Von Neumann–Bernays–Gödel set theory

Classes have several uses in NBG:

• They produce a finite axiomatization of set theory.[4]
• They are used to state a "very strong form of the axiom of choice"[5]—namely, the axiom of global choice: There exists a global choice function ${\displaystyle G}$ defined on the class of all nonempty sets such that ${\displaystyle G(x)\in x}$ for every nonempty set ${\displaystyle x.}$ This is stronger than ZFC's axiom of choice: For every set ${\displaystyle s}$ of nonempty sets, there exists a choice function ${\displaystyle f}$ defined on ${\displaystyle s}$ such that ${\displaystyle f(x)\in x}$ for all ${\displaystyle x\in s.}$[lower-alpha 1]
• The set-theoretic paradoxes are handled by recognizing that some classes cannot be sets. For example, assume that the class ${\displaystyle Ord}$ of all ordinals is a set. Then ${\displaystyle Ord}$ is a transitive set well-ordered by ${\displaystyle \in }$. So, by definition, ${\displaystyle Ord}$ is an ordinal. Hence, ${\displaystyle Ord\in Ord}$, which contradicts ${\displaystyle \in }$ being a well-ordering of ${\displaystyle Ord.}$ Therefore, ${\displaystyle Ord}$ is not a set. A class that is not a set is called a proper class, ${\displaystyle Ord}$ is a proper class.[6]
• Proper classes are useful in constructions. In his proof of the relative consistency of the axiom of global choice and the generalized continuum hypothesis, Gödel used proper classes to build the constructible universe. He constructed a function on the class of all ordinals that, for each ordinal, builds a constructible set by applying a set-building operation to previously constructed sets. The constructible universe is the image of this function.[7]

Once classes are added to the language of ZFC, it is easy to transform ZFC into a set theory with classes. First, the axiom schema of class comprehension is added. This axiom schema states: For every formula ${\displaystyle \phi (x_{1},\ldots ,x_{n})}$ that quantifies only over sets, there exists a class ${\displaystyle A}$ consisting of the ${\displaystyle n}$-tuples satisfying the formula—that is, ${\displaystyle \forall x_{1}\cdots \,\forall x_{n}[(x_{1},\ldots ,x_{n})\in A\iff \phi (x_{1},\ldots ,x_{n})].}$ Then the axiom schema of replacement is replaced by a single axiom that uses a class. Finally, ZFC's axiom of extensionality is modified to handle classes: If two classes have the same elements, then they are identical. The other axioms of ZFC are not modified.[8]

This theory is not finitely axiomatized. ZFC's replacement schema has been replaced by a single axiom, but the axiom schema of class comprehension has been introduced.

To produce a theory with finitely many axioms, the axiom schema of class comprehension is first replaced with finitely many class existence axioms. Then these axioms are used to prove the class existence theorem, which implies every instance of the axiom schema.[8] The proof of this theorem requires only seven class existence axioms, which are used to convert the construction of a formula into the construction of a class satisfying the formula.

NBG has two types of objects: classes and sets. Intuitively, every set is also a class. There are two ways to axiomatize this. Bernays used many-sorted logic with two sorts: classes and sets.[2] Gödel avoided sorts by introducing primitive predicates: ${\displaystyle {\mathfrak {Cls}}(A)}$ for "${\displaystyle A}$ is a class" and ${\displaystyle {\mathfrak {M}}(A)}$ for "${\displaystyle A}$ is a set" (in German, "set" is Menge). He also introduced axioms stating that every set is a class and that if class ${\displaystyle A}$ is a member of a class, then ${\displaystyle A}$ is a set.[9] Using predicates is the standard way to eliminate sorts. Elliott Mendelson modified Gödel's approach by having everything be a class and defining the set predicate ${\displaystyle M(A)}$ as ${\displaystyle \exists C(A\in C).}$[10] This modification eliminates Gödel's class predicate and his two axioms.

Bernays' two-sorted approach may appear more natural at first, but it creates a more complex theory.[lower-alpha 2] In Bernays' theory, every set has two representations: one as a set and the other as a class. Also, there are two membership relations: the first, denoted by "∈", is between two sets; the second, denoted by "η", is between a set and a class.[2] This redundancy is required by many-sorted logic because variables of different sorts range over disjoint subdomains of the domain of discourse.

The differences between these two approaches do not affect what can be proved, but they do affect how statements are written. In Gödel's approach, ${\displaystyle A\in C}$ where ${\displaystyle A}$ and ${\displaystyle C}$ are classes is a valid statement. In Bernays' approach this statement has no meaning. However, if ${\displaystyle A}$ is a set, there is an equivalent statement: Define "set ${\displaystyle a}$ represents class ${\displaystyle A}$" if they have the same sets as members—that is, ${\displaystyle \forall x(x\in a\iff x\;\eta \;A).}$ The statement ${\displaystyle a\;\eta \;C}$ where set ${\displaystyle a}$ represents class ${\displaystyle A}$ is equivalent to Gödel's ${\displaystyle A\in C.}$[2]

The approach adopted in this article is that of Gödel with Mendelson's modification. This means that NBG is an axiomatic system in first-order predicate logic with equality, and its only primitive notions are class and the membership relation.

