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The saturated model condition in degree $\kappa$ means that a first-order model $M$ realizes every one-variable complete type over a parameter set $A\subseteq M$ of size less than $\kappa$. Equivalently every finitely satisfiable such complete type is realized, since it can be extended to a complete type. Without a specified degree, saturated usually means $|M|$-saturated. Countable saturation means ${\aleph }_1$-saturation, including complete types over countably many parameters.

Here is a general saturated elementary extension theorem, with proof: for every infinite regular cardinal $\kappa$, every first-order structure $M$ has a $\kappa$-saturated elementary extension. At a stage $M_{\alpha }$, list all complete types over its subsets of size less than $\kappa$. There is a set of them, since their formulas lie in a set-sized language. Add a witness constant for each complete type to the elementary diagram of $M_{\alpha }$, together with every formula in that complete type evaluated at its witness constant. A finite part mentions finitely many complete types and finitely many formulas from each; each finite conjunction is realizable in $M_{\alpha }$, and the witnesses for different complete types are unconstrained relative to each other. Thus the finite part is satisfiable. The compactness theorem produces an elementary extension $M_{\alpha +1}$ realizing all those complete types. At limits take elementary unions, justified by the elementary chain theorem.

Iterate for $\kappa$ stages and put $N={\bigcup }_{\alpha <\kappa }{M}_{\alpha }$. If $A\subseteq N$ has size below $\kappa$, regularity bounds the stages at which its elements enter, so $A\subseteq M_{\alpha }$ for some $\alpha <\kappa$. A complete type over $A$ consistent with $N$ is finitely satisfiable in $M_{\alpha }$: each finite conjunction is an existential formula with parameters from $M_{\alpha }$, and elementarity transfers its truth from $N$. The complete type is therefore realized in $M_{\alpha +1}$. This provesFor a singular desired degree, apply the theorem at a larger regular cardinal. If $M$ is infinite and an uncountable regular $\kappa$ satisfies ${\kappa }^{<\kappa }=\kappa$, with $|L|<\kappa$ and $|M|\le \kappa$, the construction can keep every stage of size $\kappa$. There are then at most $\kappa$ parameter sets and complete types, and downward Löwenheim-Skolem gives that size after each extension; adding $\kappa$ distinct elements initially ensures final size exactly $\kappa$. The resulting model is saturated in the unqualified sense. The cardinal-arithmetic hypothesis belongs to this size refinement, not to the general existence theorem.

An explicit ultraproduct version is useful too. In a countable language let $U$ be a nonprincipal ultrafilter on $\omega$ and $N={\prod }_{i<\omega }{M}_i/U$, where each factor is nonempty. Enumerate a countable finitely satisfiable complete type, including its chosen parameter representatives, as ${\phi }_1(x),{\phi }_2(x),\dots$. By the Łoś theorem, the set $A_n$ of coordinates where the first $n$ formulas have a simultaneous witness belongs to $U$. PutThese are decreasing members of $U$. At coordinate $i$, let $k(i)$ be the greatest $n\le i$ with $i\in B_n$, if there is one. Choose a witness to the first $k(i)$ formulas there, and an arbitrary element when there is none. For fixed $n$, every coordinate in $B_n$ has $k(i)\ge n$, so the chosen function satisfies ${\phi }_n$ on a $U$-large set. Its ultraproduct class realizes the whole complete type. A countable language over a countable parameter set has only countably many formulas, so this proves countable saturation of a nonprincipal ultraproduct over omega.

For NFU, we give the rank-indiscernible construction of an NFU model explicitly. NFU has a unary predicate $S$ for sets, atoms have no members, extensionality applies to sets, and comprehension applies to every stratified formula. A stratification assigns integer types to variables so that equality uses equal types and $x\in y$ requires $\mathrm{type}(y)=\mathrm{type}(x)+1$. A unary set predicate adds no type difference.

