**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 recursively.So I decided to make a very strong 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 by recursion,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 a lot 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 a 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 a 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 the definition of \(\beta\) and such that it uses no ordinal \(\gamma\) if \(\gamma \notin L(\delta) \forall \delta (\delta +1 \geq \alpha)\).Now let \(L(\alpha) = \text{Sup}\{L(\beta)\}\forall \beta (\beta < \alpha)\) if \(\alpha\) is a limit ordinal

Here comes the hard part.

If we do have an ordinal that is a supremum to \(L(\gamma)\) as an initial input aka if \(\beta\) is the last term in the fundamental sequence of \(\alpha\) (aka the last part of the ordinal's normal form,such that it's broken down first when looking at the ordinal's fundamental sequence),which we denote as \(\alpha_\beta\),and \(\beta \notin L(\gamma)\) for some \(\gamma\),then:

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

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

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

I will leave the analysis for another blog.