A set is a class that belongs to at least one class: ${\displaystyle A}$ is a set if and only if ${\displaystyle \exists C(A\in C)}$. A class that is not a set is called a proper class: ${\displaystyle A}$ is a proper class if and only if ${\displaystyle \forall C(A\notin C)}$.[12] Therefore, every class is either a set or a proper class, and no class is both.

Gödel introduced the convention that uppercase variables range over classes, while lowercase variables range over sets.[9] Gödel also used names that begin with an uppercase letter to denote particular classes, including functions and relations defined on the class of all sets. Gödel's convention is used in this article. It allows us to write:

The following axioms and definitions are needed for the proof of the class existence theorem.

Axiom of extensionality.  If two classes have the same elements, then they are identical.

${\displaystyle \forall A\,\forall B\,[\forall x(x\in A\iff x\in B)\implies A=B]}$ [13]

This axiom generalizes ZFC's axiom of extensionality to classes.

Axiom of pairing.  If ${\displaystyle x}$ and ${\displaystyle y}$ are sets, then there exists a set ${\displaystyle p}$ whose only members are ${\displaystyle x}$ and ${\displaystyle y}$.

${\displaystyle \forall x\,\forall y\,\exists p\,\forall z\,[z\in p\iff (z=x\,\lor \,z=y)]}$ [14]

As in ZFC, the axiom of extensionality implies the uniqueness of the set ${\displaystyle p}$, which allows us to introduce the notation ${\displaystyle \{x,y\}.}$

Ordered pairs are defined by:

${\displaystyle (x,y)=\{\{x\},\{x,y\}\}}$

Tuples are defined inductively using ordered pairs:

${\displaystyle (x_{1})=x_{1},}$

${\displaystyle {\text{For }}n>1\!:(x_{1},\ldots ,x_{n-1},x_{n})=((x_{1},\ldots ,x_{n-1}),x_{n}).}$[lower-alpha 3]

Class existence axioms will be used to prove the class existence theorem: For every formula in ${\displaystyle n}$ free set variables that quantifies only over sets, there exists a class of ${\displaystyle n}$-tuples that satisfy it. The following example starts with two classes that are functions and builds a composite function. This example illustrates the techniques that are needed to prove the class existence theorem, which lead to the class existence axioms that are needed.

Example 1:  If the classes ${\displaystyle F}$ and ${\displaystyle G}$ are functions, then the composite function ${\displaystyle G\circ F}$ is defined by the formula: ${\displaystyle \exists t[(x,t)\in F\,\land \,(t,y)\in G].}$ Since this formula has two free set variables, ${\displaystyle x}$ and ${\displaystyle y,}$ the class existence theorem constructs the class of ordered pairs: ${\displaystyle G\circ F\,=\,\{(x,y):\exists t[(x,t)\in F\,\land \,(t,y)\in G]\}.}$ Because this formula is built from simpler formulas using conjunction ${\displaystyle \land }$ and existential quantification ${\displaystyle \exists }$, class operations are needed that take classes representing the simpler formulas and produce classes representing the formulas with ${\displaystyle \land }$ and ${\displaystyle \exists }$. To produce a class representing a formula with ${\displaystyle \land }$, intersection used since ${\displaystyle x\in A\cap B\iff x\in A\land x\in B.}$ To produce a class representing a formula with ${\displaystyle \exists }$, the domain is used since ${\displaystyle x\in Dom(A)\iff \exists t[(x,t)\in A].}$ Before taking the intersection, the tuples in ${\displaystyle F}$ and ${\displaystyle G}$ need an extra component so they have the same variables. The component ${\displaystyle y}$ is added to the tuples of ${\displaystyle F}$ and ${\displaystyle x}$ is added to the tuples of ${\displaystyle G}$: $${\displaystyle F'=\{(x,t,y):(x,t)\in F\}\,}$$ and ${\displaystyle \,G'=\{(t,y,x):(t,y)\in G\}}$ In the definition of ${\displaystyle F',}$ the variable ${\displaystyle y}$ is not restricted by the statement ${\displaystyle (x,t)\in F,}$ so ${\displaystyle y}$ ranges over the class ${\displaystyle V}$ of all sets. Similarly, in the definition of ${\displaystyle G',}$ the variable ${\displaystyle x}$ ranges over ${\displaystyle V.}$ So an axiom is needed that adds an extra component (whose values range over ${\displaystyle V}$) to the tuples of a given class. Next, the variables are put in the same order to prepare for the intersection: $${\displaystyle F''=\{(x,y,t):(x,t)\in F\}\,}$$ and ${\displaystyle \,G''=\{(x,y,t):(t,y)\in G\}}$ To go from ${\displaystyle F'}$ to ${\displaystyle F''}$ and from ${\displaystyle G'}$ to ${\displaystyle G''}$ requires two different permutations, so axioms that support permutations of tuple components are needed. The intersection of ${\displaystyle F''}$ and ${\displaystyle G''}$ handles ${\displaystyle \land }$: $${\displaystyle F''\cap G''=\{(x,y,t):(x,t)\in F\,\land \,(t,y)\in G\}}$$ Since ${\displaystyle (x,y,t)}$ is defined as ${\displaystyle ((x,y),t)}$, taking the domain of ${\displaystyle F''\cap G''}$ handles ${\displaystyle \exists t}$ and produces the composite function: $${\displaystyle G\circ F=Dom(F''\cap G'')=\{(x,y):\exists t((x,t)\in F\,\land \,(t,y)\in G)\}}$$ So axioms of intersection and domain are needed.

