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A finite Cartesian product of well-quasi-orderings is well-quasi-ordered in the componentwise order. Every infinite sequence in one well-quasi-order has an infinite nondecreasing subsequence: colour pairs by whether they are ordered, apply Ramsey's theorem, and exclude an all-bad subsequence. Restrict successively to such a subsequence in each coordinate. The same argument, or the infinite pigeonhole principle, proves closure under finite disjoint unions with no cross-component comparisons.

Fix positive integers $r,k,m$ with $m\ge r$. There is an integer $n\ge m$, depending only on $r,k,m$, such that every finite coloringhas a subset $H\subseteq [n]$ of size $m$ for which the restriction of a function $c$ to $\left(\genfrac{}{}{0ex}{}{H}{r}\right)$ is a constant function. Here $[n]=\{1,\dots ,n\}$, and $\left(\genfrac{}{}{0ex}{}{S}{r}\right)$ denotes the collection of all $r$-element subsets of $S$, rather than its binomial coefficient cardinality. The notation $[n]⟶(m{)}_k^r$ abbreviates this property. Thus $H$ is a homogeneous set for a colouring; when $r>1$, the monochromatic set consists of its $r$-element subsets, not its individual elements. This is the finite assertion in Ramsey's original Theorem B, p. 267.

For $r=1$, this is the pigeonhole principle, with sharp threshold $n=k(m-1)+1$. For instance, every red-blue coloring of five objects has three of one color; four objects can split two and two. For $r=2$, $k=2$ and $m=3$, the sharp threshold is six, as shown by the Ramsey number for triangles. Coloring pairs makes this a problem about edges of a complete graph. For larger $r$, the analogous viewpoint uses a hypergraph.

The finite Ramsey theorem lets an existence argument discard most of a large finite set while retaining a prescribed amount of uniform structure. Its threshold applies to every coloring; one does not need an assumption that colors were assigned randomly. A diagonal compactness argument deduces finite existence from Ramsey's theorem, but does not by itself identify the smallest threshold. For pairs, a Ramsey number makes that optimization explicit.

Let $K_s$ be a computable increasing finite-stage enumeration of the halting problem set $K$. Color $x<y<z$ by whether the approximations below $x$ differ at stages $y,z$. This is a recursive two-color partition of triples. An infinite homogeneous set cannot have color one: for fixed $x$, the finite approximation below $x$ eventually stabilizes. If its color is zero, choose $x$ from that set above an input $n$, then another member $y>x$; agreement with arbitrarily large later stages gives $K(n)=K_y(n)$. Thus every infinite homogeneous set computes $K$, and none is a recursive set. This makes the failure of an effective version of Ramsey's theorem explicit.

Every finite coloring of a barrier in better-quasi-order theory has an infinite restriction on which all barrier members have the same color. It extends finite-dimensional Ramsey's theorem from fixed-length subsets to variable-length initial segments. A proof proceeds by well-founded induction on the initial-segment tree and a fusion construction selecting successively nested infinite tails. It implies that every finite set with the equality order is a better-quasi-ordering. The barrier/front partition proof is given in Section 3.2 of arxiv.org/pdf/1604.05866 .

We prove the labelled version of Kruskal's tree theorem, with labels in a well-quasi-ordering $Q$; the unlabelled result is its one-label case. A label-monotone tree embedding is an injective map of finite rooted trees preserving lowest common ancestors and increasing labels. Its root may map below the host root, and an edge may map to a path. These conventions distinguish rooted-tree homeomorphic embedding from the stricter adjacency-preserving relation, which is not well-quasi-ordered.

We will use two elementary consequences of well-quasi-ordering. Every infinite sequence in a well-quasi-ordering has an infinite nondecreasing subsequence: color pairs by whether the earlier element precedes the later one, apply Ramsey's theorem, and exclude an all-bad infinite set. Restricting successively in each coordinate proves finite product closure of well-quasi-orderings.

