# Better bounds on my ordinal hyperoperators

I've defined some ordinal hyperoperators as follows:

$$a\{b\}c=\begin{cases}a+1,&b=0{\rm~or}~c=0\\\sup\{(a\{b\}x)\{y\}a:x<c,y<b\},&\text{else}\end{cases}$$

For $a$ and $c$ that satisfy $2\cdot x=x$, I believe the following computations are correct:

$$a\{1\}c=a+c\\a\{2\}c=a\cdot c\\a\{3\}c=a^c\\a\{4\}c=a^{a^c}$$

And from there, it gets a little wonky. For example, if the above is correct, we should have

$$\omega\{5\}(\omega\cdot(1+x))=\varepsilon_x$$

And by induction, I managed to prove

$$\omega\{x\}\omega\ge\varphi_{2\cdot(3+x)}(0)$$

where $\varphi$ is the Veblen function.

I recently speculated that $\omega\{k+x\}\omega$ might be closer to $\varphi_x(0)$ for some natural $k$, however I can't prove this. Assistance appreciated.

The general rule is that two increments of the hyperoperator gets you one increment of the Veblen subscript.

Your equalities are correct. More generally, let $\sigma_\alpha(\beta)$ be the smallest ordinal in the range of the function $\varphi_\alpha$ that is greater than $\beta$, and let $\tau_\alpha(\beta)$ be the ordinal $\gamma$ such that $\varphi_\alpha(\gamma) = \sigma_\alpha(\beta)$. Then for $\alpha$ an infinite ordinal, we have $\alpha \{ 5 \} \omega = \sigma_1(\alpha)$ and therefore

$$\alpha \{5\}\omega(1+\beta) = \sigma_1^{1+\beta}(\alpha) = \varepsilon_{\tau_1(\alpha)+\beta}$$

which matches your equality, since $\tau_1(\omega) = 0$.

For $b=6$ and $\alpha = \omega(1+\gamma)$, we have $\alpha \{6\} 1 = (\omega(1+\gamma)+1)\{5\}\omega(1+\gamma) = \varepsilon_{\tau_1{(\omega(1+\gamma)+1)}+\gamma}$. Each increment of the right operand will apply the function $f(x) = x \{5\} \omega(1+\gamma)$, which will add a $1+\gamma$ to the subscript each time. So $\omega(1+\gamma)\{6\}(1+\beta) = \varepsilon_{\tau_1{(\omega(1+\gamma)+1)}+\gamma+(1+\gamma)\beta}$. Since $\tau_1{(\omega(1+\gamma)+1)} = \tau_1(\gamma)$, we have

$$\omega(1+\gamma)\{6\}(1+\beta) = \varepsilon_{\tau_1{(\gamma)}+\gamma+(1+\gamma)\beta}$$

For $b = 7$ and $\alpha = \omega(1+\gamma)$, note that for $\varepsilon$-numbers we have $\gamma = \omega(1+\gamma)$, so $\omega(1+\gamma)\{7\}\beta$ will be $\beta$ successive applications of the function $\delta \mapsto \varepsilon_{\tau_1{(\delta)}+\delta+(1+\delta)(\omega(1+\gamma))}$. This function is greater than the function $\delta \mapsto \varepsilon_\delta$, so $\omega(1+\gamma)\{7\}\omega$ will be at least the next $\zeta$-number larger than $\gamma$. (Note: if $\gamma$ is itself a $\zeta$-number, we will have $\gamma \{7\}0 = \gamma+1$, so in this case we will also increase to the next $\zeta$-number.) On the other hand, the function $\delta \mapsto \varepsilon_{\tau_1{(\delta)}+\delta+(1+\delta)(\omega(1+\gamma))}$ does not take $\delta$ to the next $\zeta$-number or beyond. (Observe that $\tau_1(\delta) \le \delta+1$, and $\omega(1+\gamma) \le \delta$.) So $\omega(1+\gamma) \{7\}\omega = \sigma_2(\gamma)$, and iterating we have

$$\omega(1+\gamma) \{7\}\omega(1+\beta) = \sigma_2^{1+\beta}(\gamma) = \zeta_{\tau_2(\gamma) + \beta}$$

The pattern continues, so one can prove by induction that for $1 \le n < \omega$,

$$\omega(1+\gamma)\{2n+3\}\omega(1+\beta) = \varphi_n(\tau_n(\gamma)+\beta)$$

$$\omega(1+\gamma)\{2n+4\}(1+\beta) = \varphi_n(\tau_n(\gamma) + \gamma + (1+\gamma)\beta)$$

For $b = \omega$, we have $\alpha \{\omega\} 1 = \sup_{n < \omega} \{(\alpha+1) \{n\} \alpha\}$, and the right hand side will be the smallest ordinal greater than $\alpha$ and closed under all $\varphi_n$, namely $\sigma_\omega(\alpha)$. So for $\alpha$ infinite we have

$$\alpha \{\omega\} (1+\beta) = \sigma_\omega^{1+\beta}(\alpha) = \varphi_\omega(\tau_\omega(\alpha) + \beta)$$

For $b = \omega + 1$, we have $\alpha \{\omega+1\}1 = (\alpha+1)\{\omega\}\alpha = \varphi_\omega(\tau_\omega(\alpha)+\alpha)$. Each increment of the right operand will and $\alpha$ to the argument of $\varphi_\omega$, so we have

$$\alpha \{\omega+1\}(1+\beta) = \varphi_\omega(\tau_\omega(\alpha) + \alpha(1+\beta))$$

Then for $b = \omega+2$, we see the same phenomenon of every $\omega$ steps taking the ordinal to the next value of $\varphi_{\omega+1}$, so we have

$$\alpha \{\omega+2\}\omega(1+\beta) = \varphi_{\omega+1}(\tau_{\omega+1}(\alpha)+\beta)$$

and

$$\omega(1+\gamma) \{\omega+3\}(1+\beta) = \varphi_{\omega+1}(\tau_{\omega+1}(\gamma) + \gamma + (1+\gamma)\beta)$$

We then repeat the same pattern, and again at every limit ordinal. So for a limit ordinal $\delta$, $1 \le n < \omega$, and $\alpha$ infinite, we have

$$\alpha \{\delta\} (1+\beta) = \varphi_\delta(\tau_\delta(\alpha) + \beta)$$

$$\alpha \{\delta+1\}(1+\beta) = \varphi_\delta(\tau_\delta(\alpha) + \alpha(1+\beta))$$

$$\omega(1+\gamma) \{\delta+2n\}\omega(1+\beta) = \varphi_{\delta+n}(\tau_{\delta + n}(\gamma)+\beta)$$

$$\omega(1+\gamma) \{\delta+2n+1\}(1+\beta) = \varphi_{\delta+n}(\tau_{\delta+n}(\gamma) + \gamma + (1+\gamma)\beta)$$