What is the consistency strength of F accessibility?

Question

Let's assume all axioms of $\text{Z}- \text{Infinity}$.

Now let $F$ be any function that is definable over the whole universe of discourse by a formula in the language of $\text{Z}$.

Now we define $F$ related accessibility $``ACC^F"$ as:

$$X \ ACC^F \alpha \iff \alpha <X \wedge \exists \beta < X \ \big{[}F \big{(}\bigcup(\alpha \cup \beta)\big{)} \geq X \big {]} $$

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} )$

Define $F$ related linear accessibility $``LACC^F"$ as:

$$X \ LACC^F \alpha \iff \alpha <X \wedge \exists \beta <X \ \big{[} \beta \ ACC^F \alpha \wedge F \big{(}\bigcup (\alpha \cup \beta)\big{)} \geq X \big{]} $$

$\text{Axiom schema of Accessibility:}$ if $F$ is a binary function symbol, and $\phi(y)$ is a formula in which $x$ doesn't occur free and $y$ occurs free and only free, then all closures of:

$$\exists \alpha \forall y [\phi(y) \to y \ LACC^F \alpha] \to \exists x \forall y \ [y \in x \leftrightarrow transitive(y) \wedge \phi(y) ]$$

are axioms.

Where: $ transitive(y) \iff \forall m \in y \ (m \subset y)$

Now $\text{Z}-\text{INF.} + \text{Accessibility}$ would prove $\text{Con(ZF)}$, I assume.

Should this theory be consistent, what is the consistency strength of $\text{Z}-\text{INF.}+ \text{Accessibility}$?

Does it prove Replacement?

Afternote: a continuation of the line of thought along this theory that might help salvage this approach is present at:

What is the consistency strength of Z+ Accessibility?

Answer 1

Unless I'm missing something, this scheme is inconsistent.

Take $$F(x)=\bigcup_{y\in x}(\mathcal{P}(y)).$$ Then for all $X$ with more than one element and any $a$ with $a<X$ we have $X$ ACC$^F$ $a$: taking $b=\{\{X\}\}$, we have $\mathcal{P}(X)\subseteq F(\bigcup (a\cup b))$.

Now consider $\varphi(y)$ to be the statement "$y$ has more than two elements," and let $a=2$ ...