The completeness theorem of first-order logic

4.7. Countability and Completeness

Numbers of injections

For any finite sets X, Y, the set Inj(X,Y) of injections from X to Y is finite (subset of the finite set Y^X), and its cardinal only depends on those of X and Y. Let us denote

P(|Y|,|X|) = |Inj(X,Y)|

Then obviously ∀n∈ℕ,

P(n,0) = 1
P(n,1) = n
∀k∈ℕ, n<k ⇒ P(n,k) = 0

Now let A ⊂ X, B = X\A and π_A : Inj(X,Y) → Inj(A,Y) defined by f ↦ f_|A. Then

∀g∈ Inj(A,Y), π_A•(g) ≈ Inj(B, Y\Im g)

by the bijection f ↦ f_|B. Hence with |A|=i=|Im g|, |B|= k and |Y|= n+i,

∀n,k,i ∈ℕ, P(n+i, k+i) = P(n,k)⋅P(n+i, i)

In particular,

P(n+1, k+1) = P(n,k)⋅(n+1)
P(n+i, i+1) = n⋅P(n+i, i)

The last one can also be written

∀n,k ∈ℕ, P(n,k+1) = (n−k)⋅P(n,k)

by extension to n<k ⇒ P(n,k+1)= 0 = P(n,k).
Either formula, together with P(n,0) = 1 and the 0 value for n<k, forms a recursive definition of P(n,k). Either way, this implies coincidence with the definition of the falling factorial (among other possible notations and names) on a larger domain

∀n∈ℤ,∀k∈ℕ, P(n,k) = ∏_i<k (n−i)

where i<k abbreviates i∈V_k.

The above formulas keep validity for negative n except that P(n,k) ≠ 0. Namely, it is conveniently expressed there by the rising factorial

∀n∈ℤ,∀k∈ℕ, P⁺(n,k) = ∏_i<k (n+i) = P(n+k−1,k)
P(-n,k) = P⁺(n,k)⋅(-1)^k

The factorial comes as a particular case :

∀n∈ℕ, n! = |𝔖_n| = P(n,n) = P⁺(1,n) >0

The falling and rising factorials can be expressed from it : ∀n,k ∈ℕ,

n!⋅P⁺(n+1, k) = n!⋅P(n+k,k) = (n+k)!

Numbers of combinations

For any set X and any k∈ℕ, a k-combination of X is a subset A⊂X such that |A| = k. The number of k-combinations of V_n (and thus of any set of n elements) is denoted C(n,k). This easily implies by Inj(V_k,V_n)∋f ↦ Im f, the following formula which we extend as a definition of C on a larger domain:

∀n∈ℤ,∀k∈ℕ, k!⋅C(n,k) = P(n,k)

Indeed P(n,k) is still a multiple of k! for negative n, as it repeats its values from positive n for the same k. Namely, let us similarly define ∀n∈ℤ,∀k∈ℕ,

C⁺(n,k) = C(n+k−1,k)
C(-n,k) = C⁺(n,k)⋅(-1)^k

including C⁺(0,0) = C(0,0) = 1 from the latter formula.

As 0! = 1! = 1 we generally have C(n,0) = C⁺(n,0) = 1 and C(n,1) = C⁺(n,1) = n.
As A ↦ ∁_X A defines a bijective correspondence between k-combinations and n-combinations of V_n+k,

∀n,k∈ℕ, C(n+k,k) = C(n+k,n) = C⁺(n+1,k) = C⁺(k+1,n)

The first equality is more usually written k≤n ⇒ C(n,k) = C(n,n−k).

Given n=|X| and k,l∈ℕ, the number of possible data of disjoint A,B ⊂X with |A| = k and |B| = l, can be expressed in 3 ways according to which of A, B or A∪B is chosen first. This gives the following double equality (its validity for all n∈ℤ follows either by relating cases of negative n to those of positive n, or by deducing the general formula from numerical definitions after multiplication by k!⋅l!) :

∀n∈ℤ,∀k,l∈ℕ, C(n−k,l)⋅C(n,k) = C(n−l,k)⋅C(n,l) = C(k+l,l)⋅C(n,k+l)

In particular when l = 1

∀n∈ℤ,∀k∈ℕ, (n−k)⋅C(n,k) = n⋅C(n−1,k) = (k+1)⋅C(n,k+1)

