I am considering a *Principle of Ubiquity*, expressed as follows - for a class theory where precisely the elements are sets - with the aid of set abstracts:


For $\alpha(y,z)$ a first order condition so that $\forall y(\exists w(y\in w)\to \exists w (\{z|\alpha(y,z)\}\in w))$:


$\forall v(\exists t (v\in t)\to\exists t(\{w|\forall x(v\in x\wedge \forall y(y\in x\to \{z|\alpha(y,z)\}\in x)\to w\in x)\}\in t))$


The set abstracts can be eliminated with the following Mendelsonian abstraction schema:


$\forall x(x\in \{x|\alpha\}\leftrightarrow\exists y(x\in y)\wedge\alpha)$


It is immediate that we get a theorem of infinity, as well as the least transitive closure of all sets; moreover, several further instances of replacement will hold, though with countable co-finality.


May Z with Ubiquity, instead of just the Axiom of Infinity, justify the Recursion Theorem?



Ubiquity beyond infinity, transitive closure and the recursion theorem?
The answer seems to be affirmative.


Let an adapted version of ubiquity be as follows:


For $\alpha(y,z)$ a functional first order condition so that


$\forall y(\exists w(y\in w)\to \exists w(\{z|\alpha(y,z)\}\in w))$:


$\forall v(\exists w(v\in w)\to\exists w(\{u|\forall x(v\in x\wedge \forall n \forall y(((n,y)\in x\to (n+1,\{z|\alpha(n,z)\})\in x))\to u\in x)\}\in w))$


Let $v=(0,X)$


Let $F=\{u|\forall x(v\in x\wedge \forall n \forall y(((n,y)\in x\to (n+1,\{z|\alpha(n,z)\})\in x))\to u\in x)\}$


We see that $(0,X)\in F$.


Moreover, for any $n$ and $y$, if $(n,y)\in F$ then $(n+1,\{z|\alpha(n,z)\}\in F)$


We see that function F is the least class with these properties, and from the assumption of Infinitude that F is a set.


As F is a set, Mendelsonian abstraction gives us that


$(0,X)\in F\wedge ((n+1,y)\in F\leftrightarrow\exists w((n,w)\in F\wedge \forall u(u\in w\leftrightarrow \alpha (n,u)))$


Am I done?


Edit: Unsurprisingly, Replacement entails Ubiquity.


