and \(f_\varphi(\omega,0)\)

Next look at \(\varphi(\omega+1,0)\). It is equal to \(f_\varphi(\omega,\varphi(\omega,\cdots,\varphi(\omega,0))...))\) (with \(\omega\text{ }\varphi\)'s (in the more compact way, it is the first ordinal such that \(\alpha\) = \(\varphi(1,\alpha)\)). Hence, it behaves in the same way as, say \(\varphi(2,0) = \alpha = \varphi(1,\alpha)\). We can pretty confident that \(\varphi(\omega+1,0)\) corresponds to \(\{X,X,X+1\} = \{X,p,X+1\} = \{X,\{X,p-1,X\},X+1\}\) structure.

When the one structure represents the hyper-operator of the other, X times, we reach expandal arrays, which is surely corresponds to \(\varphi(1,0,0) = \Gamma_0\).

More generally, \(\varphi\) function with n entries behaves at the same as linear arrays with n-1 entries. Thus, \(\{X,X (1) 2\} \&\ n\} \approx f_{\vartheta(\Omega^\omega)}(n)\). Chris Bird received the same results, and after more considerations you can see that limit ordinal of the non-legion arrays is the \(\vartheta(\Omega^\Omega)\), the Large Veblen Ordinal.

To be clear, I just present the selection of some comparisons:

\(X \uparrow\uparrow X \&\ n \approx f_{\varepsilon_0}(n)\)

\(X \uparrow\uparrow\uparrow X \&\ n \approx f_{\zeta_0}(n)\)

\(X \uparrow\uparrow\uparrow\uparrow X \&\ n \approx f_{\eta_0}(n)\)

\(\{X,X,X\} \&\ n \approx f_{\varphi(\omega,0)}(n)\)

\(\{X,X,X+1\} \&\ n \approx f_{\varphi(\omega+1,0)}(n)\)

\(\{X,X,2X\} \&\ n \approx f_{\varphi(\omega 2,0)}(n)\)

\(\{X,X,X^2\} \&\ n \approx f_{\varphi(\omega^2,0)}(n)\)

\(\{X,X,X^X\} \&\ n \approx f_{\varphi(\omega^\omega,0)}(n)\)

\(\{X,X,X \uparrow\uparrow X\} \&\ n \approx f_{\varphi(\varepsilon_0,0)}(n)\)

\(\{X,X,\{X,X,X\}\} \&\ n \approx f_{\varphi(\varphi(\omega,0),0)}(n)\)

\(\{X,X,1,2\} \&\ n \approx f_{\varphi(1,0,0)}(n)\)

\(\{X,X,2,2\} \&\ n \approx f_{\varphi(1,1,0)}(n)\)

\(\{X,X,3,2\} \&\ n \approx f_{\varphi(1,2,0)}(n)\)

\(\{X,X,X,2\} \&\ n \approx f_{\varphi(1,\omega,0)}(n)\)

\(\{X,X,\{X,X,X\},2\} \&\ n \approx f_{\varphi(1,\varphi(\omega,0),0)}(n)\)

\(\{X,X,\{X,X,1,2\},2\} \&\ n \approx f_{\varphi(1,\varphi(1,0,0),0)}(n)\)

\(\{X,X,1,3\} \&\ n \approx f_{\varphi(2,0,0)}(n)\)

\(\{X,X,1,4\} \&\ n \approx f_{\varphi(3,0,0)}(n)\)

\(\{X,X,1,X\} \&\ n \approx f_{\varphi(\omega,0,0)}(n)\)

\(\{X,X,1,\{X,X,1,2\}\} \&\ n \approx f_{\varphi(\varphi(1,0,0),0,0)}(n)\)

\(\{X,X,1,1,2\} \&\ n \approx f_{\varphi(1,0,0,0)}(n)\)

\(\{X,X,1,1,1,2\} \&\ n \approx f_{\varphi(1,0,0,0,0)}(n)\)