For completeness, here is the required proof of Higman's lemma. If finite words over $Q$ admitted a bad sequence, choose one $w_0,w_1,\dots$ whose next word always has the least length admitting a bad continuation of the fixed prefix. No word is empty. Write $w_i=v_i{a}_i$, separating the last letter. The set of prefixes $\{v_i\}$ is a well-quasi-ordering. Otherwise select a bad sequence of these prefixes with strictly increasing owner indices $n_0<n_1<\cdots$; this is possible because every finite initial set of owners supplies only finitely many prefixes, and a bad sequence cannot repeat a word. Thenis bad. An earlier original word embedding into $v_{n_j}$ would also embed into $w_{n_j}$, contradicting the original badness; comparisons inside the new tail are excluded by its choice. But $v_{n_0}$ is shorter than $w_{n_0}$, contradicting minimality. Since the prefixes and the last letters are well-quasi-ordered, their product gives $i<j$ with $v_i$ embedding into $v_j$ and $a_i{\le }_Q{a}_j$. Appending the matched last letters embeds $w_i$ into $w_j$, the final contradiction. Thus finite words over a WQO are WQO.

Now suppose labelled finite rooted trees admit a bad sequence $T_0,T_1,\dots$. Choose it minimally in vertex count at each successive position. Let $\mathcal{S}$ be all immediate child subtrees of its members. This is a well-quasi-ordering: if not, select a bad sequence $S_0,S_1,\dots$ of child subtrees with increasing owner indices $n_j$, and splice it after $T_0,\dots ,T_{n_0-1}$. Every child subtree embeds into its owner, so a comparison from the prefix to the tail would contradict original badness. The splice is bad and its first replacement is strictly smaller, contradicting the minimal bad sequence choice.

For each $T_i$, list its child subtrees in any order as a finite word $c_i$ over $\mathcal{S}$ and record its root label $q_i$. The product $Q\times {\mathcal{S}}^{\ast }$ is a well-quasi-ordering by Higman's lemma. Hence some $i<j$ has $q_i{\le }_Q{q}_j$ and a subsequence embedding of $c_i$ into $c_j$. Match the roots and use the chosen subtree embeddings in distinct target branches. Their union preserves lowest common ancestors and labels, and stretches only edge paths. Thus $T_i$ embeds into $T_j$, a contradiction. We have proved

To deduce Friedman's finite form of Kruskal's theorem, fix $k\in \mathbb{N}$. At position $i=0,1,\dots$ permit unlabelled trees of at most $k+i$ vertices. Up to isomorphism only finitely many trees are possible at each position. Form the finite bad-sequence tree whose nodes are finite bad sequences satisfying these bounds, ordered by extension. It is finitely branching. If there were arbitrarily long admissible bad sequences, König infinity lemma would give an infinite branch, contradicting Kruskal's tree theorem. ThereforeOne can instead use a fixed finite set of labels, giving a correspondingly parameterized finite statement by exactly the same proof. The crucial feature is a uniform bound $N$ for each initial size allowance, not merely termination of one chosen sequence.

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.

Use the rooted-tree homeomorphic embedding relation: an injective map of vertices preserves lowest common ancestors, so edges may become paths and the source root may map below the target root. We prove the stronger labelled statement for labels in any well-quasi-ordering $Q$, with labels increasing under the embedding. Forgetting labels gives Kruskal's tree theorem.

First every infinite sequence in a well-quasi-ordering has an infinite nondecreasing subsequence. Colour its index pairs according as the earlier term is or is not below the later term. Ramsey's theorem gives an infinite constant-colour subsequence; the bad colour would contradict well-quasi-ordering. Applying this successively in each coordinate proves finite product closure of well-quasi-orderings.

We need Higman lemma with its proof. Order finite words over $Q$ by subsequence embedding with increased letters. If there were a bad sequence, choose one $w_0,w_1,\dots$ minimally: at each position choose a shortest word admitting a bad continuation of the fixed prefix. No word is empty. Write $w_n=u_n{a}_n$, separating the last letter, and choose $n_0<n_1<\cdots$ with $a_{n_0}\le a_{n_1}\le \cdots$. ConsiderAn earlier prefix word cannot embed in a later $u_{n_j}$, since it would then embed in $w_{n_j}$. Nor can $u_{n_i}$ embed in $u_{n_j}$: adding the comparable last letters would embed $w_{n_i}$ in $w_{n_j}$. Thus the displayed sequence is bad, contradicting minimality at $n_0$, since $u_{n_0}$ is shorter. Finite words over a well-quasi-order are well-quasi-ordered.