The class existence axioms are divided into two groups: axioms handling language primitives and axioms handling tuples. There are four axioms in the first group and three axioms in the second group.[lower-alpha 4]

Axioms for handling language primitives:

Membership.  There exists a class ${\displaystyle E}$ containing all the ordered pairs whose first component is a member of the second component.

${\displaystyle \exists E\,\forall x\,\forall y\,[(x,y)\in E\iff x\in y]\!}$ [18]

Intersection (conjunction).  For any two classes ${\displaystyle A}$ and ${\displaystyle B}$, there is a class ${\displaystyle C}$ consisting precisely of the sets that belong to both ${\displaystyle A}$ and ${\displaystyle B}$.

${\displaystyle \forall A\,\forall B\,\exists C\,\forall x\,[x\in C\iff (x\in A\,\land \,x\in B)]}$ [19]

Complement (negation).  For any class ${\displaystyle A}$, there is a class ${\displaystyle B}$ consisting precisely of the sets not belonging to ${\displaystyle A}$.

${\displaystyle \forall A\,\exists B\,\forall x\,[x\in B\iff \neg (x\in A)]}$ [20]

Domain (existential quantifier).  For any class ${\displaystyle A}$, there is a class ${\displaystyle B}$ consisting precisely of the first components of the ordered pairs of ${\displaystyle A}$.

${\displaystyle \forall A\,\exists B\,\forall x\,[x\in B\iff \exists y((x,y)\in A)]}$ [21]

By the axiom of extensionality, class ${\displaystyle C}$ in the intersection axiom and class ${\displaystyle B}$ in the complement and domain axioms are unique. They will be denoted by: ${\displaystyle A\cap B,}$ ${\displaystyle \complement A,}$ and ${\displaystyle Dom(A),}$ respectively.[lower-alpha 5] On the other hand, extensionality is not applicable to ${\displaystyle E}$ in the membership axiom since it specifies only those sets in ${\displaystyle E}$ that are ordered pairs.

The first three axioms imply the existence of the empty class and the class of all sets: The membership axiom implies the existence of a class ${\displaystyle E.}$ The intersection and complement axioms imply the existence of ${\displaystyle E\cap \complement E}$, which is empty. By the axiom of extensionality, this class is unique; it is denoted by ${\displaystyle \emptyset .}$ The complement of ${\displaystyle \emptyset }$ is the class ${\displaystyle V}$ of all sets, which is also unique by extensionality. The set predicate ${\displaystyle M(A)}$, which was defined as ${\displaystyle \exists C(A\in C)}$, is now redefined as ${\displaystyle A\in V}$ to avoid quantifying over classes.

Axioms for handling tuples:

Product by ${\displaystyle V}$.  For any class ${\displaystyle A}$, there is a class ${\displaystyle B}$ consisting of the ordered pairs whose first component belongs to ${\displaystyle A}$.

${\displaystyle \forall A\,\exists B\,\forall u\,[u\in B\iff \exists x\,\exists y\,(u=(x,y)\land x\in A)]}$[23]

Circular permutation.  For any class ${\displaystyle A}$, there is a class ${\displaystyle B}$ whose 3‑tuples are obtained by applying the circular permutation ${\displaystyle (y,z,x)\mapsto (x,y,z)}$ to the 3‑tuples of ${\displaystyle A}$.

${\displaystyle \forall A\,\exists B\,\forall x\,\forall y\,\forall z\,[(x,y,z)\in B\iff (y,z,x)\in A]}$[24]

Transposition.  For any class ${\displaystyle A}$, there is a class ${\displaystyle B}$ whose 3‑tuples are obtained by transposing the last two components of the 3‑tuples of ${\displaystyle A}$.