\(\{X,X (1) 2\} \&\ n \approx f_{\vartheta(\Omega^\omega)}(n)\)

\(\{X,X,2 (1) 2\} \&\ n \approx f_{\vartheta(\Omega^\omega+1)}(n)\)

\(\{X,X,3 (1) 2\} \&\ n \approx f_{\vartheta(\Omega^\omega+2)}(n)\)

\(\{X,X,X (1) 2\} \&\ n \approx f_{\vartheta(\Omega^\omega+\omega)}(n)\)

\(\{X,X,1,2 (1) 2\} \&\ n \approx f_{\vartheta((\Omega^\omega)2)}(n)\)

\(\{X,X,1,3 (1) 2\} \&\ n \approx f_{\vartheta((\Omega^\omega)3)}(n)\)

\(\{X,X,1,X (1) 2\} \&\ n \approx f_{\vartheta((\Omega^\omega)\omega)}(n)\)

\(\{X,X,1,1,2 (1) 2\} \&\ n \approx f_{\vartheta((\Omega^\omega)\vartheta(\Omega))}(n)\)

\(\{X,X,1,1,1,2 (1) 2\} \&\ n \approx f_{\vartheta((\Omega^\omega)\vartheta(\Omega^2))}(n)\)

\(\{X,X (1) 3\} \&\ n \approx f_{\vartheta((\Omega^\omega)\vartheta(\Omega^\omega))}(n)\)

\(\{X,X (1) X\} \&\ n \approx f_{\vartheta(\Omega^{\omega+1})}(n)\)

Notice that all these comparisons looks very similar with the regular arrays, where \(\{n,n (1) n\}\) represented by \(X+1\) structure. In general, if we have A structure (in BEAF), then \(A \&\ n \approx f_{\vartheta(\Omega^\alpha)}(n)\), where \(\alpha\) is the ordinal that associated with A. So, continuing onwards:

\(\{X,X (1) X,X\} \&\ n \approx f_{\vartheta(\Omega^{\omega+2})}(n)\)

\(\{X,X (1) X,X,X\} \&\ n \approx f_{\vartheta(\Omega^{\omega+3})}(n)\)

\(\{X,X (1)(1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega 2})}(n)\)

\(\{X,X (1)(1)(1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega 3})}(n)\)

\(\{X,X (2) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^2})}(n)\)

\(\{X,X (3) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^3})}(n)\)

\(\{X,X (0,1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^\omega})}(n)\)

\(\{X,X (1,1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^{\omega+1}})}(n)\)

\(\{X,X (0,2) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^{\omega 2}})}(n)\)

\(\{X,X (0,0,1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^{\omega^2}})}(n)\)

\(\{X,X (0,0,0,1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^{\omega^3}})}(n)\)

\(\{X,X ((1) 1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^{\omega^\omega}})}(n)\)

\(\{X,X ((0,1) 1) 2\} \&\ n \approx f_{\vartheta(\Omega^{\omega^{\omega^{\omega^\omega}}})}(n)\)

\(X \uparrow\uparrow X \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(1)})}(n)\)

\(X \uparrow\uparrow\uparrow X \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(2)})}(n)\)

\(X \uparrow\uparrow\uparrow\uparrow X \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(3)})}(n)\)

\(\{X,X,X\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\omega)})}(n)\)

\(\{X,X (1) 2\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^\omega)})}(n)\)

\(\{X,X (2) 2\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\omega^2})})}(n)\)

\(\{X,X (3) 2\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\omega^3})})}(n)\)

\(\{X,X (0,1) 2\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\omega^\omega})})}(n)\)

\(\{X,X ((1) 1) 2\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\omega^{\omega^\omega}})})}(n)\)

\(\{X,X ((0,1) 1) 2\} \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\omega^{\omega^{\omega^\omega}}})})}(n)\)

\(X \uparrow\uparrow X \&\ n \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\vartheta(1)})})}(n)\)