Now suppose labelled rooted trees admit a bad sequence $T_0,T_1,\dots$, chosen minimally in vertex count at each position. Let $\mathcal{A}$ consist of all proper rooted subtrees of these trees, rooted at a descendant and containing all its descendants. We claim $\mathcal{A}$ is well-quasi-ordered. Otherwise take a bad sequence $S_0,S_1,\dots$ from $\mathcal{A}$. Only finitely many subtrees come from any fixed finite prefix of the $T_n$'s, and a bad sequence cannot repeat an object. Thin it so that $S_i$ comes from $T_{m_i}$ with $m_0<m_1<\cdots$. The sequenceis bad: $T_j⪯S_i$ would imply $T_j⪯T_{m_i}$, since a rooted subtree embeds in its host; comparisons within the $S_i$'s are excluded by their choice. But $S_0$ is strictly smaller than $T_{m_0}$, contradicting minimality. This proves the claim.

Choose an arbitrary ordering of the children of each $T_n$. Record its root label and the finite word of rooted child-subtrees:By Higman lemma and finite product closure, two records, with indices $i<j$, are comparable. Map the root of $T_i$ to the root of $T_j$, and map its selected child-subtrees into the distinct selected branches of $T_j$. Within a branch the embeddings preserve lowest common ancestors; between different branches both lowest common ancestors are the roots. Labels increase everywhere. This gives $T_i{⪯}_Q{T}_j$, contradicting badness. Finite labelled rooted trees are well-quasi-ordered by label-monotone homeomorphic embedding. In particular, this proves the unlabelled theorem. A root-preserving variant follows by giving the root a special incomparable label unavailable at nonroot vertices; the general labelled theorem then forces root to root. Adjacency-preserving embeddings are a different, more restrictive relation and do not satisfy the same assertion.

For Friedman's finite form of Kruskal's theorem, fix $k\in \mathbb{N}$ and consider finite bad sequences satisfying $|T_i|\le k+i$, with indices starting at $1$. Their prefixes form a finite bad-sequence tree. At each depth there are only finitely many possible unlabelled tree isomorphism types below the size bound, so this tree is finitely branching. If bad sequences existed at every finite length, König infinity lemma would yield an infinite branch, hence an infinite bad sequence of rooted trees, contrary to what was just proved. ThereforeThe same proof works with any fixed finite label set. The finite size bounds are what make the bad-prefix tree finitely branching; the proof does not claim a small effective numerical bound.

We give a relative consistency proof: a model of ZFC yields a model of NFU. This proves the requested consistency relative to the usual background set theory; it does not claim the optimal, much weaker consistency bound for bare NFU.

Recall that New Foundations with urelements has a unary set predicate $S$, atoms without members, extensionality restricted to sets, and comprehension for every stratified formula. Stratification assigns integer types to variables, with equal types at equality and adjacent types at membership. We construct a model of simple typed set theory with atoms with a type-shifting automorphism, then identify its types using that automorphism.

Start with a model of ZFC and a Skolem expansion of it. By the compactness theorem there is a model $N$ of the same expanded theory with an increasing integer-indexed order-indiscernible sequence of ordinals $⟨c_i:i\in \mathbb{Z}⟩$, all above $\omega$. Here is the finite-satisfiability argument. A finite fragment names only finitely many $c_i$ and asks for equal truth values of finitely many formulas on increasing tuples. In the starting model choose an infinite increasing sequence of ordinals above $\omega$. Colour its finite subsets by these truth vectors, successively treating each tuple length. The Ramsey's theorem supplies an infinite homogeneous subsequence for the finitely many colours and arities. Its first required number of elements interprets all named $c_i$ and satisfies that fragment. Compactness gives the entire sequence. This argument works for the countable expanded language because each fragment uses only finitely many formulas.

Replace $N$ by the Skolem hull of the $c_i$. The mapis well defined by indiscernibility: equal terms remain equal after shifting their increasing lists of indices. The same argument preserves every formula. Shifting indices backwards gives its inverse, so $j$ is an automorphism of the hull, and $j(c_i)=c_{i+1}>c_i$. The Skolem hull is elementary by its witness functions. This is a concrete Ehrenfeucht-Mostowski model; it is not an elementary class self-embedding of the actual well-founded universe.