Start with the actual set structure $B=(V_{\omega +\omega },\in )$. This structure satisfies extensionality and every pure-membership separation instance: the subset defined in a set $a$ of rank below $\omega +\omega$ still has rank below $\omega +\omega$. We do not assert that $B$ satisfies replacement or all of ZFC. Expand it by Skolem functions. In it use the sequence of rank objects $V_{\omega +n+6}$, $n<\omega$. For two such objects $d<e$ in sequence order, we havewith quantifiers in $B$. The second statement holds because every subset of $V_{\alpha }$ belongs to $V_{\alpha +1}$, and the later rank is at least $\alpha +1$.

Introduce constants $d_i$, $i\in \mathbb{Z}$, and require them to be order indiscernibles in the Skolem language, satisfying these two assertions whenever $i<j$. Also require $V_{\omega +5}\subseteq d_i$ for every $i$; this fixed rank object is first-order definable in $B$. Every finite part of these requirements together with $\mathrm{Th}(B^{\ast })$ is satisfiable: color finite increasing tuples of the original rank sequence by the truth values of the finitely many formulas mentioned, and use Ramsey's theorem to get a homogeneous subsequence. The rank assertions hold for every increasing tuple there. The compactness theorem gives a model of the full theory. In its Skolem hull $N$ generated by the $d_i$, the integer shift extends to an automorphism $j$ with $j(d_i)=d_{i+1}$, by the term-transport proof of the Ehrenfeucht-Mostowski theorem. Thus $N$ still satisfies extensionality and all pure-membership separation instances, andThis is a set-sized first-order model; no satisfaction predicate for a proper-class universe is being assumed.

Externally let $D_i=\{x\in N:N\models x\in d_i\}$. The automorphism bijects $D_i$ onto $D_{i+1}$. Interpret a typed structure with sort $i$ equal to $D_i$. At sort $i+1$, declare $y$ a set exactly when $N\models y\subseteq d_i$; otherwise it is an atom. Define adjacent-sort membership byThe guard is necessary: a typed atom might have some ordinary $N$-members in $D_i$, and those must not become its typed members. Set-extensionality follows from extensionality in $N$, because the ordinary members of a typed set all lie in $D_i$. For any typed formula on sort $i$, its quantifiers can be restricted to the corresponding $d_l$, its set predicates replaced by subset assertions, and its memberships replaced by the guarded relation. This is an ordinary pure-membership formula of $N$ with finitely many parameters. Separation gives its extension $X\subseteq d_i$, and the displayed closure assertion gives $X\in d_{i+1}$. It is therefore a typed set witnessing comprehension. The automorphism shifts these sorted domains and preserves the set flags and guarded memberships.

Collapse the types onto $D_0$. DefineObjects not satisfying $S$ have no $E$-members. If two sets have the same $E$-members, their images under $j$ are subsets of $d_0$ with the same ordinary members, so are equal by extensionality of $N$; injectivity of $j$ gives equality of the original objects.

To verify every stratified comprehension instance, assign a variable $v$ its stratification type $t(v)$ and translate its value $a\in D_0$ to $j^{t(v)}(a)\in D_{t(v)}$. Equality is preserved. An atomic $E(x,y)$ translates to the guarded adjacent-sort membership because $t(y)=t(x)+1$; the unary predicate $S(y)$ translates to $j^{t(y)}(y)\subseteq d_{t(y)-1}$. Induction on formulas, using the domain bijections for quantifiers, preserves truth. If the free variable $x$ has type $t$, typed comprehension produces its extension $X\subseteq d_t$, with $X\in d_{t+1}$. Put $b=j^{-(t+1)}(X)\in D_0$. Then $S(b)$ holds, and for every $a\in D_0$,This is exactly the required NFU set. No formula containing the external automorphism $j$ has been used in a separation or comprehension scheme; $j$ only transports values after the pure formula is formed.

