Would this argument support logicism?

Question

Notice the following theory written in multi-sorted first order logic with equality and membership, with axioms of:

Comprehension:$\small k=1,2,..,\omega, \omega+1,..$ $\forall x^{i_1},...,x^{i_n} \exists x^k: \forall y^{<k} (\ y^{<k} \in x^k \leftrightarrow \phi(y^{<k},x^{i_1},..,x^{i_n}))$

Sorts: $for \ \small i < n, j\geq n$ $ \forall x^i \exists x^{<n} (x^{<n}=x^i) \land \forall x^j \not \exists x^{<n}(x^{<n} = x^j)$

sorts when not otherwise specified are taken to range over all metatheoretic ordinals $\{0,1,2,..,\omega,\omega+1, \omega+2,..\}$

This theory interprets Zermelo set theory + Rank; so it constructs a cumulative hierarchy of size $V_{\omega + \omega}$

The caveat here is that this is too near to logic! The sorting axiom can be considered as belonging to the underlying multi-sorted FOL with identity. We only have ONE set scheme that is comprehension. Not only that we can push this further into the realm of high order monadic logic with identity and accumulative sorting, i.e. higher sort unary predicates can take any lower sort unary predicates as their arguments, but the opposite is not allowed, so $x^i(x^k)$ is only a formula if $k<i$. Now the axioms would read as:

Comprehension:$\small k=1,2,..,\omega, \omega+1,..$ $\forall x^{i_1},...,x^{i_n} \exists x^k: \forall y^{<k} ( x^k(y^{<k}) \leftrightarrow\phi(y^{<k},x^{i_1},..,x^{i_n}))$

Sorting: $\small i < n; \ j \geq n,\ k>i, \ k>j$

$\forall x^i \exists x^{<n}: \forall x^k ( x^k(x^i) \iff x^k(x^{<n})) \\ \forall x^j \neg \exists x^{<n}: \forall x^k ( x^k(x^j) \iff x^k(x^{<n}))$

here $y^{<k}$ would be a variable that range over unary predicates of all sorts below $k$, while variables like $x^k$ denote variables ranging over all unary predicates of sort $k$.

We can push that further (go beyond metatheoretic ordinals less than $\omega+\omega$) as long as we have ordinal notations, i.e. as long as our metatheoretic ordinals (indexing sorts) are below $\omega_1^ {CK}$.

The point here is that the comprehension axiom can be understood as being about logic, and so a logical scheme.

It appears that $\sf ZFC$ and higher extensions, all can have models in sets belonging to these theories.

why that won't count as an argument in favor of logicism?

Answer 1

This is not an altogether new argument for logicism. Those who are skeptical of logicism will regard the comprehension scheme as non-logical, and one non-logical assumption means the deal is off (even Russell acknowledged that, in regard to the need for Infinity in Principia Mathematica). You actually smuggle infinity in by subverting the usual restrictions on multi-sorted logic. For some (not not for me, I have different attitudes about this on different days of the week) some abstraction principle for properties IS part of logic, and quantification over properties is part of logic...and the nonlogical character of assumptions about how many objects there are remains a problem.