From there follows

∀n∈ℤ,∀k∈ℕ*, C(n,k) = C(n−1,k−1) + C(n−1,k)

The simple proof for n,k∈ℕ*, is that a k-combination of V_n is either a k-combination of V_n−1, or ({n−1} ∪ a k−1-combination of V_n−1).
General proof :
k⋅C(n−1,k) = (n−k)⋅C(n−1,k−1) = k⋅(C(n,k) − C(n−1,k−1)) ∎
This proof has 2 variants, each failing to cover some cases due to multiplication by zero:

n⋅C(n−1,k) = (n−k)⋅C(n,k) = n⋅(C(n,k) − C(n−1,k−1))
(n−k)⋅C(n−1,k) = n⋅C(n−1,k) − k⋅C(n−1,k) = (n−k)⋅(C(n,k) − C(n−1,k−1))

The form of these versions together suggest an idea of possible failure on n=k=0 for an extension of C(n,k) to negative k. Two natural extensions may be considered. The one giving 0 for all negative k, preserves C(n,k) = C(n−1,k−1) + C(n−1,k) but breaks C(n,k) = C(n,n−k). Another, preserves the latter but breaks the former and switches the sign of the link between C(n+k−1,k) and C(-n,k) for negative k. We shall not use either.

The number of strictly monotone g:V_k → V_n is C(n,k).
Proof : g bijectively matches the k-combination Im g of V_n. ∎

The number of monotone f:V_k → V_n is C⁺(n,k).
Proof : f bijectively matches strictly monotone g:V_k → V_n+k−1 by g(x) = f(x) + x. ∎

Monotone f:V_k → V_n+1, monotone g:V_n → V_k+1, and k-combinations A of V_n+k, are so bijectively related by

∀x<k, ∀y<n, f(x) ≤ y ⇔ x < g(y) ⇔ x < |A∩V_x+y+1|

The number of h:V_n → ℕ such that ∑h = k is C⁺(n,k).
Proof : h bijectively matches monotone f:V_k → V_n by ∀x<n, h(x) = |f_•(x)|. ∎

The number of h:V_n → ℕ such that ∑h ≤ k is C⁺(n+1,k).
Proof : they are bijectively h = h'_{|V_n} for h':V_n+1 → ℕ such that ∑h' = k. ∎

The number of h:V_k → ℕ such that ∑h < n is C⁺(n,k).
Proof : h bijectively matches monotone f:V_k → V_n by ∀x<k, f(x) = ∑_i≤x h(i). ∎

Related wikipedia pages:

Binomial coefficient

Combination

Multiset

Countable sets

A set A is called countable if |A| ≤ |ℕ|, and countably infinite if |A| = |ℕ|.

The following are equivalent

A is either finite or countably infinite
A is countable
There exists a total order on A such that all <⃖(x) are finite.

Obviously 1. ⇒ 2. ⇒ 3.
Proof of 3. ⇒ 1.
It is easy to see that f = (A∋x ↦ |<⃖(x)|) is strictly monotone. Thus
∀x∈A, f[<⃖(x)] ⊂ <⃖(f(x)) = V_f(x), and (as f is injective) |f[<⃖(x)]| = |<⃖(x)| = f(x).
Thus by Dedekind-finiteness f[<⃖(x)] = V_f(x).
Finally, Im f ≠ ℕ ⇒ ∃y∈ ℕ\Im f, ∀x∈A, y∉ V_f(x) ∴ f(x) ≤ y hence A is finite.∎
Moreover with finite A, Im f = V_|A| for the same reason as with f[<⃖(x)].

Another way to prove 2. ⇒ 1. is as follows. Let A⊂ℕ, and assume A≠∅ (as ∅ is finite).
Let S_A the successor function (3.12) of A.
As we described the set of all finite subsets of ℕ in 4.6., if A is infinite then A ⊂ Dom S_A, thus |ℕ| ≤ |A|.∎

Any quotient of a countable set is countable. Namely, like already seen with finite sets, any surjection f: ℕ ↠ A has a section given by y↦ min f_•(y).

For the same reason, any product of nonempty subsets of ℕ is nonempty. However, to extend this to products of nonempty countable sets, depends on the existence of a family of choices of injections to ℕ. So, it works if a definition specifying such choices is available; otherwise, it fails since not using the axiom of choice was the point of the question.

