*sets*of things, where the things in question may be "concrete". The sentence "there is a set of chairs" is an example of a

*mixed*mathematical claim. It uses the (presumably) non-mathematical predicate "

*x*is a chair" as the defining formula $\varphi(x)$ in an instance of the

*Comprehension Scheme*:

- $\exists y \forall x(x\in y \leftrightarrow \varphi(x))$

*Science Without Numbers*) argued that the justification of the use of such (believed-to-be-untrue) sentences in reasoning involves their

*conservativeness*.

*Unrestricted*comprehension is, of course, inconsistent. For replacing $\varphi(x)$ with $x \notin x$ yields:

- $\exists y \forall x(x\in y \leftrightarrow x \notin x)$

*non-mathematical*vocabulary. Suppose that we call sentences containing only "concrete" predicates

*nom*-sentences; call instances of the Comprehension Scheme using only

*nom*-formulas

*comp*-axioms. The simplest relevant conservativeness result says:

- If there is a derivation of a
*nom*-sentence $B$ from a*comp*-axiom, there is also a derivation of $B$ from*logic*.

Here is the simplest proof of the simplest kind of conservativeness result which will give a flavour of

*why*comprehension axioms are conservative. We consider the simplest scenario: the

*nom*-language $\mathcal{L}$ is a first-order language (with identity) and has only a single unary concrete predicate $Fx$ and we consider the simplest

*comp*-axiom, $\exists y \forall x(Rxy \leftrightarrow Fx)$, where we write $Rxy$ to mean "

*x*is an element of

*y*". The conservativeness result now is: for any $\mathcal{L}$-sentence $B$,

- $ \text{If } \exists y \forall x(Rxy \leftrightarrow Fx) \vdash B \text{ then } \vdash B$.

First, note that, in general, if $\exists y \varphi(y) \vdash B$, then $\varphi(y/c) \vdash B$ (where $c$ is a new constant: it is skolem constant). So, it will be sufficient to show that we can do this given a derivation $(P_0, P_1, ..., P_n)$ of $B$ from $\forall x(Rxc \leftrightarrow Fx)$. Let $\mathcal{L}(c)$ be the result of extending $\mathcal{L}$ with the new constant $c$. The main idea of the proof is that the assumption $\forall x(Rxc \leftrightarrow Fx)$ looks just like a "

*definition*" of $Rxc$. So, the plan is to consider each formula $P_i$ in the derivation, and

**replace**any occurrence of $Rt_1t_2$ in $P_i$ by $Ft_1 \wedge t_2 = c$ (where $t_1, t_2$ are terms). This replacement therefore eliminates the symbol $R$ (i.e., the membership predicate). Let $(P_i)^{\circ}$ be the result of making this replacement.

From the definition of "derivation", each $P_i$ is either an axiom of logic, or is the assumption formula $\forall x(Rxc \leftrightarrow Fx)$, or is obtained by Modus Ponens on previous formulas. The hope is that, after the replacements, the new sequence of formulas $((P_0)^{\circ}, ..., (P_n)^{\circ})$ is, "more or less", a derivation of B in logic.

Since $B$ does not contain the symbol $R$, the replacement makes no difference to $B$. (I.e., $B^{\circ}$ is just $B$.) Next, if $P_i$ is a logical axiom containing $R$, then replacing $Rt_1t_2$ by $Ft_1 \wedge t_2 = c$ will give a

*logical*axiom in $\mathcal{L}(c)$. Next, if $P_i$ is the assumption formula $\forall x(Rxc \leftrightarrow Fx)$, then replacing $Rxc$ by $Fx \wedge c = c$ yields $\forall x((Fx \wedge c = c) \leftrightarrow Fx)$. But this is itself a logically derivable $\mathcal{L}(c)$-sentence. Finally, if $P_i$ is obtained by Modus Ponens from $P_j$ and $P_k$ (with $j, k < i$), we need to check that after we've made the replacements, then result is still an instance of Modus Ponens. The only thing to check is that applying the replacement to a conditional $P_j \rightarrow P_i$ yields the conditional of applying the replacements; and this is so. (I.e., that $(P_j \rightarrow P_i)^{\circ}$ is $(P_j)^{\circ} \rightarrow (P_i)^{\circ}$.) It follows then that, assuming we "paste in" the missing derivation of $\forall x((Fx \wedge c = c) \leftrightarrow Fx)$, the replacement yields a derivation of $B$ in

*logic*alone, in the language $\mathcal{L}(c)$. However, since $B$ does not contain $c$, if there is a logical derivation of $B$ in $\mathcal{L}(c)$, then there is a logical derivation of $B$ in $\mathcal{L}$ itself. So, $\vdash B$, as required.

This is probably the simplest case of a Field-style conservativeness result for mathematical axioms over "nominalistic" sentences. One can then build-up to more complicated cases by modifying this kind of proof. E.g., to consider

*comp*-axioms where the defining formula is $\varphi(x)$, where $\varphi(x)$ is any $\mathcal{L}$-formula (rather just the atomic formula $Fx$). Also, to consider a

*nom*-language $\mathcal{L}$ with various primitive predicates for concreta. In these latter cases, the

*comp*-axioms have the form $\forall x(Rxc \leftrightarrow \varphi(x))$, where $\varphi(x)$ is an $\mathcal{L}$-formula and $c$ is a constant (a new constant is needed for each formula $\varphi(x)$). The method is, again, to replace occurrences of the atomic formula $Rt_1t_2$ as they appear in a derivation by certain $\mathcal{L}(c_1, ..., c_n)$-formulas. This will transform a given derivation using

*comp*-axioms into a derivation in logic alone (in the language $\mathcal{L}(c_1, ..., c_n)$).

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