The construction also permits the usual infinity requirement. The ordinary $\omega$ and its successor graph $f$ are definable in $N$, hence fixed by $j$, and lie below $V_{\omega +5}$, so belong to all the required domains. Under $E$, the members of $\omega$ are still its ordinary $N$-members. The new Kuratowski pair satisfiesSince $j(f)=f$, membership of this pair in $f$ under $E$ is equivalent to $⟨x,y{⟩}_N\in f$. Thus $f$ is an internal injection of $\omega$ into itself omitting zero, giving a Dedekind-infinite set.

We have built a model of NFU, with infinity if it is included in the formulation, in ordinary ZFC metatheory:In particular $\mathrm{Con}(\mathrm{ZFC})\Rightarrow \mathrm{Con}(\mathrm{NFU})$. This proof does not assume $\mathrm{Con}(\mathrm{ZFC})$ as an additional internal axiom: its starting structure is the set $V_{\omega +\omega }$, not a model of ZFC.

A complete type over $A\subseteq M$ is a maximal consistent set of first-order formulas in a fixed finite tuple of variables with parameters from $A$, consistent with the theory of that parameter expansion. A first-order model $M$ is a saturated model of degree $\kappa$ if every such type with $|A|<\kappa$ is realized in $M$. It suffices to require this for one-variable types and realize a finite tuple successively. Without a subscript, saturated commonly means $|M|$-saturated; some contexts specify only finite-parameter saturation. We state the degree explicitly.

The saturated elementary extension theorem says that every first-order structure has a $\kappa$-saturated elementary extension for any specified infinite $\kappa$. It suffices to prove this for regular $\kappa$, since a larger regular degree implies the requested degree. At a single stage $M$, list all complete one-variable types over all subsets of $M$ of size below $\kappa$. This collection is a set. Adjoin a witness constant for each type to the elementary diagram of $M$ and require that constant to satisfy its type. Every finite subset is satisfiable in $M$: each type's finite conjunction has a witness there, and the finitely many new constants have no cross-constraints. The compactness theorem gives an elementary extension realizing all these types simultaneously.

Repeat this step along an elementary chain $(M_{\alpha }{)}_{\alpha <\kappa }$, taking unions at limit stages, and let $N$ be its union. The elementary chain theorem follows here by induction on first-order formulas: an existential statement true in the union has a witness in a later stage; elementarity brings an existential witness back to the earlier stage containing its parameters. Thus each stage is elementary in $N$.

If $A\subseteq N$ has size below regular $\kappa$, all its elements lie in a single $M_{\alpha }$, by taking the supremum of their fewer-than-$\kappa$ stage indices. Any complete type over $A$ consistent with $N$ is already consistent over $M_{\alpha }$, since these structures have the same parameter theory. It was therefore realized in $M_{\alpha +1}$. This proves $\kappa$-saturation of $N$.

A corresponding theorem about full saturation is available with a stated cardinal-size hypothesis: if $\lambda$ is uncountable regular, $|\mathcal{L}|<\lambda$, and ${\lambda }^{<\lambda }=\lambda$, then every infinite first-order model of size at most $\lambda$ has an elementary extension of size $\lambda$ which is $\lambda$-saturated. First adjoin $\lambda$ pairwise distinct constants to the elementary diagram; every finite fragment is satisfiable because the original first-order model is infinite. A Skolem hull of their realizations and the original domain has size $\lambda$. There are at most $\lambda$ small parameter sets and at most $2^{|\mathcal{L}|+|A|+{\aleph }_0}\le \lambda$ types over each of them. Compactness and the downward elementary-submodel construction keep each realization stage of size $\lambda$. Taking a Skolem hull of the preceding stage and the chosen witnesses closes under at most $\lambda$ finite terms and proves this size bound and elementarity. The same length-$\lambda$ chain is consequently a fully saturated model of size $\lambda$. The cardinal hypothesis is not silently asserted to hold for every cardinal.