\(X \uparrow\uparrow X \&\ n \&\ n \&\ n \&\ n \approx f_{\vartheta(\Omega^{\vartheta(\Omega^{\vartheta(\Omega^{\vartheta(1)})})})}(n)\)

All that limits to the Large Veblen Ordinal, \(\vartheta(\Omega^\Omega)\). Ikosarakt1 (talk ^ contribs) 20:28, March 24, 2013 (UTC)

If you are right, what ordinal may be limit of BEAF? If we reach LVO without even && operator, I wonder how big meameamealokkapoowa oompa may really be. LittlePeng9 (talk) 21:36, March 24, 2013 (UTC)

Yes, I have recently come to the same conclusion concerning \(\{X,X,1,2\}\) and \(\Gamma_0\). I haven't examined linear arrays and above yet, but your conclusions seem reasonable. Of course, we need a rigorous definition for Bowers' arrays to be sure.

TREE(3) will have to be moved down the chain, to around the Small Veblen ordinal. (We don't know the exact ordinal for TREE(3) but I guess we will assume that it is not much larger than the Small Veblen ordinal.) Deedlit11 (talk) 22:01, March 24, 2013 (UTC)

@LittlePeng9: Chris Bird analyzed Bowers' notation and concluded that it ends well below the Bachmann-Howard ordinal, and I think below \(\vartheta (\Omega^{\Omega^{\Omega}})\). Deedlit11 (talk) 00:05, March 25, 2013 (UTC)

Some measurements for legion arrays:

If {n,n/2} is at the level of the LVO, then {n,n/3} is at the level of LVO*2, {n,n/4} is at the level of {n,n/4} is at the level of LVO*3, and {n,n/1+F} is at the level of LVO*alpha, where we take alpha to be the ordinal associated to F. So {n,n / 1 / 2} will be at the level of LVO^2, {n,n / 1 / 1 / 2} will be at the level of LVO^3, {n,n (/1) 2} will be at the level of LVO^omega, and {n,n (/F) 2} will be at the level of LVO^(omega^alpha). Thus n && n && n will be at the level of LVO^LVO, and {n,n // 2} will be at the level of epsilon_{LVO+1}. We go through the same hierarchy up with //'2 so we get that {n, n /// 2} is at the level of epsilon_{LVO+2}. In general, {L, X^F} = {n,n (F)/ 2} will be at the level of epsilon_{LVO + alpha}. It follows that {L, L} = epsilon_{epsilon_{LVO+1}}, and probably {L, L, X} = phi(2, LVO+1). It's not clear to me how higher arrays are defined in terms of L - it seems like very bad notation to me - but it seems reasonable that {L, L, ..., L, X} = phi(n, LVO+1). So L2 space is at the level of Gamma_{LVO+1} = phi(1, 0, LVO+1). L3 space will be at level phi(1, 0, LVO+2), LF space will be at level phi(1, 0, LVO + alpha). LLF space will perhaps be at level phi(1, 1, LVO + alpha) (but again, I would need a precise definition of what LF is), and perhaps (L^F)G will be at phi(1, alpha, LVO + beta). So (L^L)F would be at level phi(2, 0, LVO + alpha).

Alternatively, perhaps {L, L, ..., L} is at level phi (1, 0, ..., 0, LVO+1}, with the same number of terms in each. So L2-space will be the second ordinal fixed by the Schutte Klammersymbolen, i.e. \(\theta(\Omega^{\Omega},1)\). In general, LF space will be the alphath ordinal fixed by the Schutte Kalmmersymbolen, i.e. \(\theta(\Omega^{\Omega}, \alpha)\). So perhaps LLF space will be \(\theta(\Omega^{\Omega}+1, \alpha)\), and in general (L^F)G space will be \(\theta(\Omega^{\Omega}+\alpha, \beta)\), and thus the closure space (I guess we are calling this L^L) will be at the level of \(\theta (\Omega^{\Omega} + \Omega, 0)\). So we are very far from \(\theta(\Omega^{\Omega^{\Omega}}, 0)\). Deedlit11 (talk) 01:54, March 25, 2013 (UTC)

