Choice of bound in the Laver property

Question

$\newcommand{\P}{\mathbb{P}}$ Recall the following formulation of the Laver property (Page 459, Combinatorial Set Theory: With a Gentle Introduction to Forcing (Second Edition) by Halbeisen):

Laver Property. Let $\mathcal{F}$ be the set of all functions $S : \omega \to \operatorname{fin}(\omega)$ such that for every $n < \omega$, $|S(n)| \leq \color{red}{2^n}$. A forcing notion $\P$ has the Laver property iff for all $f \in {}^\omega\omega \cap V$ in the ground model and every $\P$-name $\dot{g}$ for a function in ${}^\omega\omega$ such that $1 \Vdash_\P \forall n < \omega(\dot{g}(n) \leq f(n))$, we have $1 \Vdash_\P \exists S \in \mathcal{F} \cap V \, \forall n < \omega(g(n) \in S(n))$.

Is it possible to modify the bound $\color{red}{2^n}$ above? For instance, is the Laver property equivalent to the statement above with the bound replaced by $r^n$ for some $r < \omega$, or even by $h(n)$ where $h \in {}^\omega\omega \cap V$?

Answer 1

You can replace $2^n$ by any function $h: \omega \to \omega \setminus \{0\}$ in $V$ so that $h(n)$ diverges to infinity. For instance if you want to get the Laver property for $h_0$ from the Laver property for $h_1$, let $i_n$, for each $n$, be the least $m$ so that $$h_0(m') \geq h_1(n),$$ for every $m' \geq m$.

Instead of $\dot g$ simply consider a name for $$g'(n) = g \restriction [i_n, i_{n+1}).$$

The Laver property for $h_1$ gives $h_1(n)$ guesses for the values of $g$ on the interval $[i_n, i_{n+1})$ which is less than the number of guesses asked by $h_0$ on that interval (and in fact ever after $i_n$). The only additional guess we have to make is on $[0, i_0)$, but that's of course fine.