We now prove consistency of NFU by the permitted alternative method of indiscernibles and a domain-shifting structure automorphism. This can start from an explicit set structure available in ZFC, so we need not assume that there is a first-order model of all ZFC. Letand give $F$ the value $\varnothing$ off the natural numbers. Regard $(H,\in ,F,\omega )$ as a first-order structure, with a named object $\omega$. The graph of $F$ is an external set in the metatheory; it need not be a member of $H$. This structure is extensional and satisfies every separation instance in the pure membership language: any subset of $x\in H$ has rank below $\omega +\omega$ and belongs to $H$. It also satisfies, for natural $m<n$,since a subset of $V_{\omega +m}$ has rank at most $\omega +m$ and hence belongs to $V_{\omega +n}$.

Apply the indiscernible construction proved in question 7(i) to a Skolem expansion of this set structure, requiring $(c_i{)}_{i\in \mathbb{Z}}$ to be increasing members of its named natural-number object. For every finite fragment use an increasing sequence of ordinary natural numbers in the initial structure in the Ramsey argument. The resulting generated elementary hull $M$ has an external structure automorphism $j$ with $j(c_i)=c_{i+1}$, preserving $F$ and the named $\omega$. Put $D_i=F^M(c_i)$ and $D=D_0$. Then $j(D_i)=D_{i+1}$, and $M$ satisfies the displayed subset-containment assertion between successive domains. No claim that $M$ satisfies replacement is required.

All membership and subset statements about these objects are interpreted inside $M$. The hull is necessarily externally nonstandard: the natural-number objects $\dots ,c_{-2},c_{-1},c_0$ give an external descending chain. The external structure automorphism is not an object that must be definable within that first-order model.

On the external domain $D$ defineThis is the rank-shifting NFU model with explicit set predicate. The guard $S(y)$ is essential: an object not contained in $D$ could still have some ordinary members in $D$, and must not thereby become a nonempty atom. The objects failing $S$ have no $E$-members. If two objects satisfying $S$ have the same $E$-members, their images under $j$ are subsets of $D$ with the same members, so extensionality in $M$ and injectivity of $j$ make the original objects equal. Thus set-extensionality and the atom condition hold.

It remains to prove every comprehension instance. Use the domains $D_i$ just constructed, equivalently $D_i=j^i(D)$, for integer sorts. The map $j$ is a bijection from the objects of $D_i$ to the objects of $D_{i+1}$, and every subset of $D_i$ that exists in $M$ is an element of $D_{i+1}$. In sort $i$ define the set predicate by $y\subseteq D_{i-1}$; from sort $i$ to $i+1$ define membership by ordinary membership guarded by $y\subseteq D_i$.

For a stratified formula, assign integer types to its variables with equal types for equality and adjacent types for membership. Shift all types uniformly if necessary so the finitely many used types are nonnegative. Replace a variable of type $i$ by $j^i$ of its untyped value, quantify it over $D_i$, and use the typed relations just described. Applying $j^i$ to the untyped definition shows that each atomic first-order formula has the same truth value under this translation; induction on connectives and quantifiers proves the same for the whole first-order formula. Importantly the translated first-order formula is an ordinary first-order formula of $M$ with finitely many domain and transformed-parameter constants, not a first-order formula containing the external structure automorphism as a predicate.

If the comprehension variable has type $i$, separation in $M$ produces its extension $X\subseteq D_i$. The subset-containment assertion gives $X\in D_{i+1}$. Put $y=j^{-(i+1)}(X)\in D$. Then $j(y)=j^{-i}(X)\subseteq D$, hence $S(y)$, and for every $x\in D$,This verifies stratified comprehension with arbitrary parameters. The empty extension gives a distinguished empty set; the always-true first-order formula gives a universal set. The unstratified Russell predicate is not an allowed comprehension instance. ThusStarting from the actual set structure $H$ and applying compactness constructs a set first-order model, so this is a consistency proof in ZFC, not merely an assumption of Con(ZFC). It proves the requested consistency with an explicit interpretation; it does not depend on treating the much stronger full extensionality of NF as already consistent. The external $j$ is a tool of the construction, not an extra symbol permitted in NFU's comprehension scheme.