${\displaystyle \forall A\,\exists B\,\forall x\,\forall y\,\forall z\,[(x,y,z)\in B\iff (x,z,y)\in A]}$[25]

By extensionality, the product by ${\displaystyle V}$ axiom implies the existence of a unique class, which is denoted by ${\displaystyle A\times V.}$ This axiom is used to define the class ${\displaystyle V^{n}}$ of all ${\displaystyle n}$-tuples: ${\displaystyle V^{1}=V}$ and ${\displaystyle V^{n+1}=V^{n}\times V.\,}$ If ${\displaystyle A}$ is a class, extensionality implies that ${\displaystyle A\cap V^{n}}$ is the unique class consisting of the ${\displaystyle n}$-tuples of ${\displaystyle A.}$ For example, the membership axiom produces a class ${\displaystyle E}$ that may contain elements that are not ordered pairs, while the intersection ${\displaystyle E\cap V^{2}}$ contains only the ordered pairs of ${\displaystyle E}$.

The circular permutation and transposition axioms do not imply the existence of unique classes because they specify only the 3‑tuples of class ${\displaystyle B.}$ By specifying the 3‑tuples, these axioms also specify the ${\displaystyle n}$-tuples for ${\displaystyle n\geq 4}$ since:

$${\displaystyle (x_{1},\ldots ,x_{n-2},x_{n-1},x_{n})=((x_{1},\ldots ,x_{n-2}),x_{n-1},x_{n}).}$$

The axioms for handling tuples and the domain axiom imply the following lemma, which is used in the proof of the class existence theorem.

One more axiom is needed to prove the class existence theorem: the axiom of regularity. Since the existence of the empty class has been proved, the usual statement of this axiom is given.[lower-alpha 6]

Axiom of regularity.  Every nonempty set has at least one element with which it has no element in common.

$${\displaystyle \forall a\,[a\neq \emptyset \implies \exists u(u\in a\land u\cap a=\emptyset )].}$$

This axiom implies that a set cannot belong to itself: Assume that ${\displaystyle x\in x}$ and let ${\displaystyle a=\{x\}.}$ Then ${\displaystyle x\cap a\neq \emptyset }$ since ${\displaystyle x\in x\cap a.}$ This contradicts the axiom of regularity because ${\displaystyle x}$ is the only element in ${\displaystyle a.}$ Therefore, ${\displaystyle x\notin x.}$ The axiom of regularity also prohibits infinite descending membership sequences of sets:

$${\displaystyle \cdots \in x_{n+1}\in x_{n}\in \cdots \in x_{1}\in x_{0}.}$$

Gödel stated regularity for classes rather than for sets in his 1940 monograph, which was based on lectures given in 1938.[26] In 1939, he proved that regularity for sets implies regularity for classes.[27]

Class existence theorem — Let ${\displaystyle \phi (x_{1},\dots ,x_{n},Y_{1},\dots ,Y_{m})}$ be a formula that quantifies only over sets and contains no free variables other than ${\displaystyle x_{1},\dots ,x_{n},Y_{1},\dots ,Y_{m}}$ (not necessarily all of these). Then for all ${\displaystyle Y_{1},\dots ,Y_{m}}$, there exists a unique class ${\displaystyle A}$ of ${\displaystyle n}$-tuples such that:

$${\displaystyle \forall x_{1}\cdots \,\forall x_{n}[(x_{1},\dots ,x_{n})\in A\iff \phi (x_{1},\dots ,x_{n},Y_{1},\dots ,Y_{m})].}$$

The class ${\displaystyle A}$ is denoted by ${\displaystyle \{(x_{1},\dots ,x_{n}):\phi (x_{1},\dots ,x_{n},Y_{1},\dots ,Y_{m})\}.}$[lower-alpha 7]

The theorem's proof will be done in two steps:

1. Transformation rules are used to transform the given formula ${\displaystyle \phi }$ into an equivalent formula that simplifies the inductive part of the proof. For example, the only logical symbols in the transformed formula are ${\displaystyle \neg }$, ${\displaystyle \land }$, and ${\displaystyle \exists }$, so the induction handles logical symbols with just three cases.
2. The class existence theorem is proved inductively for transformed formulas. Guided by the structure of the transformed formula, the class existence axioms are used to produce the unique class of ${\displaystyle n}$-tuples satisfying the formula.

Transformation rules.  In rules 1 and 2 below, ${\displaystyle \Delta }$ and ${\displaystyle \Gamma }$ denote set or class variables. These two rules eliminate all occurrences of class variables before an ${\displaystyle \in }$ and all occurrences of equality. Each time rule 1 or 2 is applied to a subformula, ${\displaystyle i}$ is chosen so that ${\displaystyle z_{i}}$ differs from the other variables in the current formula. The three rules are repeated until there are no subformulas to which they can be applied. This produces a formula that is built only with ${\displaystyle \neg }$, ${\displaystyle \land }$, ${\displaystyle \exists }$, ${\displaystyle \in }$, set variables, and class variables ${\displaystyle Y_{k}}$ where ${\displaystyle Y_{k}}$ does not appear before an ${\displaystyle \in }$.

