1 Lecture 08 - University of Pennsylvaniaweinstei/006Memoirs.pdf · PHIL 006/506, Fall, 2016 Scott Weinstein 1 1 Lecture 08.30 We began with a survey of some of the highlights of

PHIL 006/506, Fall, 2016 Scott Weinstein 1

1 Lecture 08.30

We began with a survey of some of the highlights of logic, mathematics, andphilosophy from Pythagoras, Plato, Diophantus, and Euclid, to the Godel Com-pleteness Theorem, the Church-Turing Theorem (which together indicate thescope and limits of the execution of Leibniz’s dream of a charcteristica uni-versalis and a calculus ratiocinator), Tarski’s proof that the theory of the realfield is decidable, and the “DPRM” Theorem that the solvability of Diophantineequations in integers is algorithmically undecidable, thereby resolving Hilbert’sTenth Problem negatively and completing a mathematical arc from antiquity:geometry is decidable, while arithmetic is not. (We observed, as an aside, thatPlato’s awe-struck response to the Pythagoreans’ demonstration that the Dio-phantine equation p2 − 2q2 = 0 has no solution in positive integers is the originof Western philosophy.) In a future edition of this memoir, I may include furtherdetail concerning these matters, but for the nonce, I want to get down to brasstacks and review the concrete problem of axiomatizability we started to engagetoward the end of class.

Let N = 〈N, s, 0〉, the structure consisting of the non-negative integers Ntogether with the successor function s and the distinguished element 0. Westarted to provide an axiomatization of the first-order theory of this structure.What does this mean?

Definition 1 Let A be a relational structure (for example N). If θ is a first-order sentence, we write A |= θ for θ is true in A, alternatively, A satisfies θ. IfΣ is a set of first-order sentences we write A |= Σ if and only if A |= θ, for everyθ ∈ Σ. Th(A) = {θ | A |= θ}. We say two structures A and B are elementarilyequivalent (and write A ≡ B) if and only if Th(A) = Th(B). We say that a setof sentences (it could be a singleton) axiomatizes (the theory of) a structure Aif and only if A |= Σ and for every structure B, if B |= Σ, then B ≡ A. The ideais that no better description of A can be given by first-order sentences than isalready given by Σ.

We wrote down a few sentences that describe basic properties of the structureN.

1. (∀x)(sx 6= 0) (0 has no predecessor);

2. (∀x)(x 6= 0→ (∃y)sy = x) (everything but 0 has a predecessor);

3. (∀x)(∀y)(sx = sy → x = y) (the successor function is injective)

Let’s call the conjunction of these sentences ϕ. We noted that if A |= ϕ thenA has a substructure (perhaps all of A) that is isomorphic to N; in particularevery such structure is infinite. Indeed, ϕ essentially says that the universe ofany satisfying structure is Dedekind infinite: it supports a function which isinjective but not surjective. It took a (mere?) moment to recognize that ϕdoes not axiomatize N. For example, the structure which consists of N togetherwith an additional element a with s(a) = a satisfies ϕ, so we realized we’d


need to rule out the existence of fixed points under successor. We added thesentence α1 : (∀x)sx 6= x. And it didn’t take long to realize that we’d need toadd infinitely many sentences αn the nth of which rules out successor cycles oflength n. If we write sn for n iterations of s we may write αn as (∀x)snx 6= x.

Let Σ = {ϕ} ∪ {αn | n ≥ 1}. We ended with the following questions:

• Does Σ axiomatize N?

• Is something even stronger true, namely, does Σ categorize N? (We say aset of sentences Σ categorizes a structure A just in case for every structureB, B |= Σ if and only if B is isomorphic to A.)

We’ll pursue these questions at the start of class on Thursday.

2 Lecture 09.01

We’d mentioned “isomorphism” last time, but hadn’t defined it.

Definition 2 Suppose A = 〈A,EA, sA, 0A〉 and B = 〈B,EB , sB , 0B〉. A func-tion f is an isomorphism from A onto B if and only f is a bijection of A ontoB and

• f(0A) = 0B;

• for all a ∈ A, f(sA(a)) = sB(f(a));

• for all a1, a2 ∈ A, 〈a1, a2〉 ∈ EA if and only if 〈f(a1), f(a2)〉 ∈ EB.

We say A is isomorphic to B (and write A ∼= B) if and only if there is anisomorphism from A onto B.

We attacked the questions we’d asked at the end of the last class meeting.In particular, we began by showing that

Lemma 1 Σ does not categorize N, that is, there is a structure A such thatA |= Σ but A 6∼= N.