Set $D_i=V_{c_i}^N$, interpreted externally as the objects belonging to that internal rank segment, for every integer $i$. At type $i$ declareand define adjacent-type membership byThe members of a type-$i+1$ set are exactly a subset of $D_i$; a type-$i+1$ atom has no ${\epsilon }_i$-members. Every internal subset of $D_i$ is an object of $D_{i+1}$, since $c_{i+1}>c_i$. Extensionality for sets is inherited from $N$. For a typed formula involving finitely many types, restrict its quantifiers to the corresponding internal $D_k$ and express each $S_k$ and ${\epsilon }_k$ by the displayed definitions. Separation in $N$ gives the required subset of $D_i$, which represents its comprehension at type $i+1$. This verifies the typed theory even though many higher-type objects are atoms. The automorphism $j$ carries $D_i,S_i,{\epsilon }_i$ to the next type.

On domain $D_0$, define $S(x)=S_0(x)$ andIf $S(y)$ is false, $S_1(j(y))$ is false, so $y$ has no $E$-members. For sets $y,z$, equality of their $E$-extensions gives equality of the typed sets $j(y),j(z)$ and hence $y=z$. To check stratified comprehension, translate a variable of type $k$ into $j^k(x)$. The automorphism laws make membership and equality translate exactly. A stratified predicate on a variable of type zero defines a typed subset $B$ of $D_0$, represented by a set at type one. Then $b=j^{-1}(B)\in D_0$ has $E$-extension exactly that predicate. A uniform type shift handles any other type assigned to the comprehended variable. The resulting untyped structure is a model of NFU.

The arrow in the original PDF permits partial functions. A partial computable function ${\mathbb{N}}^k\rightharpoonup \mathbb{N}$ is computed by a finite program which halts with the specified value on every input in its domain and diverges on inputs outside the domain. A total computable function is the special case with full domain. Coding finite tuples by a primitive recursive pairing function identifies the different finite arities effectively, without identifying total and partial computability.

One precise machine model is a Turing machine. Its finite transition rules and configurations admit effective numerical coding. Bounded simulation is primitive recursive, and unbounded execution gives the partial result. Equivalently, start from zero, successor and projections and close under composition, primitive recursion and unbounded minimization. The Kleene normal form theorem makes the equivalence concrete:where $T$ is a primitive recursive predicate checking a coded halting computation and $U$ reads its output. If no computation exists, the minimum and hence the value are undefined. Checking a finite history is effective because every consecutive configuration must obey one of the finitely many transition rules.

The universal partial computable function evaluates the program indexed by $e$ on an input. The S-m-n theorem says that fixing finitely many input parameters can be done by a total computable operation on program Gödel numbers: a new program stores those parameters and calls the original universal evaluator. These facts explain both universal simulation and uniform reductions. A set is computably enumerable when it is the domain of a partial computable function, equivalently the range of an effective enumeration. It is recursive when its characteristic function is total computable. A set is recursive precisely when it and its complement are computably enumerable: dovetail the two enumerations to decide each input.

The diagonal halting problem $K=\{e:{\phi }_e(e)\downarrow \}$ is computably enumerable but not recursive. A total decider would yield a program which, on $e$, diverges if the decider says $e\in K$ and halts otherwise. Applying this program to its own index gives a contradiction. There is likewise no total computable universal listing of all total computable functions: the diagonal function $n\mapsto U(n,n)+1$ would be total and absent from the listing.

Here is a full proof of the Rice theorem. Let $\mathcal{P}$ be a nonempty proper class of partial computable functions whose property depends only on the function, not its program. First suppose the everywhere-undefined function $\perp$ is not in $\mathcal{P}$, and choose a program for a function $g\in \mathcal{P}$. Given an index $e$, construct uniformly a program which on $x$ first waits for ${\phi }_e(e)$ to halt and, only then, simulates $g(x)$. The S-m-n theorem gives a total computable index map $h$. If $e\notin K$, then ${\phi }_{h(e)}=\perp$; if $e\in K$, then ${\phi }_{h(e)}=g$. ThusA decision procedure for the property would decide $K$, impossible. If $\perp \in \mathcal{P}$, apply the same argument to its nontrivial complement. Every nontrivial extensional property of partial computable functions has an undecidable index set. Totality, having finite domain, and taking a particular value somewhere are examples. A syntactic property such as program length is not covered; nor does the theorem say that no individual program's behaviour can be established.

Self-reference can also be made effective. For a total computable index transformation $f$, let $d(x)$ be an index for the program which first computes ${\phi }_x(x)$ and then runs the program with that resulting index on its own input. The S-m-n theorem makes $d$ total computable. Choose an index $e$ for the total function $x\mapsto f(d(x))$, and put $n=d(e)$. ThenThis proves the recursion theorem: every effective transformation of programs has a fixed point at the level of the function computed, though not necessarily at the level of its numerical code.