Sums of total orders

The sum of a totally ordered family of totally ordered sets ((A_i, ≤_i)_i∈I with a total order on I), is their disjoint union U with the total order defined as follows. Assuming these sets pairwise disjoint to simplify notations as U = ⋃_i∈I A_i, let

≤_U = (⋃_i∈I ≤_i) ∪ (⋃_i<j∈I A_i×A_j)

If I and all A_i are well-ordered then their sum is also well-ordered.
Proof: for any nonempty B⊂U, the least element of B∩A_i for the least i such that B∩A_i ≠ ∅, is the least element of B.∎

However, the property of finiteness of all <⃖(x) is not always preserved. For example it holds in V_n ⊔ ℕ, but not in ℕ ⊔ V_n if n ≠ 0.
For any sequence a∈ℕ^ℕ, the sum U = ∐_n∈ℕ V_{a_n} satisfies

∀(n,x)∈U, |<⃖(n,x)| = x + ∑_k<n a_k

forming a bijection from U to V_{∑_n∈ℕ a_n} if a has finite support, or to ℕ otherwise.
If a is a constant function with value b∈ℕ*, this gives the bijection from ℕ×V_b to ℕ

∀(q,r)∈ℕ×V_b, |<⃖(q,r)| = b⋅q + r

whose inverse is called the Euclidean division by b.

More generally, any set of the form U = ⋃_n∈ℕ A_n where A_n is finite with an invariant (specifiable) total order (thus specified injection to V_{|A_n|}) is countable.

x\y	0	1	2	3	4
0	0	1	4	9	16
1	2	3	5	10	17
2	6	7	8	11	18
3	12	13	14	15	19

Proof. The union is a quotient of the disjoint union, which is countable.∎

In particular, ℕ² is countable because it is ⋃_n∈ℕ V_n².
This gives a bijection from ℕ² to ℕ defined by (x,y) ↦ (x < y ? y⋅y + x : x⋅x + x + y).
Hence the countability of any U = ⋃_n∈ℕ A_n where each A_n is countable with an invariant injection to ℕ (as they define a surjection from a subset of ℕ² to U).

Another bijection, with a generalization to ℕ^k for any k > 0, will be introduced next.

Lexicographic order

A lexicographically ordered set is a set of functions F ⊂ Y^X where X, Y are totally ordered sets, such that for all f≠g ∈F, the set {x∈X| f(x) ≠ g(x)} has a greatest element m_f,g : this lets a total order on F be defined by f<g ⇔ f(m_f,g) < g(m_f,g).

(Another definition uses the least element, which amounts to reversing the order of X, but that is less useful)

The positional numeral systems, of which the decimal system is a particular case, come depending on a choice of base b > 1, as the use of the lexicographically ordered set V_b^(ℕ).
Indeed

∀g∈V_b^(ℕ),∀x∈ℕ, {f∈V_b^(ℕ)|f<g ∧ m_f,g = x} ≈ V_b^x×V_g(x)

from which comes

|{f∈V_b^(ℕ)|f<g}| = ∑_x∈ℕ b^x ⋅ g(x)

hence V_b^(ℕ) is countably infinite. The particular case V₂^(ℕ) is identifiable with ℘_fin(ℕ),

∀B∈℘_fin(ℕ),|{A∈℘_fin(ℕ)|A<B}| = ∑_x∈B 2^x

and this bijection from ℕ to ℘_fin(ℕ) is a curried form of the BIT predicate.

Now for k > 0 let F_k ⊂ ℕ^k the set of monotone k-tuples of natural numbers. For all g∈F_k and x<k the set {f∈F_k|f<g ∧ m_f,g = x} is in bijection (by restriction) with the set of monotone f:V_x+1 → V_g(x), with cardinal C⁺(g(x),x+1) = C(g(x)+x,x+1). So

|{f∈F_k|f<g}| = ∑_x<k C⁺(g(x),x+1)

hence F_k is countably infinite. Finally, ℕ^k is also countably infinite as h∈ℕ^k bijectively defines f∈F_k by

∀x<k, f(x) = ∑_i≤x h(i).