But in BEAF there are array spaces such absurdal as L & L. Also, is there any reading about Klammersymbolen? I understand general concept, but nothing more LittlePeng9 (talk) 06:33, March 25, 2013 (UTC)


 * L & L? My analysis went well past the "&" operator.
 * I couldn't find much on the web on the Klammersymbolen, except for:


 * http://www.cs.man.ac.uk/~hsimmons/ORDINAL-NOTATIONS/FromBelow.pdf
 * http://www.cs.swan.ac.uk/~csetzer/articles/ordsyscor010124.ps


 * Both of those papers generalize the Klammersymbolen so they are harder to read than need be.
 * But, the notation is not that complicated; it's just an extension of the Extended Veblen notation into the transfinite.
 * To do that, we need to explicitly name the place values of the variables. For example, instead of phi(8, 0, 0, 0, 3, 0, 5, 6, 4), represent it as (8 @ 8, 3 @ 4, 5 @ 2, 6 @ 1, 4 @ 0). We can represent all Extended Veblen notations this way, but we can also say (1 @ omega, c @ 0), which is defined as the cth ordinal that satisfies (x @ n, 0 @ 0) = x for all n < omega.
 * More generally, define (a_1 @ b_1, a_2 @ b_2, ..., a_n @ b_n, c @ 0) as the cth ordinal that satisfies the system of equalities (a_1 @ b_1, a_2 @ b_2, ..., a_{n-1} @ b_{n-1}, d @ b_n, x @ e) = x for all d < a_n, e < b_n. Deedlit11 (talk) 03:50, March 26, 2013 (UTC)

Array of
We found out that the & is the array of. And i am sorry for the confusion that we caused in BEAF. $Jiawhein$\(a\)\(l\)\(t\) 04:19, April 14, 2013 (UTC)

But you never said is array of... $Jiawhein$\(a\)\(l\)\(t\) 04:24, April 14, 2013 (UTC)


 * Sorry, I thought you knew. Deedlit11 (talk) 06:00, April 14, 2013 (UTC)

SDA forums?
Just curious, are you the same person as "Deedlit" on the Speed Demos Archive forums? --Ixfd64 (talk) 19:51, April 16, 2013 (UTC)


 * Yep, that's me. I speedran FF12 a few years back. I go by Deedlit in a bunch of places.


 * Nice! It's really a small Internet after all... --Ixfd64 (talk) 20:17, April 16, 2013 (UTC)


 * If I not miscalculated, even if you turn all 0's and 1's in the Internet to elementary particles and crumple them in one ball, this ball will be so small that you will barely able to see it with the naked eye! Internet contains roughly = 33.6 zettabits (that is, 0's and 1's). Human body contains about 64 octillion elementary particles, and then \({6.4 \times 10^{28} \over 3.36 \times 10^{22}} \approx 1904762\). In other words, this ball would be roughly 1904762 times smaller than body of average human, and we're speaking about entire Internet! Ikosarakt1 (talk ^ contribs) 20:42, April 16, 2013 (UTC)
 * One of my favorite cartoonists computed that storing the Internet into a bunch of modern hard drives would at most fill up an oil tanker. FB100Z &bull; talk &bull; contribs 20:49, April 16, 2013 (UTC)


 * I guess it's not surprising that the internet would be about a drop of water (1/18 of a gram in atoms, definitely visible), since the internet has to be stored, and we aren't anywhere near 3D atomic level of storage. But of course this stuff is increasing exponentially - it won't take all that long before we surpass a human in size. (well, it should be within our lifetimes.) Deedlit11 (talk) 20:56, April 16, 2013 (UTC)
 * It has been calculated that if we take all electrons moving due to global Internet transfer, mass-energy we are giving them for the flow is equivalent to 50 grams! Nicely explained here. LittlePeng9 (talk) 05:23, April 17, 2013 (UTC)

