Take the language $\mathcal L(=,\in)\_{\omega\_1,\omega\_1}$, if we restrict infinite quantification strings, i.e. of the forms $``(\forall v\_i)\_{i \in \omega}"; ``(\exists v\_i)\_{i \in \omega}"$, to bounded forms, that is $``\forall v\_0 \in x, \forall v\_1 \in x,..."$ and $``\exists v\_0 \in x , \exists v\_1 \in x,..."$ abbreviated as $``(\forall v\_i \in x)\_{i \in \omega}"; ``(\exists v\_i \in x)\_{i \in \omega}"$ respectively, or even if we further demand the bound to be a definable constant or the value of a function. Call the resulting language as "*bounded*-$\mathcal L(=,\in)\_{\omega\_1, \omega\_1}$", and denote it by $\mathcal L(=,\in)\_{\omega\_1, [\omega\_1]}$



> 
> How would $\mathcal L(=,\in)\_{\omega\_1, [\omega\_1]}$ differ from $\mathcal L(=,\in)\_{\omega\_1, \omega\_1}$ as regards basic properties of logic like Completeness, Compactness, etc..
> 
> 
> 


As a clear example, if to $\mathcal L(=,\in)\_{\omega\_1, [\omega\_1]}$ we add all axioms of $\sf ZFC$ written, as usual, in the fragment $\mathcal L(=,\in)\_{\omega,\omega}$ and suppose we add an axiom of Foundation but in a bounded form, that is:


$\textbf{Foundation: } \\\forall \alpha \, (\forall v\_i \in V\_\alpha)\_{i \in \omega} \, \exists x: \bigvee\_{i \in \omega} (x=v\_i) \land \bigwedge\_{i \in \omega} (v\_i \not \in x)$


Would that theory be different from theory written in $\mathcal L(=,\in)\_{\omega\_1,\omega\_1}$, axiomatized by $\sf ZFC$ axioms in $\mathcal L(=,\in)\_{\omega,\omega}$ , and Foundation written as above but unbounded.



Is bounded-$\mathcal L_{\omega_1,\omega_1}$ significantly different from $\mathcal L_{\omega_1,\omega_1}$?
Your new formulation of foundation is equivalent to the corresponding unbounded form. The reason is that every failing instance of the unbounded formulation will lead to a failing instance of the bounded formulation, since if there is a countable collection of $v\_i$ with no $\in$-minimal element, then by the usual ZFC axioms it follows that $v\_0$ is in some $V\_\alpha$, and so the $v\_i$ that appear within this $V\_\alpha$ will fall under the bounded form, and an $\in$-minimal for that part of the family will be $\in$-minimal for the whole family since $V\_\alpha$ is transitive.


