# Axiom schema of specification (formula arguments)

Some sources define the formula like this:

$$\forall w_1,\ldots,w_n \, \forall A \, \exists B \, \forall x \, ( x \in B \Leftrightarrow [ x \in A \wedge \varphi(x, w_1, \ldots, w_n , A) ] )$$

Why are all the w* arguments necessary? My logic tells me that this is equivalent:

$$\forall A \, \exists B \, \forall x \, ( x \in B \Leftrightarrow [ x \in A \wedge \varphi(x) ] )$$

EDIT: I'll clarify what I'm asking.

If we have a formula of 2 variables, then for a given value of one of them we can always express it as a formula of 1 variable.

So in the first schema we have one instance for the formula $$\varphi(x, z) = \exists{u}(x \in u \wedge u \in z)$$ and this allows us to construct certain sets. The same sets can be constructed with the second schema if we have one instance for each of the formulas $$\varphi_z(x) = \exists{u}(x \in u \wedge u \in z)$$ there are as many formulas as there are values of z.

So each set defined using the first schema can be defined using only one-variable formulas. Then we don't need the more complex definition. Am I wrong?

• In general, you may need more than one parameter; we can use an example from K.Kunen, The Foundations of Mathematics (2009) : "In Comprehension [$\forall z (\exists y \forall x (x \in y \leftrightarrow x \in z \land \phi(x)))$, i.e.our "Selection", or Separation Axiom], we can even have $z$ free - for example, it's legitimate to form $z^* = {x \in z : \exists u ( x \in u \land u \in z)}$; so once we have officially defined $2$ as $\{O, 1 \}$, we'll have $2^* = \{O, 1 \}^* = \{O \}$. The proviso that $\phi$ cannot have $y$ free avoids self-referential definitions such as the Liar Paradox." – Mauro ALLEGRANZA Jan 10 '14 at 7:49

The necessity of the universal quantifiers is an artifact of the fact that in the classical formulations of logic the emphasis falls on propositions that could have truth values assigned to them, and therefore these propositions are necessarily represented by formulas with no free variables. In this set-up, axioms, theorems, etc. are all necessarily propositions, hence represented by formulas with no free variables. Hence, to formulate the axioms in the axiom schema as propositions, you need the universal quantifiers.

In less classical approaches (ones concerned more with the proof calculus like type theory, or category theory), the emphasis falls on hypothetical assertions that something is true. When formulating classical logic, such an assertion may be denoted by $\phi\vdash_{\vec x}\psi$, for example, where $\phi$ and $\psi$ are arbitrary formulas, and $\vec x$ is a list containing all the free variables occurring in $\phi$ and $\psi$. This assertion is to be read as: in the context that the symbols in list $\vec x$ are elements of the universe so that $\phi$ is satisfied, $\psi$ is also satisfied.

This formulation is unnecessary (though I find it convenient) for classical logic, since the assertion $\phi\vdash_{\vec x}\psi$ is, by the inference rule of implication equivalent to $\vdash_{\vec x}\phi\rightarrow\psi$ (in the context $\vec x$, the formula $\phi$ implies $\psi$ is satisfied), which is equivalent from the definition of the universal quantifierto $\vdash\forall\vec x(\phi\rightarrow\psi)$ (where $\forall\vec x$ is an abbreviation for $\forall x_1\forall x_2\forall\dots\forall x_n$ if $\vec x=[x_1,\dots,x_n]$). This assertion says that in the empty context, and with no hypotheses, the (free variable free) formula $\forall\vec x(\phi\rightarrow\psi)$ is satisfied.

(Note, however, that for non-classical logics, or fragments of classical logic for which we are granted only restricted access to quantifiers and connectives, the above equivalence may simply not be available, hence this alternative formulation).

So in particular, in this alterative formulation of classical logic, the assertion $$\vdash_{\vec w}\forall A \, \exists B \, \forall x \, ( x \in B \leftrightarrow [ x \in A \wedge \varphi] )$$ can be taken as an axiom, and is by definition of $\forall$ equivalent to the assertion $$\vdash\forall\vec w(\forall A \, \exists B \, \forall x \, ( x \in B \leftrightarrow [ x \in A \wedge \varphi]$$

Notice that we are still keeping track of the free variables. As far as I understand, you can't not keep track of the free variables, and part of the non-classical approach is that the tracking is built in (since if you want to deal with multi-sorted logic or type theory, you have to keep track not just of which symbols are variables, but also of what sort or types those variables are).

To answer the clarified question, it is simply not true in general that any two-variable formula can be replaced by an infinite family of single-variable formulas. The reason is that elements of the domain are simply not part of the alphabet over which the formulas are defined as strings of symbols. In particular for set theory, formulas are strings of symbols made out of variables ($x,y,z,\dots$), the logical connectives ($\wedge,\vee,\neg,\rightarrow$), the two quantifiers ($\exists,\forall$), parentheses ($(,)$), the equality sign ($=$), and the non-logical relation symbol $\in$.