Finally, consider the Jockusch triple coloring computing the halting set. Fix computable finite approximations $K_0\subseteq K_1\subseteq \cdots$ with union $K$. For $x<y<z$, defineThis is a total recursive two-coloring because both stages are finite computations. Suppose $H$ is infinite and homogeneous. Its color cannot be one: fix $x\in H$, wait until every enumeration below $x$ has stabilized, and choose later $y<z$ from $H$ beyond that stage. Their color is zero.

If $H$ has color zero, it computes $K$. On input $n$, use $H$ to find $x>n$ and $y>x$ in $H$. For arbitrarily large $z\in H$ with $z>y$, homogeneity gives $K_y\cap x=K_z\cap x$. Taking stages sufficiently large to reach the final finite approximation gives $K_y\cap x=K\cap x$. Consequently membership of $n$ in $K$ is decided by the finite stage $K_y$. A recursive $H$ would therefore make $K$ recursive, a contradiction. There is a recursive partition of triples with no infinite recursive homogeneous set. This is the nontrivial finite-color meaning of the partition of an infinite domain; if “infinite partition” meant infinitely many colors, the coloring by the least member would already give a trivial example. Ramsey's theorem still guarantees infinite homogeneous sets for the displayed two-coloring, but every such set is computationally nontrivial.

A useful precise form of the Ehrenfeucht-Mostowski theorem is this: for an infinite structure $M$ and any linear order $I$, a model of $\mathrm{Th}(M)$ can be generated as a Skolem hull by distinct order indiscernibles $(a_i{)}_{i\in I}$. Every order automorphism of $I$ extends to an automorphism of that hull in the chosen Skolem language. One may first name all elements of $M$, giving an elementary extension containing the original structure. The generation and uniqueness assertions concern this specified expansion, not arbitrary automorphisms of its reduct.

Choose a Skolem expansion $M^{\ast }$: for each existential formula add a function selecting a witness whenever one exists. Introduce constants $c_i$ for $i\in I$. Take the theory of $M^{\ast }$, distinctness of these constants, and all equivalencesThe formulas here are in the fixed expanded language, without the newly added $c_i$ as extra parameters. If naming $M$, use its complete elementary diagram and allow those original names in the formulas.

Every finite fragment has a model by Ramsey's theorem. Only finitely many formulas and constants occur. Take a sequence of distinct elements of $M$. For each arity appearing, color the increasing tuples from this sequence by the finite vector of truth values of all formulas of that arity. Successive Ramsey thinning gives an infinite subset homogeneous for each of these finitely many colorings. Assign the finitely many constants in increasing order to distinct elements of this subset. This satisfies every displayed equivalence in the fragment; the original theory is satisfied because the underlying structure remains $M^{\ast }$. Compactness now gives a model of the full theory, with the required order-indiscernible sequence.

The substructure generated by the constants under the Skolem functions is elementary: any existential formula with parameters which are terms in the generators has its Skolem witness in that same hull. If $g$ is an order automorphism of $I$, defineIndiscernibility makes this well-defined, by applying it to the equation between any two term representations using the increasing tuple of all their indices. The same argument preserves every relation and function; $g^{-1}$ supplies the inverse. It is unique in the generated Skolem structure. This completes the Ramsey proof, rather than stopping at the existence of indiscernibles.

For simple typed set theory with atoms, use sorts $D_0,D_1,\dots$ and membership only from $D_i$ to $D_{i+1}$. A set predicate distinguishes sets from atoms, extensionality compares sets of the same positive type, and comprehension produces in type $i+1$ every subset of $D_i$ defined by a well-typed formula. Atoms have no members. First-order models use Henkin semantics; they are not required to have their next domain equal to the full external power set. Typical ambiguity is the schema $\varphi \leftrightarrow {\varphi }^{+}$ for each closed sentence, where every sort index in ${\varphi }^{+}$ is raised by one.