Proof: Let Z=〈Z, sZ〉 be the structure consisting of all the integers with theirusual successor function. For any set I, let ZI be the disjoint union of I manycopies of Z, that is, the universe of ZI is the cartesian product I × Z, andsZI (〈i, p〉) = 〈i, p+ 1〉. Finally, let AI be the union of the structures N and ZI .We observed that for every set I, AI |= Σ. Moreover, for all I and J , AI ∼= AJif and only if there is a bijection from I onto J (that is, the cardinality of Iequals the cardinality of J). It follows at once that Σ does not categorize N.

That leaves the question, “Does Σ axiomatize N?” (Recall that a set ofsentences Γ axiomatizes a structure C if and only if for every structure B, B |=Γ if and only if B ≡ C.) We made a start on answering this question byestablishing:

Lemma 2 For every structure B, B |= Σ if and only if for some set I, B ∼= AI .


Proof: As just noted in the proof of Lemma 1, for every I, AI |= Σ. Thus, itsuffices to show that for every B, if B |= Σ, then B ∼= AI for some I.

Suppose B |= Σ. Define an equivalence relation ≈ on the universe of B asfollows. For all a, b ∈ B

a ≈ b if and only if for some n, (sB)n(a) = b or (sB)n(b) = a.

That is, a ≈ b just in case one of a or b is reachable from the other by finitelymany applications (perhaps 0) of the successor function in B. We argued that≈ factors B into equivalence classes of exactly two isomorphism types: theequivalence class of 0B (which we called the standard part of B) is isomorphicto N, and is the unique such equivalence class; all the other equivalence classes(if any) are isomorphic to Z.

Next, we introduced one of the fundamental concepts of logic, the notion oflogical consequence.

Definition 3 Let Γ be a set of first-order sentences and θ a first-order sentence.θ is a logical consequence of Γ (written Γ |= θ) if and only if for every structureB, if B |= Γ, then B |= θ. Cn(Γ) = {θ | Γ |= θ}.

We gave an alternative characterization of axiomatizability in terms of conse-quence .

Lemma 3 A set of sentences Γ axiomatizes a structure A if and only if Cn(Γ) =Th(A).

(Make sure you can prove this; ask about it next time, if you can’t.)Next time, we will establish

Proposition 1 Σ axiomatizes N.

as a corollary to Lemmas 2 and 3, and the following fundamental result con-cerning first-order logic.

Theorem 1 (Lowenheim-Skolem) If Γ is a countable set of first-order sen-tences such that B |= Γ for some infinite structure B, then for every infinite setC there is a structure C with universe C such that C |= Γ. In other words, if acountable set of first-order sentences is satisfied by an infinite model, then it issatisfied by a model of cardinality κ for every infinite cardinal κ.

It follows immediately from Lemma 1 and Proposition 1 (why?) that

Corollary 1 Th(N) does not categorize N (and thus there is no set of first-ordersentences that categorizes N).

We observed that there are sentences of other logical languages that docategorize N. Recall the basic properties of successor we wrote down in our firstclass meeting:

1. (∀x)(sx 6= 0) (0 has no predecessor);


2. (∀x)(x 6= 0→ (∃y)sy = x) (everything but 0 has a predecessor);

3. (∀x)(∀y)(sx = sy → x = y) (the successor function is injective)

Let’s call the conjunction of these sentences ϕ. The conjunction of ϕ with thefollowing sentence of the infinitary language Lω1ω categorizes N:

(∀x)(x = 0 ∨ x = s0 ∨ x = ss0 ∨ . . . ∨ x = sn0 ∨ . . .).

Also, ϕ conjoined with the following second-order axiom of induction categorizesN:

(∀P )[(P (0) ∧ (∀x)(P (x)→ P (sx)))→ (∀x)P (x)].

3 Lecture 09.06

We proceeded with the proof of Proposition 1. The next definition introducessome of the important concepts we will use in the proof.

Definition 4 Mod(Γ) = {B | B |= Γ}. Γ is satisfiable if and only if Mod(Γ) 6=∅. Γ is complete if and only if for all structures B,C ∈ Mod(Γ), B ≡ C. Γ isκ-categorical if and only if for all structures B,C ∈ Mod(Γ), if |B| = |C| = κ,then B ∼= C.

Since Σ ⊆ Th(N), in order to establish Proposition 1, it suffices to show that

Lemma 4 Σ is complete.