1. ${\displaystyle \,Y_{k}\in \Gamma }$ is transformed into ${\displaystyle \exists z_{i}(z_{i}=Y_{k}\,\land \,z_{i}\in \Gamma ).}$
2. Extensionality is used to transform ${\displaystyle \Delta =\Gamma }$ into ${\displaystyle \forall z_{i}(z_{i}\in \Delta \iff z_{i}\in \Gamma ).}$
3. Logical identities are used to transform subformulas containing ${\displaystyle \lor ,\implies ,\iff ,}$ and ${\displaystyle \forall }$ to subformulas that only use ${\displaystyle \neg ,\land ,}$ and ${\displaystyle \exists .}$

Transformation rules: bound variables.  Consider the composite function formula of example 1 with its free set variables replaced by ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$: ${\displaystyle \exists t[(x_{1},t)\in F\,\land \,(t,x_{2})\in G].}$ The inductive proof will remove ${\displaystyle \exists t}$, which produces the formula ${\displaystyle (x_{1},t)\in F\land (t,x_{2})\in G.}$ However, since the class existence theorem is stated for subscripted variables, this formula does not have the form expected by the induction hypothesis. This problem is solved by replacing the variable ${\displaystyle t}$ with ${\displaystyle x_{3}.}$ Bound variables within nested quantifiers are handled by increasing the subscript by one for each successive quantifier. This leads to rule 4, which must be applied after the other rules since rules 1 and 2 produce quantified variables.

4. If a formula contains no free set variables other than ${\displaystyle x_{1},\dots ,x_{n},}$ then bound variables that are nested within ${\displaystyle q}$ quantifiers are replaced with ${\displaystyle x_{n+q}}$. These variables have (quantifier) nesting depth ${\displaystyle q}$.

Example 2:  Rule 4 is applied to the formula ${\displaystyle \phi (x_{1})}$ that defines the class consisting of all sets of the form ${\displaystyle \{\emptyset ,\{\emptyset ,\dots \},\dots \}.}$ That is, sets that contain at least ${\displaystyle \emptyset }$ and a set containing ${\displaystyle \emptyset }$ — for example, ${\displaystyle \{\emptyset ,\{\emptyset ,a,b,c\},d,e\}}$ where ${\displaystyle a,b,c,d,}$ and ${\displaystyle e}$ are sets. $${\displaystyle {\begin{aligned}\phi (x_{1})\,&=\,\exists u\;\,[\,u\in x_{1}\,\land \,\neg \exists v\;\,(\;v\,\in \,u\,)]\,\land \,\,\exists w\;{\bigl (}w\in x_{1}\,\land \,\exists y\;\,[(\;y\,\in w\;\land \;\neg \exists z\;\,(\;z\,\in \,y\,)]{\bigr )}\\\phi _{r}(x_{1})\,&=\,\exists x_{2}[x_{2}\!\in \!x_{1}\,\land \,\neg \exists x_{3}(x_{3}\!\in \!x_{2})]\,\land \,\,\exists x_{2}{\bigl (}x_{2}\!\in \!x_{1}\,\land \,\exists x_{3}[(x_{3}\!\in \!x_{2}\,\land \,\neg \exists x_{4}(x_{4}\!\in \!x_{3})]{\bigr )}\end{aligned}}}$$ Since ${\displaystyle x_{1}}$ is the only free variable, ${\displaystyle n=1.}$ The quantified variable ${\displaystyle x_{3}}$ appears twice in ${\displaystyle x_{3}\in x_{2}}$ at nesting depth 2. Its subscript is 3 because ${\displaystyle n+q=1+2=3.}$ If two quantifier scopes are at the same nesting depth, they are either identical or disjoint. The two occurrences of ${\displaystyle x_{3}}$ are in disjoint quantifier scopes, so they do not interact with each other.

Proof of the class existence theorem.  The proof starts by applying the transformation rules to the given formula to produce a transformed formula. Since this formula is equivalent to the given formula, the proof is completed by proving the class existence theorem for transformed formulas.

Proof of the class existence theorem for transformed formulas

The following lemma is used in the proof.