We show finite satisfiability of this entire typed theory with ambiguity. Choose a list of ranks ${\rho }_0<{\rho }_1<\cdots$ above $\omega$, spaced by at least two. For a finite increasing selection of ranks $r_{-1}<r_0<\cdots <r_m$, letwhere $S_i$ is the set of objects declared to be sets at type $i$. Membership from $D_i$ to $D_{i+1}$ is ordinary membership restricted to objects of $S_{i+1}=\mathcal{P}(D_i)$; all other type-$i+1$ objects are atoms. The spacing puts $\mathcal{P}(D_i)$ inside $D_{i+1}$. Thus comprehension and set extensionality hold exactly, and atoms have no typed members. There are infinitely many objects at every visible type, so any usual typed infinity requirement also holds.

The auxiliary rank $r_{-1}$ provides a uniform interpretation if a set predicate is allowed at type zero; no type-minus-one domain is used in the theory. This avoids artificially declaring all type-zero objects to be atoms, which would not be invariant under type-raising. If the language has no type-zero set predicate, simply ignore this auxiliary interpretation.

A finite set of ambiguity instances uses finitely many closed sentences of bounded highest type, say at most $m$. Color increasing $(m+2)$-tuples of ranks by their finite truth vector in the structure with visible domains $D_0,\dots ,D_m$ and the auxiliary preceding rank. Ramsey's theorem gives an infinite homogeneous rank set. Choose $m+3$ consecutive ranks from it and build the hierarchy on all these ranks. The first window evaluates each original sentence; shifting the window one place evaluates its uniform type-raise. Homogeneity makes the truth vectors equal. All finite typed axioms and the selected ambiguity instances therefore hold together. Choosing $m$ large enough covers every sort needed by the finite fragment. Many-sorted compactness, equivalently ordinary first-order compactness after encoding sorts, yieldsThis is a model existence argument in the ambient set theory; no equality of adjacent type cardinalities is claimed. The gaps between ranks supply atoms, and the ambiguity is about truth of sentences, not a bijection between domains.

We prove the Ehrenfeucht-Mostowski theorem by ordered products of ultrafilters, without using Ramsey's theorem. Choose a Skolem expansion $M^{\ast }$ of the infinite structure and distinct elements $b_0,b_1,\dots$. Fix a nonprincipal ultrafilter $U$ on $\mathbb{N}$.

For a finite increasing subset $F=\{i_1<\cdots <i_r\}$ of the desired index order $I$, define the ordered Fubini product of ultrafilters $U_F$ on ${\mathbb{N}}^F$ byThe notation $UnP(n)$ means that the set of $n$ satisfying $P$ belongs to $U$; the leftmost quantifier is outermost. Upward closure and intersection follow successively at each coordinate, and applying complement at each step proves the ultrafilter dichotomy. Products are ordered; no false claim of symmetry is needed.

Put $M_F=(M^{\ast }{)}^{{\mathbb{N}}^F}/U_F$, with $M_{\varnothing }=M^{\ast }$. If $F\subseteq G$, pullback of a function by coordinate projection inducesFor a cylinder condition involving only coordinates in $F$, deleting the unused ultrafilter quantifiers makes its membership in $U_G$ exactly its membership in $U_F$. The Łoś theorem therefore makes ${\sigma }_{FG}$ an elementary embedding, not merely a homomorphism. The projection identities give ${\sigma }_{GH}{\sigma }_{FG}={\sigma }_{FH}$.

Take the directed limit of this system: elements from two stages are identified if their images agree at a common larger finite stage, and formulas are evaluated at such a stage. Elementary coherence makes this well-defined. It also makes each stage elementary in the limit: a formula uses finitely many parameters, and an existential witness at a later stage can be pulled back at the level of existential truth because the transition maps are elementary. Denote the limit by $N^{\ast }$.

For each $i\in I$, let $a_i$ be the limit image of the coordinate function $n_i\mapsto b_{n_i}$. These elements are distinct. For $i<j$, after fixing $n_i$, equality $b_{n_i}=b_{n_j}$ holds for at most one $n_j$, a set not in $U$; hence equality fails in the two-coordinate ultrapower and therefore in the limit.

For any increasing $i_1<\cdots <i_r$, Łoś evaluatesThe right-hand side depends on the formula and the number of increasing coordinates, not their particular indices. Thus $(a_i{)}_{i\in I}$ is order-indiscernible for every formula of the expanded language simultaneously; no preliminary homogeneous thinning is used. If elements of $M$ were named before expanding, the same computation gives indiscernibility over those names and the initial stage embeds the original structure into the limit.