We prepared ourselves for the proof of Lemma 4 by studying a little set theory.

Set Theory Supplement

For further background consult Chapters 1-5 of Jech, Set Theory, available onCanvas.

Let X and Y be sets. X is equipollent to Y if and only if there is a bijectionfrom X onto Y. We write X ∼ Y for X is equipollent to Y.

X � Y, if and only if, there is an injection from X into Y.

Theorem 2 (Cantor-Schroeder-Bernstein) If X � Y and Y � X, thenX ∼ Y.

We write X ≺ Y if and only if X � Y and X 6∼ Y . Let N be the set ofnatural numbers (aka non-negative integers), Q be the set of rational numbers(ratios of integers) and R be the set of real numbers. Let X × Y = {〈a, b〉 |a ∈ X and b ∈ Y }, the cartesian product of X and Y . It is easy to see that forevery set X, X � X ×X. Define f : N×N 7→ N, by f(m,n) = 2m · 3n. Since fis injective, N× N ∼ N, by Theorem 2. It follows that Q ∼ N.

Let X2 = {f | f : X 7→ {0, 1}}, the set of 0-1 valued functions with domainX.


Theorem 3 (Cantor Diagonal Theorem) X ≺ X2.

Proof: Let X be a set and let G : X −→ X2. It suffices to show that G is notsurjective. Define f : X −→ {0, 1} as follows: for every a ∈ X,

f(a) = 1− [G(a)](a).

It is easy to see that for all b ∈ X,G(b) 6= f.We write P(X) for the powerset of X, that is, {Y | Y ⊆ X}. Observe that

P(X) ∼ X2.

Theorem 4 R ∼ N2.

Proof: For r ∈ R define D(r) = {q ∈ Q | q < r} (the left, open Dedekind cutin Q determined by r). It follows at once from the fact that for all r, s ∈ Rwith r < s, there is a q ∈ Q such that r < q < s, that D is an injection fromR into P(Q). Hence, R � N2. On the other hand, for each f ∈ N{0, 2} defineC(f) =

∑∞i=0 f(i) · 3i+1. It follows from the least upper bound principle (which

allows us to show that the summation of an infinite series is well-defined) thatC is an injection. Thus, N2 � R.

In light of the fact that N ≺ R the following question is quite natural. Is therean infinite set Y ⊂ R such that Y 6∼ N and Y 6∼ R? Cantor conjectured that theanswer to this question is negative (Cantor’s Continuum Hypothesis - CH) andHilbert placed the question first on the famous list of problems he promulgated inhis address to the 1900 International Congress of Mathematicians. The questionis yet to be resolved. Indeed, the solution will require new insights into thenature of sets, since the current definitive axiomatization of set theory, Zermelo-Fraenkel set theory with the axiom of choice (ZFC), fails to settle the question,as the following results show.

Theorem (Godel, 1939): CH is not refutable in ZFC.Theorem (Cohen, 1963): CH is not provable in ZFC.

Ordinals and Cardinals

Let X be a set and R be a binary relation on X, that is, R ⊆ X × X. R isa well-ordering of X, if and only if, R is transitive and asymmetric, and everynonempty subset of X has an R-least element.

Ordinal numbers provide representatives for well-orderings. Suppose X is aset of sets. We say X is a transitive set, if and only if, for every y ∈ X, y ⊆ X. Aset of sets X is an (von Neumann) ordinal number, if and only if, X is transitiveand X is well-ordered by ∈ (restricted to X). The (proper class of) ordinals isitself, in the obvious sense, well-ordered by ∈ . Every ordinal is equal to its setof predecessors. The successor of an ordinal α is α∪{α}; the supremum of a setof ordinals is its union. For every well-ordering 〈X,R〉 there is a unique ordinalnumber α such that 〈X,R〉 is isomorphic to 〈α,∈〉.

Some ordinals: 0, 1, 2, . . . , ω, ω + 1, . . . , ω + ω, . . . , ω × ω, . . .Zermelo’s Well-Ordering Principle (a version of the Axiom of Choice): For

every set X, there is an ordinal α such that X ∼ α.


We call the least ordinal number α such that X ∼ α the cardinality of X(and write |X| = α). An ordinal number α is a cardinal number, if and only if,|α| = α. We define addition, multiplication, and exponentiation of cardinals asfollows. Let κ and λ be cardinals.

• κ+ λ = |({0} × κ) ∪ ({1} × λ)|.

• κ · λ = |κ× λ|.

• κλ = |λκ|.

