Work in progress!!!

I was thinking of making some strong ordinal notation to use it to compare functions.I was gonna define it using simpe recursion,but then I thought of Taranovsky's notation which is very strong BECAUSE it's not defined using simple recursion. So I decided to make a very strong ordinal notation using a different aproach than what I usually do.If you're about to comment,please bare in mind that this is just an early version and the reason why I am making this post is because I want to know if the way I described it is well-defined and if it's really what I'm trying to do. Also we should note that it is defined recursively,so it's limit must be equal to or below \(\omega^{\text{CK}}_1\) and it is probably weaker than Taranovsky's notation. 

I wasn't sure how to define it but now I realized it doesn't take that many of definitions.

Let \(\sigma(0,\alpha) = \alpha+1\) and \(\sigma(\alpha)\) will be short for \(\sigma(\alpha,0)\).

Let \(\sigma(\alpha+1) = \text{Sup}\{\sigma(\alpha,\sigma(\alpha)),\sigma(\alpha,\sigma(\alpha,\sigma(\alpha))),\sigma(\alpha,\sigma(\alpha,\sigma(\alpha,\sigma(\alpha)))),........\}\) (if \(\alpha\) is not a supremum to \(L(\beta)\))

Now let (if \(\alpha\) is not the supremum to \(L(\gamma)\)) \(\sigma(\alpha +1,\beta +1) = \\ \text{Sup}\{\sigma(\alpha,\sigma(\alpha+1,\beta)+1),\sigma(\alpha,\sigma(\alpha,\sigma(\alpha+1,\beta)+1)),\sigma(\alpha, \sigma(\alpha,\sigma(\alpha,\sigma(\alpha+1,\beta)+1)))........\}\).

If \(\alpha\) is a limit ordinal (that is not supremum of \(L(\beta)\)) then \(\sigma(\alpha) =  \text{Sup}\{\sigma(\gamma),\sigma(\gamma,\sigma(\gamma)),\sigma(\gamma,\sigma(\gamma,\sigma(\gamma)))........\}\forall \gamma (\gamma < \alpha)\)

And \(\sigma(\alpha,\beta+1) = \text{Sup}\{\sigma(\gamma,\sigma(\alpha,\beta)+1),\sigma(\gamma,\sigma(\gamma,\sigma(\alpha,\beta)+1)).........\}\forall \gamma (\gamma < \alpha)\)

And if \(\beta\) is a limit ordinal then \(\sigma(\alpha,\beta) = \text{Sup}\{\sigma(\alpha,\gamma)\}\forall \gamma (\gamma < \beta)\) (if \(\alpha\) is not the supremum to \(L(\gamma)\))

Let \(L(\alpha)\) be the set of all ordinals \(\beta\) such that \(\beta\) is definable in \(\sigma()\) using \(\sigma\) no more than \(\alpha\) times in it's definition and such that it uses no ordinal \(\gamma\) iff \(\gamma \notin L(\delta) \forall \delta<\alpha\).Now let \(L(\alpha) = \bigcup_{\beta<\alpha}L(\beta)\) if \(\alpha\) is a limit ordinal and let \(L(0)=\{0\}\).

Here we say that \(\alpha\) is the supremum of \(L(\beta)\) for some \(\beta\) iff it's the minimal ordinal not in \(L(\beta)\).

Here comes the more difficult part.

If we do have an ordinal \(\alpha\) that is the supremum of \(L(\beta)\) for some \(\beta\) then:

\(\forall \alpha (\alpha \notin L(\beta)) : \sigma(\alpha +1) = \sigma(\alpha, \sigma(\gamma +1)) \forall \gamma (\beta \in L(\delta) \land \gamma \notin L(\delta) \Rightarrow \beta < \delta \vdash \alpha < \gamma)\)

\(\forall \alpha_1,\alpha_2 (\alpha_1,\alpha_2 \in L(\beta)) :\alpha_1 < \alpha_2 \Leftrightarrow \exists \gamma ((\gamma \in L(\delta)\forall \delta < \gamma) \land (\alpha_1 \leq \sigma(\gamma) < \alpha_2 \lor \alpha_1 < \sigma(\gamma) \leq \alpha_2))\) 

It should be clear that \(\forall \alpha,\beta (\alpha < \beta) L(\alpha) \subset L(\beta)\) and that this definition of \(\sigma(\alpha)\) only works \(\forall \alpha (\alpha < \omega_1)\) (for now).

I will leave the analysis for another blog.