The Skolem hull of these generators is elementary by the witness property. Transporting the generators by any order automorphism of $I$ transports every term, with well-definedness and relation preservation supplied by the formula equivalence just proved. The inverse transport supplies an inverse automorphism. The ordered-ultrapower construction gives the full Ehrenfeucht–Mostowski model and its order automorphisms without Ramsey's theorem.

For the Ehrenfeucht-Mostowski theorem, let $M$ be an infinite first-order structure and $I$ any linear order. Choose a full Skolem expansion $M^{\ast }$: add witness functions for existential formulas, repeat in the expanded languages, and take their union. We will construct a model of its theory containing distinct generators $c_i$, $i\in I$, forming an order-indiscernible sequence.

Adjoin the elementary diagram of $M^{\ast }$, the inequalities $c_i\ne c_j$ for $i\ne j$, and for every expanded-language formula $\varphi (x_1,\dots ,x_r)$ and every two increasing $r$-tuples of indices, the sentencesEvery finite subcollection is satisfiable in $M^{\ast }$. To prove this, choose a countably infinite sequence of distinct elements of $M$, and for each of the finitely many relevant arities color its increasing index tuples by the finite vector of truth values of the formulas of that arity. Repeated applications of the Ramsey's theorem give an infinite subsequence homogeneous for all these finitely many finite colorings. Interpret the finitely many mentioned $c_i$ by its elements in their index order. All indiscernibility and distinctness constraints hold, and the finite diagram fragment holds because the ambient structure is still $M^{\ast }$.

By compactness theorem, a model $N^{\ast }$ of the full enlarged theory exists. Its Skolem hull $H$ generated by the $c_i$ is elementary: whenever an existential formula with parameters in $H$ holds in $N^{\ast }$, its chosen witness function returns a value in $H$, so the Tarski-Vaught test applies. Its reduct is a model of $\mathrm{Th}(M)$. If the index order has size $\kappa$, the hull has size at most $\kappa +|L|+{\aleph }_0$, and it contains $|I|$ distinct generators.

Every order automorphism $h$ of $I$ induces the automorphismTo check well-definedness, write any two terms being compared over the same increasing list of their generators. Their equality is an expanded-language formula; indiscernibility of the order-indiscernible sequence preserves its truth under $h$. The same argument preserves each relation, and the term construction preserves the functions. Applying $h^{-1}$ gives the inverse. The extension is unique in the specified Skolem language since every hull element is a term in the generators; no uniqueness among all automorphisms of the reduct is being asserted. This proves the Ehrenfeucht-Mostowski theorem, including its usual automorphism consequence.

For simple typed set theory with atoms, use sorts $0,1,\dots$, predicates $S_i$ for sets, and typed membership from sort $i$ to sort $i+1$. Atoms have no members, extensionality compares sets, and every well-typed formula defines a set of the next sort. Typical ambiguity is the schema $\varphi \leftrightarrow {\varphi }^{+}$ for every closed typed sentence, where the superscript raises all types by one.

Here are concrete full typed models before imposing the ambiguity schema. Choose increasing natural numbers $n_{-1}<n_0<n_1<\cdots$ and setThe auxiliary $D_{-1}$ merely interprets the set predicate at the lowest sort. Define typed membership byThen the other objects of $D_{i+1}$ are atoms, regardless of their old membership. Every subset of $D_i$ has its representative in $S_{i+1}$, giving comprehension even for formulas with higher-sort quantifiers and parameters. Extensionality holds for the set representatives and atoms have no typed members. The rank gaps leave infinitely many atoms at each sort, and all sorts are infinite.

Consider finitely many ambiguity instances with unraised sentences ${\varphi }_1,\dots ,{\varphi }_t$, and choose a fixed window length $r$ sufficient to include every sort used and its auxiliary predecessor. Color every increasing $r$-tuple of natural numbers by the vector of truth values of these unraised sentences in its finite rank-window interpretation. There are only $2^t$ colors. The Ramsey's theorem gives an infinite homogeneous set. Choose $r+1$ successive points from it. The first $r$ points and last $r$ points have the same color, while the latter interpretation of ${\varphi }_j$ is exactly the former model's interpretation of ${\varphi }_j^{+}$. The auxiliary predecessor shifts too, so this remains true for sentences mentioning the lowest set predicate. Extend the chosen rank sequence arbitrarily to interpret the remaining sorts. This yields a model of typed set theory satisfying all the specified ambiguity instances.