Expansion lemma — Let ${\displaystyle 1\leq i<j\leq n,}$ and let ${\displaystyle P}$ be a class containing all the ordered pairs ${\displaystyle (x_{i},x_{j})}$ satisfying ${\displaystyle R(x_{i},x_{j}).}$ That is, ${\displaystyle P\supseteq \{(x_{i},x_{j}):R(x_{i},x_{j})\}.}$ Then ${\displaystyle P}$ can be expanded into the unique class ${\displaystyle Q}$ of ${\displaystyle n}$-tuples satisfying ${\displaystyle R(x_{i},x_{j})}$. That is, ${\displaystyle Q=\{(x_{1},\ldots ,x_{n}):R(x_{i},x_{j})\}.}$

Proof:

1. If ${\displaystyle i=1,}$ let ${\displaystyle P_{1}=P.}$
   Otherwise, ${\displaystyle i>1,}$ so components are added in front of ${\displaystyle x_{i}{\text{:}}}$ apply the tuple lemma's statement 1 to ${\displaystyle P}$ with ${\displaystyle z=(x_{1},\dots ,x_{i-1}).}$ This produces a class ${\displaystyle P_{1}}$ containing all the ${\displaystyle (i+1)}$-tuples
   $${\displaystyle ((x_{1},\dots ,x_{i-1}),x_{i},x_{j})=(x_{1},\dots ,x_{i-1},x_{i},x_{j})}$$

   
   satisfying ${\displaystyle R(x_{i},x_{j}).}$
2. If ${\displaystyle j=i+1,}$ let ${\displaystyle P_{2}=P_{1}.}$
   Otherwise, ${\displaystyle j>i+1,}$ so components are added between ${\displaystyle x_{i}}$ and ${\displaystyle x_{j}{\text{:}}}$ add the components ${\displaystyle x_{i+1},\dots ,x_{j-1}}$ one by one using the tuple lemma's statement 2. This produces a class ${\displaystyle P_{2}}$ containing all the ${\displaystyle j}$-tuples
   $${\displaystyle (((\cdots ((x_{1},\dots ,x_{i}),x_{i+1}),\cdots ),x_{j-1}),x_{j})=(x_{1},\dots ,x_{j})}$$

   
   satisfying ${\displaystyle R(x_{i},x_{j}).}$
3. If ${\displaystyle j=n,}$ let ${\displaystyle P_{3}=P_{2}.}$
   Otherwise, ${\displaystyle j<n,}$ so components are added after ${\displaystyle x_{j}{\text{:}}}$ add the components ${\displaystyle x_{j+1},\dots ,x_{n}}$ one by one using the tuple lemma's statement 3. This produces a class ${\displaystyle P_{3}}$ containing all the ${\displaystyle n}$-tuples
   $${\displaystyle ((\cdots ((x_{1},\dots ,x_{j}),x_{j+1}),\cdots ),x_{n})=(x_{1},\dots ,x_{n})}$$

   
   satisfying ${\displaystyle R(x_{i},x_{j}).}$
4. Let ${\displaystyle Q=P_{3}\cap V^{n}.}$ Extensionality implies that ${\displaystyle Q}$ is the unique class of ${\displaystyle n}$-tuples satisfying ${\displaystyle R(x_{i},x_{j}).}$

Class existence theorem for transformed formulas — Let ${\displaystyle \phi (x_{1},\ldots ,x_{n},Y_{1},\ldots ,Y_{m})}$ be a formula that:

1. contains no free variables other than ${\displaystyle x_{1},\ldots ,x_{n},Y_{1},\ldots ,Y_{m}}$;
   
2. contains only ${\displaystyle \in }$, ${\displaystyle \neg }$, ${\displaystyle \land }$, ${\displaystyle \exists }$, set variables, and the class variables ${\displaystyle Y_{k}}$ where ${\displaystyle Y_{k}}$ does not appear before an ${\displaystyle \in }$ ;
   
3. only quantifies set variables ${\displaystyle x_{n+q}}$ where ${\displaystyle q}$ is the quantifier nesting depth of the variable.

Then for all ${\displaystyle Y_{1},\dots ,Y_{m}}$, there exists a unique class ${\displaystyle A}$ of ${\displaystyle n}$-tuples such that:

$${\displaystyle \forall x_{1}\cdots \,\forall x_{n}[(x_{1},\ldots ,x_{n})\in A\iff \phi (x_{1},\ldots ,x_{n},Y_{1},\ldots ,Y_{m})].}$$

Proof: Basis step: ${\displaystyle \phi }$ has 0 logical symbols. The theorem's hypothesis implies that ${\displaystyle \phi }$ is an atomic formula of the form ${\displaystyle x_{i}\in x_{j}}$ or ${\displaystyle x_{i}\in Y_{k}.}$

Case 1: If ${\displaystyle \phi }$ is ${\displaystyle x_{i}\in x_{j}}$, we build the class ${\displaystyle E_{i,j,n}=\{(x_{1},\ldots ,x_{n}):x_{i}\in x_{j}\},}$ the unique class of ${\displaystyle n}$-tuples satisfying ${\displaystyle x_{i}\in x_{j}.}$