Template:wedge
Vote now! $Jiawhein$\(a\)\(l\)\(t\) 12:09, April 26, 2013 (UTC)

Loader.c/Calculus of Constructions
Few days ago me together with Ikosaract had a short conversation about Loader's function and CoC. He asked me if I could make a user blog post about this calculus, so everyone would understand it better, but I think you would be a lot better person to ask for this. So I'm asking you - would you be able to write a blog post (or maybe a whole wiki article?) about CoC and how Loader's program incorporates it? Thanks in advance. LittlePeng9 (talk) 19:53, April 8, 2014 (UTC)
 * You might want to check out the Coq proof assistant to familiarize yourself with the system. I tried my hand at figuring out CoC several times, and I still don't fully understand it. you're.so.pretty! 20:02, April 8, 2014 (UTC)
 * Actually, I don't understand the Calculus of Constructions that well either. I could try writing something on it, but I would have to base it on what I can glean from the internet. Perhaps it would be best to start the simple typed lambda calculus, then work our way up to System T, System F, then the CoC. Deedlit11 (talk) 20:37, April 9, 2014 (UTC)

Computer Turn-offage
Please don't tell me you haven't turned off your computer in 10 days. King2218 (talk) 16:06, July 27, 2014 (UTC)

I think I killed you
[15:13]  !attack Deedlit11 [15:13]  Wojowu creeps up behind Deedlit11 and swings at them with a bat! [15:13]  Thud! Deedlit11 falls to the ground unconscious! [15:14] == Deedlit11 [~Deedlit@ip68-111-90-23.oc.oc.cox.net] has quit [Ping timeout: 263 seconds] [15:14]  Well

Sorry LittlePeng9 (talk) 17:14, September 10, 2014 (UTC)
 * ahahahahahaha you're.so.pretty! 21:26, September 10, 2014 (UTC)

Ramsey numbers
I did it . LittlePeng9 (talk) 12:38, May 4, 2015 (UTC)

Reminder
Just a reminder than you have a few solutions left to write down. LittlePeng9 (talk) 19:29, May 20, 2015 (UTC)

Fundamental sequences for extended Veblen-function
Hello, Deedlit, I will be very grateful If you'd read and commented this my post. Did I correctly understand FS up to SVO? If there are not mistakes, I'd like to write an article about Veblen function.Denis Maksudov (talk) 13:21, February 25, 2017 (UTC)

I published my article about Veblen function.

I have a question: let p=(...,a@b,g-1@0) then for case a-limit and b-successor

here Veblen defines FS as(...,a@b,g@0)[n]=(...,a[n]@b,p+1@0)

(I translated Veblen to zero-indexced).

and here Ranzi defines (...,a@b,g@0)[n]=(...,a[n]@b,p+1@b-1)

This diffence would be for all possible limit cases for a and b

Are both definitions correct? --Denis Maksudov (talk) 20:01, March 18, 2017 (UTC)


 * Yes, both look fine. Deedlit11 (talk) 07:57, March 19, 2017 (UTC)

Ordinal hierarchy
 I studied your googology and want to write a popular article about this in Russian. Because in Russian in the Internet there is almost no material on googology. I want to use the most visual comparisons.

And I want to forgive you for help. Could you check the comparisons in this table?

http://lihachevss.ru/ordinal.html

If possible, write in which step I was wrong.

Thank you. And sorry for my English.

Scorcher007 (talk) 02:04, March 23, 2017 (UTC)


 * I would just like to point out it is not true that \(I_n\) (the \(n\)-th inaccessible cardinal) is not the least \(\Pi^0_n\)-indescribable cardinal. LittlePeng9 (talk) 05:21, March 23, 2017 (UTC)


 * Hello Scorcher! Some comments:


 * I'm suspicious of the claim that \(\Pi^1_0\)-indescribable is equivalent to being Mahlo, but I don't really know. Can someone find a reference confirming or denying this?