x\y	0	1	2	3	4
0	0	1	3	6	10
1	2	4	7	11
2	5	8	12
3	9	13

In particular with k=2 we get b₂ : ℕ×ℕ → ℕ defined as b₂(x,y) = T_x+y+x where the triangular numbers are defined by T_n = C⁺(n,2) or equivalently by T₀ = 0 ∧ ∀n∈ℕ, T_n+1 = T_n+n+1.
This gives ℕ² the ground (0,S)-term algebra structure (recursively defining b₂^-1(z))

0_ℕ×ℕ = (0,0)
S(x,0) = (0,Sx)
S(x,Sy) = (Sx,y)

Another way to express b₂^-1 (to infer x and y from b₂(x,y)) first calculates x+y as the only n such that T_n ≤ b₂(x,y) < T_Sn.

Countable term algebras

Constructing ground term algebras over any language, suffices to also give term algebras with any set of variables, by treating those variables as more constants in the language. This construction remains to be specified for any language L₀∪L₁ where the set L₀ of constants and the set L₁ of other symbols (with nonzero arity), are both non-empty (beyond the one-one case for which it gave ℕ).
Obviously, such ground term algebras are also Dedekind-infinite. We shall prove their existence, starting with cases of countable L, for which they are also countably infinite.
Indeed we can define a ground term L₀∪L₁-algebra structure on ℕ as follows.
^L₀∪L₁ℕ = L₀∪ ^L₁ℕ, while L₀ is countable, and since L₁ is countable,

^L₁ℕ = ∐_s∈L₁ ℕ^n_s ≈ L₁×ℕ ≈ ℕ

By such a bijection of a natural kind as above defined, any (s,u) ∈ ^L₁ℕ matches an n ∈ ℕ such that Im u ⊂ V_n+1.
But, after composing this with a natural bijection of L₀∪ ℕ with ℕ starting with a constant (from L₀), to give ℕ a bijective L₀∪L₁-structure, any (s,u) ∈ ^L₁ℕ finally matches there an n ∈ ℕ such that Im u ⊂ V_n.
In other words, the relation R from 4.1 is included in the strict order of ℕ and is thus well-founded. This gives ℕ a role of ground L₀∪L₁-term algebra. ∎

Generally, one may form a ground term L-algebra out of an injective L-algebra, by taking its minimal subalgebra.

Aside this construction of countable term algebras based on the axiom of infinity, another idea to form injective algebras starts by assuming the existence of a model U of a very simple kind of set theory. Namely, the excerpt of set theory which describes tuples, makes U an injective algebra over the language made of the tuples definers. From this, term algebras come as subalgebras generated by sets whose elements are not tuples. Beyond the mere tuples definers, for any language L of algebraic symbols which belong to U, the set U is an injective L-algebra. If U is standard, this takes the simple form ^LU ⊂ U, making (U, Id_^LU) an injective L-algebra.

The below needs rework

The Completeness Theorem

Let us finally prove the Completeness theorem previously commented in 1.1, 1.9, 1.10, 1.B and 1.C.

Theorem. First-order logic has a proving system both sound and complete in the following equivalent senses

Any consistent first-order theory T with countable language has a countable model.
Any formula true in all countable models of such a theory is deducible from its axioms.

Here is a rather short sketch of proof of the first statement (implicitly suggesting a possible form of a proving system, ignoring efficiency issues) - an even shorter version is contained in that other page.

Reduce T to a single-type theory T₁ simulating the use of types by relevant unary predicate symbols and axioms, but without requiring all objects to belong to some type. Thus, adding the axiom (∃x, 1) to T₁ brings no contradiction (it does not prevent empty types).
This allows to re-write axioms of T₁ in prenex form, that is chains of quantifiers applied to quantifier-free formulas, by equivalences adapted from 2.5 and 2.7, such as

(C ∧ ∃_Ax, R(x)) ⇔ (∃x, C ∧ A(x) ∧ R(x))
(C ∨ ∃_Ax, B(x)) ⇔ (∃x, C ∨ (A(x) ∧ B(x)))
(C ∧ ∀_Ax, B(x)) ⇔ (∀x, C ∧ (A(x) ⇒ B(x)))

Interpret the symbol = in axioms as an ordinary predicate symbol, with the axioms of equality [1.11] (those only using the structures named in the language, which suffice, are already in prenex form and do not use ∃).