Theorem 5 For all infinite cardinals κ and cardinals λ,

κ+ λ = κ · λ = max{κ, λ}.

By Cantor’s Diagonal Theorem, for every cardinal number κ, κ < 2κ. Theinfinite cardinals form a well-ordered class. For each ordinal α, we denote theαth infinite cardinal by ℵα. Thus, ℵ0 = ω = {0, 1, 2, . . .} is the smallest infinitecardinal. We say a set X is countable if and only |X| ≤ ℵ0. Note that ℵ1,the first uncountable cardinal, is the set of countable ordinals. We can restateCantor’s Continuum Hypothesis (CH) as 2ℵ0 = ℵ1.


4 Lecture 09.08

We proceeded with the proof of Lemma 4. It is a corollary to the next tworesults. The first of these, a corollary to Theorem 1 (the Lowenheim-SkolemTheorem), provides a useful completeness criterion.

Corollary 2 ( Los-Vaught) If Γ is a countable set of first-order sentenceswhich has no finite models and which is κ-categorical for some infinite cardinalκ, then Γ is complete.

Proof: Let Γ be a countable set of first-order sentences which has no finitemodels and let κ be an infinite cardinal such that Γ is κ-categorical. Supposemoreover, toward a contradiction, that Γ is not complete. Then there are infinitestructures A,B ∈ Mod(Γ) such that A 6≡ B. It follows from Theorem 1 thatthere are structures C and D such that |C| = |D| = κ, C ≡ A, and D ≡ B. Butthen, C ∼= D and C 6≡ D, which contradicts the fact that isomorphic structuressatisfy the same first-order sentences..

Lemma 5 For every uncountable cardinal κ, Σ is κ-categorical.

Proof: Recall that the universe of AI (call it UI) is equal to {0, 1, 2, . . .} ∪ (I ×{. . .− 1, 0, 1 . . .}). By Theorem 5, we thus have

|UI | = ℵ0 + max{|I|,ℵ0}.

It follows at once that |UI | = ℵ0, if |I| ≤ ℵ0, and |UI | = |I|, if |I| > ℵ0. Recallfrom the proof of Lemma 1 that for all I and J , AI ∼= AJ if and only if |I| = |J |,and from Lemma 2 that for every structure B, B |= Σ if and only if for some setI, B ∼= AI . It follows at once that there are countably many countable models ofΣ up to isomorphism, and for every uncountable cardinal κ exactly one modelof Σ of cardinality κ up to isomorphism.

This concludes the proof that Σ axiomatizes N. We noted that a sea-changein the study of model theory took place with Michael Morley’s 1963 result that

Theorem 6 If Γ is a countable set of first-order sentences and Γ is κ-categoricalfor some κ > ℵ0, then Γ is κ-categorical for every κ > ℵ0.

The material below was not covered in Lecture 4 and serves as a preview ofthings to come. The questions posed may be suitable Bring-Back Examinationproblems.

Let δ (theory of dense linear order without end points) be the conjunctionof the following sentences.

• (∀x)x 6< x (< is irreflexive)

• (∀x)(∀y)(∀z)((x < y ∧ y < z)→ x < z) (< is transitive)

• (∀x)(∀y)(x 6= y → (x < y ∨ y < x)) (< is linear)


• (∀x)(∀y)(x < y → (∃z)(x < z ∧ z < y)) (< is dense)

• (∀x)(∃y)x < y (< has no maxima)

• (∀x)(∃y)y < x (< has no minima)

Try to prove the following result (due to Georg Cantor).

Theorem 7 δ is ℵ0-categorical.


5 Lecture 09.13

The following Compactness Theorem expresses a fundamental property of first-order logic.

Theorem 8 (Godel-Mal’cev) If every finite subset of a set of first-order sen-tences Γ is satisfiable, then Γ itself is satisfiable. Equivalently, if a first-ordersentence θ is a logical consequence of a set of first-order sentences Γ, then θ isa logical consequence of a finite subset of Γ.

Definition 5 The signature (also called language) of a structure A (writtenL(A)) is the collection of relation, function, and constant symbols interpreted byA. If L(A) ⊆ L(B) and B�L(A) = A we say that A is a reduct of B, alternatively,B is an expansion of A.

The proof of the following proposition illustrates the use of the CompactnessTheorem in conjunction with the expansion of a signature by the addition ofnew constant symbols.

Proposition 2 If Γ is a set of first-order sentences and Γ has arbitrarily largefinite models, then Γ has an infinite model.