 * I'm not familiar with your usage of \(\Omega_{(\alpha_n,\ldots,\alpha_0)}\), but if you mean that \(\Omega_{(1,\alpha)}\) is the \(\alpha\)th fixed point of \(\beta \rightarrow \Omega_\beta\), and so on, it looks fine.


 * The major problem occurs when you go from M to K. This is something that I guess I didn't make clear enough when I wrote Ordinal Notations VI; the weakly compact notation is actually more powerful than people are thinking. When we get to Mahlo cardinals, we can define Mahlo(0) to be the Mahlo cardinals, then define Mahlo(1) to be the Mahlo cardinals that are limits of Mahlo cardinals, then define Mahlo(1) to be Mahlo(1) cardinals that are limits of Mahlo(1) cardinals, and so on. Then we can define Mahlo(1,0) to be cardinals \(\kappa\) that are in Mahlo(\(\kappa\)), and continue the hierarchy in the usual fashion. Then we can define \(M(\alpha_n,\ldots,\alpha_0,\beta)\) to be the \(\beta\)th cardinal in Mahlo(\(\alpha_n,\ldots,\alpha_0)\). Thus we get a hierarchy similar to the inaccessible hierarchy, just like you have in your list. But, the cardinal that diagonalizes over this is not K! Rather, it is the first 2-Mahlo cardinal. This is the key thing to notice about the weakly compact ordinal notation: there are now different levels of collapsing. In the collapsing function \(\Psi_\pi(\alpha,\beta)\), \(\alpha\) repesents which type of collapsing we are doing. If \(\alpha = 0\), we collapse to a strongly critical ordinal, which is what we normally collapse to anyway, so this is our normal collapsing function. If \(\alpha = 1\), we collapse to a regular cardinal, so this is where Mahlo cardinals collapse to get us a hierarchy if inaccessible cardinals. If \(\alpha = 2\), we collapse to a Mahlo cardinal, so we can collapse 2-Mahlo cardinals to get a hierarchy of Mahlo cardinals. But we can also collapse a 2-Mahlo at the \(\alpha = 0\) and \(\alpha = 1\) levels, so we get finer and finer structures as \(\alpha\) increases.


 * In short, what you call K should actually be \(\Xi(3)\), the smallest 2-Mahlo cardinal, and what you call S(n) should actually be \(\Xi(n+2)\), the smallest (n+1)-Mahlo cardinal. The cardinal K will then diagonalize over this grander hierarchy of hierachies. So it's bigger than you think!


 * I don't think I will find the time to go over all of your inequalities, but some of them are making me wonder, like 219 and 220; why are those true? But overall, it seems like you have the right general idea. (except the major probelm above) Deedlit11 (talk) 07:42, March 24, 2017 (UTC)


 * As for the 219 and 220 steps (now it's other step, I changed table), I came to this conclusion on the basis of the following comparisons:


 * As for the 2-Mahlo cardinal, etc., as well as the diagonalization of collapsing functions with K, I believe that in my notation this will look like this:
 * 02.jpg

Scorcher007 (talk) 02:33, March 25, 2017 (UTC)

Hello, Deedlit!

I noticed several similar expressions:

$$\theta(\Omega^{n}\times \alpha_n + \cdots + \Omega^{2} \times \alpha_2 + \Omega \times \alpha_1 + \alpha_0, 0) = \varphi(\alpha_n, ... ,\alpha_2, \alpha_1, \alpha_0, 0)$$

$$\Psi(\Iota^{n}\times \alpha_n + \cdots + \Iota^{2} \times \alpha_2 + \Iota \times \alpha_1 + \alpha_0, 0) = \theta(\Omega_{(\alpha_n, ... ,\alpha_2, \alpha_1, \alpha_0, 0)},0)$$