Replace each occurrence of ∃ in an axiom by the use of an additional operator symbol K, for example

(∀x,x', ∃y, ∀z, A(x,x',y,z)) ⇢ (∀x,x',∀z, A(x,x',K(x,x'), z)).

(The axioms of equality over these are not needed)
Let M be the ground term algebra over the language of operator symbols enriched as just described.
Read the chain of ∀ in each axiom as declaring an axiom schema: one copy of the axiom per possible replacement of its variables by a tuple of elements of M. This gives a propositional theory T₂ (its axioms are composed of logical connectives over Boolean variables), whose set of Boolean variables is ^PM where P is the set of predicate symbols in the theory plus the "equality" symbol.
Observe that T₂ is still consistent, as any contradiction in T₂ (= finite set of axioms not satisfied by any tuple of Boolean values of their variables) would provide a contradiction in T₁, as follows:

From all sub-terms occurring in used axioms of T₂, list without repetition all those whose root is a symbol S which was added to the language when converting some axiom, for example ∀x,∃y,∀z,∃u, A(x,y,z,u) of T₁, into the axiom ∀x,z, A(x,K(x),z, S(x,z)) in T₂.
Successively replace these symbols by variables in T₁: for terms S(t₀,t₁) and S(t₀,t₂) (where t₀, t₁, t₂ ∈ M), the reasoning in T₁ will say
"Let y such that ∀z,∃u, A(t₀,y,z,u);
let u₁ such that A(t₀,y,t₁,u₁);
let u₂ such that A(t₀,y,t₂,u₂);"
then use the contradiction in T₂ where y replaces K(t₀), u₁ replaces S(t₀,t₁) and u₂ replaces S(t₀,t₂).

Recursively by a chosen bijection from ℕ to ^PM (or anyway a ground functional {0,S}-draft structure on ^PM), add one by one to the axioms, each of these Boolean variables if it is consistent with previously accepted axioms, so that all get values without contradiction.
Then a model is formed by quotienting M by the "equality" relation, and removing the elements with no type. ∎

As this construction depends on multiple choices (between equivalent formalizations of a theory, possible bijections from ℕ to ^PM, and variations are possible on the method itself), it can give different models, which may be non-isomorphic and even have different truth values of the theory's undecidable statements.

Deduction of the second statement from the first :

T⊢ A
⇔ T ∪ {¬A} is inconsistent
⇔ T ∪ {¬A} has no countable model
⇔ A is true in all countable models of T. ∎

Skolem's Paradox

Taking the completeness theorem with Cantor's theorem gives this paradox :
Any consistent set theory has a countable model U, including a set P defined by its term "℘(ℕ)" it sees as uncountable.

P being a part of U is meta-countable, but it is interpreted as uncountable, thus any of its meta-bijections with ℕ "does not exist" (has no representative) as an element of U.

Being meta-countable, P must differ from the meta-interpretation of ℘(ℕ) which is meta-uncountable: P contains all objects in U which play roles of subsets of ℕ, but not all meta-subsets of ℕ are so represented.

Indeed, when the powerset axiom declares the class of subsets of ℕ to be a set ℘(ℕ), it only determines ℘(ℕ) as given by the range of subsets of ℕ which happen to exist (be represented) in the current universe, thus remains relative to this universe.

In the other interpretation of what it means for a class to be a set ("strong" version in "Classes and sets in expanding universes"), this would mean that our universe already contains "really all" subsets of ℕ, forming the supposedly "true" set ℘(ℕ) which will stay fixed in any further expansion of the universe. However, this wish cannot be expressed in first-order logic nor any other conceivable mathematical formalism: there is no way to talk about meta-sets that "cannot be found" in this universe but would «exist elsewhere» (in any mysterious bigger universe) to exclude them from there. (Only one aspect of this idea can be formalized as the axiom schema of specification).

Set theory and foundations of mathematics
1. First foundations of mathematics
2. Set theory
3. Algebra 1
4. Arithmetic and first-order foundations

4.1. Algebraic terms
4.2. Quotient systems
4.3. Term algebras
4.4. Integers and recursion
4.5. Presburger Arithmetic
4.6. Finiteness
4.7. Countability and Completeness
4.8. More recursion tools
4.9. Non-standard models of Arithmetic
4.10. Developing theories : definitions
4.11. Constructions
4.A. The Berry paradox

5. Second-order foundations
6. Foundations of Geometry