Informally the axiom *schema* of accessibility states that for each unary function $F$ that is definable over the whole universe of discourse "in the language of set theory", like the powerset function $P$, or the singleton function $\iota$, or the set Union function $\bigcup$, etc.. for each such a function $F$ we can coin an accessibility notion [dependent on $F$] whereby a set $x$ is said to be reached (i.e. is accessible) from a set $\alpha$ if and only if there is a set $\beta$ that is hereditarily strictly subnumerous to $x$ such that $F(\beta)$ is supernumerous to $x$ and $\alpha \subseteq \beta$; or otherwise $x$ is hereditarily subnumerous to $\alpha$ . Now the scheme states that per each definable unary function $F$ for every set $x$ there exists a set of all sets that are hereditarily $F$ accessible from $x$. The real intention is to have a theory that can prove the existence of large cardinals. Infinity is provable since all hereditarily finite sets are seen as hereditarily power accessible from the set $\emptyset$. Now it can be proved that the set of all cardinals that are hereditarily Power (of Set Union) accessible from $\omega\_0$ exists and this would be a regular limit of regular cardinals, i.e. the first inaccessible cardinal, and from that one can prove the existence of a set of all sets that are hereditarily strictly subnumerous to this first inaccessible, and this would serve as a domain of a model of $\text{ZF}$. To get to that, we only need to add this axiom scheme on top of axioms of Zermelo set theory minus infinity and add an axiom of Transitivity that asserts that every set is a subset of some transitive set, this way we can define transitive closures and define the hereditary notions. The issues are: what would be the consistency strength of this theory? how much this theory is stronger than the theory with the $F$ function fixed to be the Power of union set function, i.e. $\forall x,y \ [F(x)=y \leftrightarrow y=P(\bigcup(x))]$? what kinds of inaccessible cardinals can this theory prove? Would this theory prove Replacement?


Formal workup:


$\text{Axiom schema of Accessibility:}$ if $``F"$ is a symbol that denotes a unary function that is ***definable*** in the language of set theory over the whole universe of discourse, and if $\phi(Y)$ is a formula in which $X$ doesn't occur free and $Y$ occurs free and only free, then all closures of: 


$$[\exists \alpha \forall Y (\phi(Y) \to Y \ ..ACC^F \ \alpha)] \to \exists X \forall Y (Y \in X \leftrightarrow \phi(Y))$$ are axioms.


Where $ACC^F$ is defined as:


$$Y \ ACC^F \ \alpha \iff Y\ ..\leq \ \alpha \lor \exists \beta \ [\beta \ ..< \ Y \wedge \alpha \subseteq \beta \wedge F(\beta) \geq Y] $$


Where generally $``..R"$ denotes "hereditarily $R$" relation defined as:


$$ X \ ..R \ \ Y \iff X \ R \ Y \wedge \forall m \in TC(X) [m \ R \ Y]$$


Where $TC(X)$ is defined in the customary manner as the minimal transitive superset of $X$.


Where: $ x < y \iff \exists f (f:x\to y \wedge f \text{ is an injection}) \wedge \not \exists g (g: y \to x \wedge g \text{ is an injection})$


and: $ x \geq y \iff \exists f (f: y \to x \wedge f \text { is an injection} )$;


and: $ x \leq y \iff y \geq x$


Now the question is:


What is the consistency strength of the theory whose axioms are the axioms of $[\text{Z} - \text{INF.}] + \text{Transitivity} + \text{Accessibility}?$


Where Axiom of Transitivity is the axiom stating that every set is a subset of some transitive set.


In particular would Replacement be provable in this theory?



What is the consistency strength of Z+ Accessibility?
A limited form of Replacement is provable in this theory! That of Replacing elements of a set of ordinals by ordinals, where an ordinal is defined in the usual manner after von Neumann.


Lemma: $\forall ordinals \ \alpha, \beta \ (\alpha \ ..\leq \ \beta \lor \beta \ ..< \ \alpha)$


The proof of this lemma is obvious, and it is the main tool that would enable the following proof of ordinal replacement.


Let $S$ be an infinite set of ordinals, then $\cup(S)$ is a ordinal such that $\forall m \in S [ m \ ..\leq \cup(S)]$
Let $F$ by any function that sends ordinals to ordinals.
Now because of schema (see prior answer) we have the set $$A=\{x|\exists m \in S (x=F(m)) \wedge x \ .. \leq \cup(S)\}$$
Now what remains are the $F$-images of elements of $S$, that are strictly bigger than $\cup(S)$, call those the "outside of $A$ ordinals", and thus $\cup(S)$ is hereditarily strictly smaller than each outside of $A$ ordinal, now we take the subset $S\_1$ of $S$ that contains all elements of $S$ that are sent by $F$ to those outside of $A$ ordinals, i.e. $$S\_1 = \{x \in S| \cup(S) \ ..< F(x)\}$$ 


Take the set $D$ of all ordinals that are equinumerous with $\cup(S)$ [schema in prior answer], take any injective function $g$ from $S\_1$ to $D$, now take the converse function $g^{-1}:rng(g) \to S\_1$, let $\kappa= min(rng(g))$, now define a new relation $h(x)=F(g^{-1}(x))$, now clearly we have:


$$\forall y (\exists x \in rng(g) [y=h(x)] \to y \ ..ACC^{h} \ \kappa)$$


The reason is because every such $y$ is an outside of $A$ ordinal so each element $x$ in $rng(g)$ would be hereditarily strictly subnumerous to $y$, and
since $h(x)=y$ and $\kappa$ is a subset of each element of $rng(g)$ then by definition we have $y \ ACC^h \ \kappa$, now each element in $y$ [thus in $TC(y)$] is either strictly supernumerous to $x$ and thus $ACC^h$ from $\kappa$ [through y], or is subnumerous to $\kappa$ thus hereditarily subnumerous to $\kappa$ thus $ACC^h$ from $\kappa$!


Then from axiom of Accessibility we get the set $B=\{y| \exists x \in S\_1 (y=F(x))\}$ exists. Now $A \cup B$ is our Replacement set. The case when $S$ is a finite set of ordinals can in a similar manner be proved, we just take $D$ to be any set of infinite countable ordinals, such that $S$ equinumerous to $D$, then take $g$ to be a bijective function that witnesses this equinumerousity, and then carry the rest of the arguemnt as above. QED


Now I'm not sure if that can lead to an argument that can prove full Replacement in this theory, but I guess so. I think if Choice is assumed, and the definition of $ACC^F$ is changed to: $$ X \ ACC^F \alpha \iff X \ ..\leq \alpha \lor \exists \beta (\beta \ ..< TC(X) \wedge \alpha \subseteq \beta \wedge F(\beta) \geq X)$$


Then I think by a modified form of the above argument full Replacement can be proven. Instead of $\cup(S)$ we take $TC(S)$, and use the same argument, but let $h(x)=TC(F(g^{-1}(x)))$. 