$$\chi(\Mu^{n}\times \alpha_n + \cdots + \Mu^{2} \times \alpha_2 + \Mu \times \alpha_1 + \alpha_0, 0) = \Psi(\Iota_{(\alpha_n, ... ,\alpha_2, \alpha_1, \alpha_0, 0)},0)$$

$$\Xi(\Kappa^{n}\times \alpha_n + \cdots + \Kappa^{2} \times \alpha_2 + \Kappa \times \alpha_1 + \alpha_0, 0) = \chi(\Mu_{(\alpha_n, ... ,\alpha_2, \alpha_1, \alpha_0, 0)},0)$$

Why not to create common ordinal notation? Something in follow kind:

I propose to use $$\Omega^{(1)}$$ instead $$\Iota$$, $$\Omega^{(2)}$$ instead $$\Mu$$ and so on,

$$\varphi_0^0(\gamma)=\omega^\gamma$$,

$$\varphi_k^0(\gamma)=\Omega_k^\gamma$$, (Let $$\Omega_0=\omega$$),

$$\varphi_0^n(\gamma)=\Omega^{(n)}_{\gamma}$$ (Let $$\Omega_\gamma^{(0)}=\Omega_\gamma$$),

$$\varphi_k^n(\gamma)=\Omega^{(n)\gamma}$$,

$$\varphi_k^n(0,\alpha_1, ...,\alpha_n)=\varphi_k^n(\alpha_1, ...,\alpha_n)$$,

$$\varphi_k^n(s,\alpha_k,z,\gamma)=\gamma$$th common fixed point of functions $$\xi \mapsto\varphi_k^n(s,\beta,\xi,z)$$ for all $$\beta<\alpha_k$$, where $$s$$ is a string of arguments $$s=\alpha_1,\alpha_2,...,\alpha_{k-1}$$, including $$\alpha_1>0$$, and $$z$$ is string of $$n-k-1$$ zeros,

$$\theta_{k-1}^i(\Omega_k^{(i)\beta_1}\alpha_1+\cdots+\Omega_k^{(i)\beta_n}\alpha_n,\gamma)=\varphi_{k-1}^i(\alpha_1,...,\alpha_n,\gamma)$$, where $$\beta_1>\cdots>\beta_n\geq 0$$.

Then, for example:

$$\theta_0^1(\Omega^{(1)\beta_1}\alpha_1+\cdots+\Omega^{(1)\beta_n}\alpha_n,\gamma)$$ is equivalent of $$\Psi(\Iota^{\beta_1}\times \alpha_1 + \cdots + \Iota^{\beta_{n}} \times \alpha_{n}, \gamma)$$,

$$\theta_0^2(\Omega^{(2)\beta_1}\alpha_1+\cdots+\Omega^{(2)\beta_n}\alpha_n,\gamma)$$ is equivalent of $$\chi(\Mu^{\beta_1}\times \alpha_1 + \cdots + \Mu^{\beta_{n}} \times \alpha_{n}, \gamma)$$,

$$\theta_0^3(\Omega^{(3)\beta_1}\alpha_1+\cdots+\Omega^{(3)\beta_n}\alpha_n,\gamma)$$ is equivalent of $$\Xi(\Kappa^{\beta_1}\times \alpha_1 + \cdots + \Kappa^{\beta_{n}} \times \alpha_{n}, \gamma)$$

and so on.