Thus every finite subset of the typed axioms together with all ambiguity instances has a model. Applying multisorted compactness theorem givesThe final model satisfies first-order typed comprehension, with quantification over its own sorts and sets as in Henkin semantics.

Pull the given finite coloring $\chi :\mathbb{N}\to [k]$ back to a finite coloring of unordered pairs by puttingThe ordering $a<b$ makes this a well-defined function on the edges of a complete graph. By Ramsey's theorem, some infinite set $X=\{x_1<x_2<\cdots \}$ has all its pairs of one color $r$ under $\tilde{\chi }$. ConsequentlyPossible coincidences between these products cause no difficulty, since the original finite coloring assigns each positive integer a single color.

Write $\mathbb{N}=\{1,2,\dots \}$ and let $\chi$ be the given finite coloring of the edges of the complete graph on the positive integers. Construct an increasing sequence $(v_i)$ and nested infinite sets $A_0\supset A_1\supset \cdots$ as follows. Set $A_0=\mathbb{N}$. Having chosen $A_{i-1}$, take $v_i=\min A_{i-1}$, partition $A_{i-1}\setminus \{v_i\}$ by the color of $\{v_i,w\}$, and use the infinite pigeonhole principle to choose an infinite set $A_i$ on which this color is constant, say $r_i$. Every element of $A_i$ exceeds $v_i$.

The infinite pigeonhole principle applied again to $(r_i)$ gives an infinite set of indices $I$ and a color $r$ such that $r_i=r$ for all $i\in I$. For $i<j$ in $I$, nesting gives $v_j\in A_i$, so $\chi (\{v_i,v_j\})=r$. Thus the infinite set $X=\{v_i:i\in I\}$ is a monochromatic complete graph:This proves the required infinite case of Ramsey's theorem; finiteness of the color set is used at both applications of the infinite pigeonhole principle.

Ramsey's theorem for $r$-sets says that every finite coloring of $[\mathbb{N}{]}^{(r)}$ has an infinite monochromatic set. We prove it by mathematical induction on $r$. The case $r=1$ is the infinite pigeonhole principle. Suppose the result holds for $r-1$, and let $c:[\mathbb{N}{]}^{(r)}\to [k]$. Choose $a_1$, then use the induction hypothesis on the coloring $F\mapsto c(F\cup \{a_1\})$ to obtain an infinite set $M_1$ on which this color is constant, say $d_1$. Inductively chooseso that $c(F\cup \{a_j\})=d_j$ for every $F\in [M_j{]}^{(r-1)}$. Some color $d$ occurs for infinitely many $d_j$. If $j_1<j_2<\cdots$ are the corresponding indices, every $r$-set from $\{a_{j_1},a_{j_2},\dots \}$ has color $d$: take its least-indexed element $a_{j_s}$, after which its other $r-1$ elements lie in $M_{j_s}$. This proves the theorem.

Suppose the Finite Ramsey theorem failed for fixed positive integers $r,k,m$. For every $n$ choose a $k$-coloring $c_n:[n{]}^{(r)}\to [k]$ with no monochromatic $m$-set. There are only finitely many colorings of $[r{]}^{(r)}$, so an infinite subsequence of the $c_n$ agrees there. Pass to a further infinite subsequence agreeing on $[r+1{]}^{(r)}$, and continue. The diagonal argument produces compatible colorings $d_N:[N{]}^{(r)}\to [k]$ such that $d_{N+1}{|}_{[N{]}^{(r)}}=d_N$ and no $d_N$ has a monochromatic $m$-set.

Define $d(F)=d_N(F)$ whenever $N\ge \max F$. Compatibility makes this a well-defined finite coloring of $[\mathbb{N}{]}^{(r)}$. By Ramsey's theorem it has an infinite monochromatic set, whose first $m$ elements contradict the defining property of a sufficiently large $d_N$. This compactness argument proves the finite statement.

Let $c:\mathbb{N}\to [k]$ be the given finite coloring. Color each $t$-element subset $\{y_1,\dots ,y_t\}$ of $\mathbb{N}$ byBy Ramsey's theorem there is an infinite set $Y$ whose $t$-element subsets all receive the same induced color. Enumerate it increasingly as $x_1<x_2<\cdots$. Then every sum $x_{i_1}^2+\cdots +x_{i_t}^2$ with $i_1<\cdots <i_t$ has that color. The argument works for every positive integer $t$; primality is not needed for this part.