Notice that not even the empty set, which we informally denote as $\{\}$, occurs as an expression in the above symbols. Rather, if we have a formula $\phi(u)$, and we want to instantiate it with $\{\}$, what we obtain is $\phi_{\{\}}\equiv\left(\forall x\neg(x\in u)\right)\wedge\phi(u)$. Consequently, the only formulas $\phi_z$ for $z$ an element of the domain are those which are definable, i.e. ones which are defined by the derivability (from the axioms) of the truth of a formula $\exists z\left(x\in z\leftrightarrow\phi(x)\right)$ is true where $\phi$ uses only the symbols defined above.

However, since there are countably many formulas, and uncountably many sets (this is a metatheoretical result), replacing $\phi(x,z)$ with $\phi_z(x)$, which you can only do for definable $z$, does not result in the same axiom schema: it results in the weaker schema that concerns only definable sets.

• Thank you for this detailed answer but my question was different. I updated it to clarify what I'm asking. – martinkunev Jan 12 '14 at 11:03
• I have edited in response. – Vladimir Sotirov Jan 12 '14 at 17:49
• I think it's worth noting that you can add constants for all the members of some model, but this doesn't really fix anything; you've just expanded the range of sentences that can define elements of your model, and there's still nothing you can do to rule out undefinable elements (by Lowenheim-Skolem). Not to mention you've made the language uncountable and broken a lot of the nice model theory. – Malice Vidrine Jan 12 '14 at 18:22
• The question here is not an issue of how we formalize first order logic. – Andrés E. Caicedo Jan 13 '14 at 1:12

You do not need parameters.

Ralf Schindler wrote a short note where you can see the details (and Ralf and I discussed it last year around May). Eventually we found an early reference, that proves an even stronger theorem. See

Azriel Levy, Parameters in the comprehension axiom schemas of set theory, in Proceedings of the Tarski symposium, Proceedings of Symposia in Pure Mathematics, vol. 25, American Mathematical Society, Providence, RI, pp. 309–324.

The theorem in Ralf's note is stronger than you are asking: We can formulate $\mathsf{ZFC}$ without requiring parameters in either comprehension (specification) or replacement.

As you can see, the argument is less than a page long. Let me give a quick summary: Note first that we can prove the existence of $0$ and $1$, and therefore of ordered pairs of the form $(a,0)$ or $(a,1)$ for any $a$. From (an instance of parameter-free) replacement, we get that $a\times\{0\}$ and $a\times\{1\}$ exist for any $a$. Also, for any $a$ and $b$, $(a\times\{0\})\cup\{(b,1)\}$ exists. From this we can prove that $$\{((u,0),(b,1))\mid u\in a\}$$ exists for any $a$ and $b$: First, $d=\mathcal P(\mathcal P((a\times\{0\})\cup\{(b,1)\}))$ exists, and the set we want is $$\{x\in d\mid \exists u\,\exists v\,(x=((u,0),(b,1)))\},$$ which exists by applying an instance of parameter-free specification.

We can now prove specification with parameters. For this, note that using pairing, we can code finitely many parameters into a single one, so it is enough to show the result for formulas with one parameter, say $\phi(x,v)$. That is, given $a,b$, we must show that $$\{x\in a\mid \phi(x,b)\}$$ exists.

We use the parameter-free instance of replacement given by the class function $F$ defined by $F(x)=0$ unless $x$ has the form $((u,0),(c,1))$ for some $u,c$, in which case we set $F(x)=u$. We see that $$F''\{((u,0),(b,1))\mid u\in a\}= \{x\in a\mid \phi(x,b)\}\cup\{0\},$$ and from this your question follows (by applying a parameter-free instance of specification to remove $0$ from the set, if needed).

To conclude, we prove replacement for formulas with one parameter (which again, by pairing, is enough). Accordingly, let $\phi(x,y,v)$ be a formula, let $b$ be a set, and suppose that for every $x$ there is a unique $y$ such that $\phi(x,y,b)$. We must show that, for any $a$, $$\{y\mid\exists x\in a\,(\phi(x,y,b))\}$$ exists.

We use the parameter-free instance of replacement given by the class function $F$ defined by $F(z)=0$ unless there are $x,c$ with $z=((x,0),(c,1))$, and there is a unique $y$ such that $\phi(x,y,c)$ holds, in which case we set $F(z)=y$. We then see that $$F''\{((x,0),(b,1))\mid x\in a\}=\{y\mid\exists x\in a\,(\phi(x,y,b))\}\cup\{0\}$$ and, as before, we are done by a last appeal to a (parameter-free) instance of specification, if removing $0$ from the set is needed.

• Why would $F''\{((u,0),(b,1))\mid u\in a\}= \{x\in a\mid \phi(x,b)\}\cup\{0\}$? The LHS is independent of $\phi$. – Achilles Sep 7 '16 at 1:01
• @Achilles Obviously it is not independent of $\phi$, that's the whole point. – Andrés E. Caicedo Sep 7 '16 at 1:30

The $w_i$ are necessary. The selection criteria $\varphi$ can refer to previously introduced free variables (the $w_i$) in addition to $A$.

Example:

Let $A$ be a set.

Suppose $P(w_1,w_2,w_3)$ (introducing free variables $w_1,w_2, w_3$ using predicate $P$)

Then there exists $B\subset A$ such that $\forall x [x\in B \iff x\in A \land\varphi(x,w_1,w_2,w_3,A)]$.

Note that the subset selected depends on the free variables $w_1,w_2,w_3$.