I don't know how to write common collapsing function for this notation. I see something in this kind:

$$C_0(\alpha, \beta)=\{0,\beta\}$$,

$$C_{n+1}(\alpha, \beta)=\{\gamma+\delta,\varphi_{\gamma}^i(\gamma,\delta),\Omega_{\gamma}^{(i)},\theta(\eta,\gamma)|\gamma,\delta,\eta \in C_{n}(\alpha, \beta),\eta<\alpha,i<\omega\}$$,

$$C(\alpha,\beta)=\cup_{n=1}^\omega C_n(\alpha,\beta)$$,

$$\theta(\alpha,\beta)=$$ the $$\beta$$th ordinal in $$\{\beta|\beta\notin C(\alpha,\beta)\}$$

What do you think Deedlit, does this idea have any perspectives?--Denis Maksudov (talk) 09:01, March 23, 2017 (UTC)

The problem with that approach is that those elements don't actually form a sequence, so there is no obvious next step. (so beyond a unification of the symbols, there is no gain) Chronolegends (talk) 17:44, March 23, 2017 (UTC)


 * Hello Denis! I have thought about something like this; it would be nice to organize varying levels of diagonalizers into an overall system. I have in mind something that I am calling "the diagonalizer hierarchy"; the main problem is that I am not sure whether my description of it is enough to uniquely specify the system. I will try to dredge it up and put it on a blog post sometime.


 * With regards to your specific suggestions, Chronolegends is right that I,M,K don't quite form a sequence, or at least not a trivially obvious one. After \(\Omega_\alpha\), we use I as a diagonalizer for the fixed point hierarchy of \(\Phi(\alpha) = \Omega_\alpha\). Then I(1,0) diagonalizes the fixed point hierarchy of \(\alpha \rightarrow I(\alpha)\), I(2,0) diagonalizes the fixed point hierarchy of \(\alpha \rightarrow I(1,\alpha)\), and so on. So we build up an inaccessible hiearachy of arbitrary arity, which is a different sort of thing then we had before, and it is this hierarchy that M diagonalizes. Note that we don't then go from M to K; after we build up a hierarchy of Mahlo cardinals, the next step is to go the the smallest 2-Mahlo cardinal (let's call it M_2) and it is this cardinal M_2 that diagonalizes over the hierarchy of 1-Mahlo cardinals. Then we go to 3-Mahlo cardinals, alpha-Mahlo cardinals, (1,0)-Mahlo cardianls, and so on, and it is this hierarchy of hierarchies that K diagonalizes over. So we do have a sequence I,M,M_2,M_3... where each one diagonalizes over the hierarchy of the previous one. And I think we can talk about the sequence M,K,... where the third member would diagonalize over a hierarchy of hierarchies of hierarchies, and so forth; the natural setting for this would be to use indescribable cardinals. I'm hoping that the diagonalizer hierarchy would extend beyond this, however. Deedlit11 (talk) 06:08, March 24, 2017 (UTC)

I am officially hyped for that :D the strength of such a thing would be absolutely bonkers Chronolegends (talk) 16:14, March 24, 2017 (UTC)

Thanks for answer, Deedlit. I meaned something like this:

\(\theta(\varphi^{1}(1,0))=\theta(\theta^{1}(1,0))=\theta(\Omega_{\Omega...})\), where \(\varphi^1\) is Veblen function on base \(\alpha \mapsto \Omega_\alpha\),

\(\theta(\varphi^{1}(1,0,0))=\theta(\theta^{1}(\Omega^{(1)}))=\)$$\psi(\psi_\Iota(\Iota^\Iota))=\Psi(\Iota)$$,

\(\theta(\theta^{1}(\Omega^{(1)}_{\Omega^{(1)}...}))=\)$$\Psi(\Iota_{\Iota...})=\Psi(\Iota_{(1,0)})$$\(=\theta(\theta^{1}(\theta^{2}(1,0)))=\theta(\theta^{1}(\varphi^{2}(1,0)))\), where \(\varphi^1\) is Veblen function on base \(\alpha \mapsto \I_\alpha\),

\(\theta(\theta^{1}(\varphi^{2}(1,0,0)))=\theta(\theta^{1}(\theta^{2}(\Omega^{(2)})))=\Psi(\I_{(1,0,0)})\)

and so on (about M and K my knowledge is so small that I will not even try to write any examples).--Denis Maksudov (talk) 19:44, March 24, 2017 (UTC)