Proof: Let L be the set of relation, function, and constant symbols appearingin Γ. Introduce an infinite set of constant symbols {ci | i ∈ N} disjoint fromL and let Λ = {ci 6= cj | i < j}. Suppose Γ has arbitrarily large finite models.Then every finite subset of Γ ∪ Λ is satisfiable. Therefore, by the CompactnessTheorem, there is a structure B with B |= Γ ∪ Λ. It follows at once that theuniverse of B is infinite. Let A be the reduct of B to L. Then A is an infinitemodel of Γ.

We observed that essentially the same argument can be used to establish an“upward” version of the Lowenheim-Skolem Theorem.

Theorem 9 If Γ has an infinite model, then for every infinite cardinal κ, Γ hasa model of cardinality at least κ.

Proof: Repeat the proof of Proposition 2 with a set of new constant symbols{cβ | β < κ} of cardinality κ.

Definition 6 A structure A is finitely axiomatizable if and only if there is asentence θ which axiomatizes A.

We remarked that the Compactness Theorem can be applied to show that

Proposition 3 N is not finitely axiomatizable.

Proof: Forthcoming as a solution to a Bring-Back Exam problem.We next introduced the notion of first-order definability.


Definition 7 Let θ be a first-order formula with free variables among v1, . . . , vn,and let a1, . . . , an be members of the universe of a structure A. We write A |=θ[a1, . . . , an] for A |= θ[s], where s(vi) = ai for 1 ≤ i ≤ n. The set {〈a1, . . . , an〉 |A |= θ[a1, . . . , an]} is the n-ary relation defined by θ on A; we denote thisrelation by θ[A]. We say that a relation R ⊆ (UA)n is definable on A if and onlyif R = θ[A] for some first-order formula θ.

Definition 8 If f is an isomorphism from A onto itself, we say f is an au-tomorphism of A, and we write Aut(A) for the set of automorphisms of A. Astructure A is rigid if and only if |Aut(A)| = 1, that is, the only automorphismof A is the identity map.

The following result, a corollary to the “isomorphism preservation property,”establishes an important connection between automorphisms and definability.Recall that f [X] = {f(a) | a ∈ X}.

Corollary 3 If A is a structure, X ⊆ UA is definable, and f ∈ Aut(A), thenf [X] = X.

Therefore, in order to show that a set X is not definable on a given structureA, it suffice to exhibit f ∈ Aut(A) with f [X] 6= X. On finite structures, thistechnique is of universal applicability.

Proposition 4 If A is a finite structure, X ⊆ UA, and for every f ∈ Aut(A),f [X] = X, then X is definable on A.

Proof: Forthcoming as a solution to a Bring-Back Exam problem.Alas, this technique is not generally applicable on infinite structures. For

example, consider N, an infinite structure of countable signature. Since theidentity map fixes all subsets, the technique cannot be used to show the unde-finability of any subset of N. But there are sets of positive integers that are notdefinable on N. This follows by a “cardinality argument”: there are uncount-ably many sets of positive integers, but only countably many sets definable onN. The following proposition characterizes those sets which are definable on N.

Proposition 5 If X ⊆ {0, 1, 2, . . .} is definable on N, then either X or itscomplement is finite.

Proof: Forthcoming as a solution to a Bring-Back Exam problem.


6 Lecture 09.15

Recall that the theory of dense linear order was discussed in class last Thursdayand that an account of that discussion is included in the memoir of Lecture 4.We formulated the theory as a finite conjunction of first-order sentences δ andstated the following theorem, which we now prove.

Theorem 10 (Cantor) δ is ℵ0-categorical.

Theorem 10 is an immediate corollary to the next two results, which involvethe notion of partially isomorphic structures. A function f is a partial isomor-phism from A to B if and only if f is an isomorphism from a substructure of Aonto a substructure of B. The next definition introduces the key notion in theproof of Theorem 10.

Definition 9 A is partially isomorphic to B (A ∼=p B) if and only if there is anonempty set P of partial isomorphisms of A to B with the following properties:

Forth Property: for every a ∈ UA and f ∈ P, there is a g ∈ P such thatf ⊆ g and a ∈ dom(g).

Back Property: for every b ∈ UB and f ∈ P, there is a g ∈ P such thatf ⊆ g and b ∈ ran(g).

We write P : A ∼=p B, if P witnesses that A ∼=p B.

Theorem 10 is a corollary of the following two results.

Theorem 11 If A and B are dense strict linear orders without endpoints, thenA ∼=p B.