Case a: ${\displaystyle \phi }$ is ${\displaystyle x_{i}\in x_{j}}$ where ${\displaystyle i<j.}$ The axiom of membership produces a class ${\displaystyle P}$ containing all the ordered pairs ${\displaystyle (x_{i},x_{j})}$ satisfying ${\displaystyle x_{i}\in x_{j}.}$ Apply the expansion lemma to ${\displaystyle P}$ to obtain ${\displaystyle E_{i,j,n}=\{(x_{1},\ldots ,x_{n}):x_{i}\in x_{j}\}.}$

Case b: ${\displaystyle \phi }$ is ${\displaystyle x_{i}\in x_{j}}$ where ${\displaystyle i>j.}$ The axiom of membership produces a class ${\displaystyle P}$ containing all the ordered pairs ${\displaystyle (x_{i},x_{j})}$ satisfying ${\displaystyle x_{i}\in x_{j}.}$ Apply the tuple lemma's statement 4 to ${\displaystyle P}$ to obtain ${\displaystyle P'}$ containing all the ordered pairs ${\displaystyle (x_{j},x_{i})}$ satisfying ${\displaystyle x_{i}\in x_{j}.}$ Apply the expansion lemma to ${\displaystyle P'}$ to obtain ${\displaystyle E_{i,j,n}=\{(x_{1},\ldots ,x_{n}):x_{i}\in x_{j}\}.}$

Case c: ${\displaystyle \phi }$ is ${\displaystyle x_{i}\in x_{j}}$ where ${\displaystyle i=j.}$ Since this formula is false by the axiom of regularity, no ${\displaystyle n}$-tuples satisfy it, so ${\displaystyle E_{i,j,n}=\emptyset .}$

Case 2: If ${\displaystyle \phi }$ is ${\displaystyle x_{i}\in Y_{k}}$, we build the class ${\displaystyle E_{i,Y_{k},n}=\{(x_{1},\ldots ,x_{n}):x_{i}\in Y_{k}\},}$ the unique class of ${\displaystyle n}$-tuples satisfying ${\displaystyle x_{i}\in Y_{k}.}$

Case a: ${\displaystyle \phi }$ is ${\displaystyle x_{i}\in Y_{k}}$ where ${\displaystyle i<n.}$ Apply the axiom of product by ${\displaystyle V}$ to ${\displaystyle Y_{k}}$ to produce the class ${\displaystyle P=Y_{k}\times V=\{(x_{i},x_{i+1}):x_{i}\in Y_{k}\}.}$ Apply the expansion lemma to ${\displaystyle P}$ to obtain ${\displaystyle E_{i,Y_{k},n}=\{(x_{1},\ldots ,x_{n}):x_{i}\in Y_{k}\}.}$

Case b: ${\displaystyle \phi }$ is ${\displaystyle x_{i}\in Y_{k}}$ where ${\displaystyle i=n>1.}$ Apply the axiom of product by ${\displaystyle V}$ to ${\displaystyle Y_{k}}$ to produce the class ${\displaystyle P=Y_{k}\times V=\{(x_{i},x_{i-1}):x_{i}\in Y_{k}\}.}$ Apply the tuple lemma's statement 4 to ${\displaystyle P}$ to obtain ${\displaystyle P'=V\times Y_{k}=\{(x_{i-1},x_{i}):x_{i}\in Y_{k}\}.}$ Apply the expansion lemma to ${\displaystyle P'}$ to obtain ${\displaystyle E_{i,Y_{k},n}=\{(x_{1},\ldots ,x_{n}):x_{i}\in Y_{k}\}.}$

Case c: ${\displaystyle \phi }$ is ${\displaystyle x_{i}\in Y_{k}}$ where ${\displaystyle i=n=1.}$ Then ${\displaystyle E_{i,Y_{k},n}=Y_{k}.}$

Inductive step: ${\displaystyle \phi }$ has ${\displaystyle k}$ logical symbols where ${\displaystyle k>0}$. Assume the induction hypothesis that the theorem is true for all ${\displaystyle \psi }$ with less than ${\displaystyle k}$ logical symbols. We now prove the theorem for ${\displaystyle \phi }$ with ${\displaystyle k}$ logical symbols. In this proof, the list of class variables ${\displaystyle Y_{1},\dots ,Y_{m}}$ is abbreviated by ${\displaystyle {\vec {Y}}}$, so a formula—such as ${\displaystyle \phi (x_{1},\dots ,x_{n},Y_{1},\dots ,Y_{m})}$—can be written as ${\displaystyle \phi (x_{1},\dots ,x_{n},{\vec {Y}}).}$

Case 1: ${\displaystyle \phi (x_{1},\ldots ,x_{n},{\vec {Y}})=\neg \psi (x_{1},\ldots ,x_{n},{\vec {Y}}).}$ Since ${\displaystyle \psi }$ has ${\displaystyle k-1}$ logical symbols, the induction hypothesis implies that there is a unique class ${\displaystyle A}$ of ${\displaystyle n}$-tuples such that:

$${\displaystyle \quad (x_{1},\ldots ,x_{n})\in A\iff \psi (x_{1},\ldots ,x_{n},{\vec {Y}}).}$$