Proof: Let A and B be dense strict linear orders without endpoints, and let Pbe the set of finite partial isomorphisms from A to B. It is easy to verify that Pwitnesses A ∼=p B.

Theorem 12 If A and B are countable and A ∼=p B, then A ∼= B.

Proof: Suppose that A and B are countable and that P : A ∼=p B. LetUA = {a0, a1, . . .} and UA = {b0, b1, . . .}. We construct a sequence of partialisomorphisms f0, f1, . . . such that for all i ∈ ω

1. fi ∈ P

2. fi ⊆ fi+1

3. ai ∈ f2i and bi ∈ f2i+1.

The construction is by induction, alternately using the forth property and theback property at even and odd stages, respectively. Let f =

⋃i∈ω fi. It is easy

to verify that f is an isomorphism of A onto B.The following result is an immediate corollary of Theorem 10 via the Los-

Vaught Theorem.


Corollary 4 δ is complete.

Let R be the structure consisting of the collection of real numbers with itsusual ordering and let Q be the structure consisting of the collection of rationalnumbers with its usual ordering. It is follows immediately from Corollary 4 thatR ≡ Q, and therefore that the least upper-bound principle cannot be expressedin first-order logic.


7 Lecture 09.20

Recall the Compactness Theorem.

Theorem 13 (Satisfaction Version) If every finite subset of a set of sentencesΣ is satisfiable, then Σ itself is satisfiable.

(Consequence Version) If Σ |= ϕ, then for some finite ∆ ⊆ Σ, ∆ |= ϕ.

We showed in class, via an application of the Compactness Theorem, that

Proposition 6 There is an infinite structure A such that A |= Th(F).

Proof: Let c0, c1, . . . be an infinite list of constant symbols and let Γ = {ci <cj | i < j}. Let Σ = Th(F) ∪ Γ. Observe that every finite subset of Σ canbe satisfied in an expansion of a large enough structure A ∈ F. Hence, by theCompactness Theorem, there is a structure B∗, with B∗ |= Σ. The reduct Bof B∗ to the language of order is an infinite model of Th(F). Observe that Bcontains a substructure order-isomorphic to 〈ω,<〉 whose universe is the set{cB∗

i | i ∈ ω}. (We called attention to the relevance of this observation to thesolution to exam problem 5(b).)

Next, we discussed a potential axiomatization ϕ of Th(F), namely, the con-junction of the following sentences.

• (∀x)x 6< x (< is irreflexive)

• (∀x)(∀y)(∀z)((x < y ∧ y < z)→ x < z) (< is transitive)

• (∀x)(∀y)(x 6= y → (x < y ∨ y < x)) (< is linear)

• (∀x)((∃y)(x < y)→ (∃y)(x < y∧ (∀z)¬(x < z∧ z < y)) (< has immediatesuccessors)

• (∀x)((∃y)(y < x)→ (∃y)(y < x∧ (∀z)¬(y < z∧ z < x)) (< has immediatepredecessors)

• ((∃y)(∀x)¬(y < x) (< has a maximum)

• (∃y)(∀x)¬(x < y) (< has a minimum)

We undertook an analysis of the infinite models of ϕ and argued that theyconsist of an ordered sum of a copy of 〈ω,<〉 followed by an ordered productO ⊗ 〈Z, <〉 of an arbitrary linear ordering O with an ordered Z-chain, followedby a copy of 〈ω,<〉∗, that is, an ordering isomorphic to . . .− 2,−1, 0, in sum astructure AO:

〈ω,<〉 ⊕ (O⊗ 〈Z, <〉)⊕ 〈ω,<〉∗.

We argued, via use of the Ehrenfeucht game (see below), that all the structuresAO are elementarily equivalent, and we observed that this sufficed to show thatCn(ϕ) = Th(F).

The Ehrenfeucht game has two players, Spoiler and Duplicator; the equip-ment for the game consists of “boards” corresponding to the graphs G and


H and pebbles a1, a2, . . . and b1, b2, . . . . The game is organized into roundsr1, r2, . . . . At each round ri the Spoiler plays first and picks one of the pair ofpebbles ai or bi to play onto a vertex of G or H, respectively; the Duplicatorthen plays the remaining pebble of the pair onto a vertex of the structure intowhich the Spoiler did not play. This completes the round. Let vi (resp. wi)be the vertex of G (resp. H) pebbled at round i, let Gi and Hi be the sub-graphs of G and H induced by {v1, . . . , vi} and {w1, . . . , wi}, respectively, andlet Ri = {〈vj , wj〉 | 1 ≤ j ≤ i}. The Duplicator wins the game at round ri if therelation Ri is an isomorphism from Gi onto Hi. The Duplicator has a winningstrategy for the i-round game, if she has a method of play which results in awin for her no matter how the Spoiler plays. In this case, we write G ∼i H.Next time, we will see that G ≡ H if and only if for all i G ∼i H.