By the complement axiom, there is a class ${\displaystyle \complement A}$ such that ${\displaystyle \forall u\,[u\in \complement A\iff \neg (u\in A)].}$ However, ${\displaystyle \complement A}$ contains elements other than ${\displaystyle n}$-tuples if ${\displaystyle n>1.}$ To eliminate these elements, use ${\displaystyle \complement _{V^{n}}A=\,}$${\displaystyle \complement A\cap V^{n}=\,}$${\displaystyle V^{n}\setminus A,}$ which is the complement relative to the class ${\displaystyle V^{n}}$ of all ${\displaystyle n}$-tuples.[lower-alpha 5] Then, by extensionality, ${\displaystyle \complement _{V^{n}}A}$ is the unique class of ${\displaystyle n}$-tuples such that:

$${\displaystyle {\begin{alignedat}{2}\quad &(x_{1},\ldots ,x_{n})\in \complement _{V^{n}}A&&\iff \neg [(x_{1},\ldots ,x_{n})\in A]\\&&&\iff \neg \psi (x_{1},\ldots ,x_{n},{\vec {Y}})\\&&&\iff \phi (x_{1},\ldots ,x_{n},{\vec {Y}}).\end{alignedat}}}$$

Case 2: ${\displaystyle \phi (x_{1},\ldots ,x_{n},{\vec {Y}})=\psi _{1}(x_{1},\ldots ,x_{n},{\vec {Y}})\land \psi _{2}(x_{1},\ldots ,x_{n},{\vec {Y}}).}$ Since both ${\displaystyle \psi _{1}}$ and ${\displaystyle \psi _{2}}$ have less than ${\displaystyle k}$ logical symbols, the induction hypothesis implies that there are unique classes of ${\displaystyle n}$-tuples, ${\displaystyle A_{1}}$ and ${\displaystyle A_{2}}$, such that:

${\displaystyle {\begin{aligned}\quad &(x_{1},\ldots ,x_{n})\in A_{1}\iff \psi _{1}(x_{1},\ldots ,x_{n},{\vec {Y}}).\\&(x_{1},\ldots ,x_{n})\in A_{2}\iff \psi _{2}(x_{1},\ldots ,x_{n},{\vec {Y}}).\end{aligned}}}$

By the axioms of intersection and extensionality, ${\displaystyle A_{1}\cap A_{2}}$ is the unique class of ${\displaystyle n}$-tuples such that:

$${\displaystyle {\begin{alignedat}{2}\quad &(x_{1},\ldots ,x_{n})\in A_{1}\cap A_{2}&&\iff (x_{1},\ldots ,x_{n})\in A_{1}\land (x_{1},\ldots ,x_{n})\in A_{2}\\&&&\iff \psi _{1}(x_{1},\ldots ,x_{n},{\vec {Y}})\land \psi _{2}(x_{1},\ldots ,x_{n},{\vec {Y}})\\&&&\iff \phi (x_{1},\ldots ,x_{n},{\vec {Y}}).\end{alignedat}}}$$

Case 3: ${\displaystyle \phi (x_{1},\ldots ,x_{n},{\vec {Y}})=\exists x_{n+1}\psi (x_{1},\ldots ,x_{n},x_{n+1},{\vec {Y}}).}$ The quantifier nesting depth of ${\displaystyle \psi }$ is one more than that of ${\displaystyle \phi }$ and the additional free variable is ${\displaystyle x_{n+1}.}$ Since ${\displaystyle \psi }$ has ${\displaystyle k-1}$ logical symbols, the induction hypothesis implies that there is a unique class ${\displaystyle A}$ of ${\displaystyle (n+1)}$-tuples such that:

$${\displaystyle \quad (x_{1},\ldots ,x_{n},x_{n+1})\in A\iff \psi (x_{1},\ldots ,x_{n},x_{n+1},{\vec {Y}}).}$$

By the axioms of domain and extensionality, ${\displaystyle Dom(A)}$ is the unique class of ${\displaystyle n}$-tuples such that:[lower-alpha 8]

$${\displaystyle {\begin{alignedat}{2}\quad &(x_{1},\ldots ,x_{n})\in Dom(A)&&\iff \exists x_{n+1}[((x_{1},\ldots ,x_{n}),x_{n+1})\in A]\\&&&\iff \exists x_{n+1}[(x_{1},\ldots ,x_{n},x_{n+1})\in A]\\&&&\iff \exists x_{n+1}\,\psi (x_{1},\ldots ,x_{n},x_{n+1},{\vec {Y}})\\&&&\iff \phi (x_{1},\ldots ,x_{n},{\vec {Y}}).\end{alignedat}}}$$