Preview 09.22

Roland Fraısse employed partial isomorphisms to provide a useful characteriza-tion of elementary equivalence (an alternative formulation of this characteriza-tion was given by Andrzej Ehrenfeucht in terms of a combinatorial game).

Definition 10 Let A, B be structures for a finite relational language (no func-tion symbols) and let P0 ⊇ . . . ⊇ Pn be a sequence of non-empty sets of partialisomorphisms from A to B. We say that P0, . . . ,Pn is an n-EF-sequence (andwrite A ∼=n B) if and only if for every 0 ≤ m < n,

Limited Forth Property: for every a ∈ UA and f ∈ Pm+1, there is ag ∈ Pm such that f ⊆ g and a ∈ dom(g).

Limited Back Property: for every b ∈ UB and f ∈ Pm+1, there is a g ∈ Pmsuch that f ⊆ g and b ∈ ran(g).

Definition 11 The quantifier rank of a first-order formula is defined by thefollowing induction.

1. qr(ϕ) = 0, if ϕ is atomic.

2. qr(ϕ ∧ ψ) = qr(ϕ ∨ ψ) = qr(ϕ→ ψ) = max(qr(ϕ), qr(ψ)).

3. qr(¬ϕ) = qr(ϕ).

4. qr((∃x)ϕ) = qr((∀x)ϕ) = qr(ϕ) + 1.

We say structures A and B are n-equivalent (and write A ≡n B) if and only iffor all sentences ϕ, if qr(ϕ) ≤ n, then A |= ϕ if and only if B |= ϕ.

Theorem 14 (Ehrenfeucht-Fraısse) If A and B are structures for a finiterelational language, then

A ≡n B if and only if A ∼=n B.

Corollary 5 (Ehrenfeucht-Fraısse) If A and B are structures for a finite re-lational language, then

A ≡ B if and only if for all n, A ∼=n B.


8 Lecture 09.22

Encouraged by your comments, we struck out in a new direction. We beganto contextualize our earlier detailed investigations of logical theories and relatethem to two of the milestones of mathematical logic: Kurt Godel’s results of1930 and 1931 which bear on the scope and limitations of the mechanization ofreasoning, respectively.

Definition 12 Let A be a finite alphabet of symbols and let A∗ be the set ofstrings over A.

1. X ⊆ A∗ is decidable if and only if there is a mechanical procedure M suchthat for every s ∈ A∗, if s ∈ X, then M with input s outputs 1 (for “In”)and if s 6∈ X, then M with input s outputs 0 (for “Out”). That is, X isdecidable if and only if there is a mechanical procedure that computes thecharacteristic function of X.

2. X ⊆ A∗ is semi-decidable if and only if there is a mechanical procedureM such that for every s ∈ A∗, if s ∈ X, then M with input s outputs 1(for “In”) and if s 6∈ X, then M with input s generates no output, thatis, the computation of M with input s fails to terminate. Thus, a semi-decision procedure yields complete and accurate positive information aboutmembership in X and yields no such negative information.

3. A set Γ of first-order sentences is a theory if and only if Cn(Γ) = Γ. Atheory Γ is axiomatizable if and only if there is a decidable set of sentencesΛ such that Γ = Cn(Λ).

The following result is an “abstract” formulation of Godel’s Completeness the-orem for first-order logic.

Theorem 15 If Γ is a decidable set of first-order sentences, then Cn(Γ) issemi-decidable.

Let N∗ = 〈ω,+,×〉 be the standard model of arithmetic. The next result is acomputation theoretic formulation of a corollary to Godel’s first incompletenesstheorem.

Theorem 16 Th(N∗) is not axiomatizable.

The next result is a corollary to Theorem 15.

Corollary 6 If a theory is complete and axiomatizable then it is decidable.

Proof: Suppose the theory Γ is both complete and axiomatizable. It followsfrom Theorem 15 that there is a semi-decision procedure M for membership inΓ. To mechanically determine membership of θ in Γ execute M on both theinputs θ and ¬θ. Since Γ is complete, one of these computations must terminate.


9 Lecture 09.29

We began with a discussion of the notion of proof and traced the arc from theearly days of Euclid through the quest for rigor in the late nineteenth-century, upto the twentieth-century era of formalization. We noted that a formal system Fof logical deduction is characterized by a decidable proof relation ΠF(δ, θ) whichholds when δ is a finite sequence (or a finite tree, or a finite . . .) which is a proofof θ in F. The following less abstract version of the completeness correspondsmore closely to Godel’s original result.

Theorem 17 There is a formal system F such that

• (Soundness of F) If (∃δ)ΠF(δ, θ), then θ is valid.

• (Completeness of F) If θ is valid, then (∃δ)ΠF(δ, θ).

We showed that Theorem 17 implies Theorem 15, and began to lay thegroundwork for establishing the reverse implication.


10 Lecture 10.04

We “flipped the classroom” and Michael led a discussion of one of the examproblems. We collaboratively succeeded in establishing that

Theorem 18 The collection of ordinals is strictly well-ordered by membership.

Apropos this result, April asked, “What’s a class, anyway?”

11 Lecture 10.11

We decided to delve a little more deeply into set theory, by way of answeringthe question posed at the end of the last class. We began with another question,“how are sets formed?” We first considered an intuitively compelling answer,formulated explicitly by Frege as the Comprehension Principle: a set is givenas the extension of a property, that is, for every property P , there is a set Ssuch that an object o is a member of S if and only if o possesses the propertyP . The set S is called the extension of P.

We looked at some problematic instances of this principle. First we consid-ered the extension of the property is an ordinal number. By Theorem 18 thiscollection, call it Ord is strictly well-ordered by membership. Moreover, by anearlier result, every member of an ordinal is an ordinal, hence Ord is transitive.It follows that Ord is an ordinal, and thus, Ord ∈ Ord, but we have now arrivedcontradiction, the Burali-Forti Paradox, since membership orders the ordinalsstrictly. What’s gone wrong? One analysis is that the Comprehension Principleis at fault – somehow the extension of the property is an ordinal is not a set.How very odd!

The Burali-Forti paradox is rather technical, and one wonders whether theremight be some escape that avoids scrapping the intuitively attractive Compre-hension Principle. Not so with Russell’s Paradox, formulated in a letter toFrege in 1901. Consider the property of being a non-self-membered set. Callthe extension of this property, Veronica. Is Veronica a member of itself? Eitheranswer, yes or no, leads to a contradiction! In this case, the conclusion that theComprehension Principle is faulty seems inescapable. Apparently, sets cannotbe understood as freely formed via conceptual extraction from a comprehensiveuniverse of objects. We discussed an alternative conception, the cumulative hi-erarchy, also called the rank hierarchy, of sets, which gives a satisfying view ofset formation that explains the failure of the Comprehension Principle.

We define the hierarchy by recursion in stages indexed by ordinals as follows.P(X) is the power set of X.

• V0 = ∅

• Vβ+1 = P(Vβ)

• Vγ =⋃β<γ Vβ


The proper class V is defined by the formula (∃α)x ∈ Vα, and is thoughtof as the universe of all sets. We define the rank of a set x (written rk(x)) asthe least α such that x ∈ Vα+1. It is easy to see from the construction that forevery set x and y ∈ x, rk(y) < rk(x). It follows at once that for every set x,x 6∈ x, and that if the members of a class U are unbounded in rank, that is, forevery α, there is an x such that U(x) and α < rk(x), then U is not a set (orsince we’re thinking of classes as given by properties, the extension of U is nota set, or better still U has no extension). Since both the ordinals and x 6∈ xare unbounded in rank, we have an explanation of the Paradoxes of Burali-Fortiand Russell.

Next time we’ll discuss the Continuum Hypothesis.


12 Lecture 10.13

We began our study of the Continuum Problem, the first on the famous listof problems posed by David Hilbert in his 1900 address to the InternationalCongress of Mathematicians:

The investigations of Cantor on such assemblages of points suggesta very plausible theorem, which nevertheless, in spite of the moststrenuous efforts, no one has succeeded in proving. This is the theo-rem: Every system of infnitely many real numbers, i.e., every assem-blage of numbers (or points), is either equivalent to the assemblageof natural integers, 1, 2, 3, . . . or to the assemblage of all real num-bers and therefore to the continuum, that is, to the points of a line;as regards equivalence there are, therefore, only two assemblages ofnumbers, the countable assemblage and the continuum. From thistheorem it would follow at once that the continuum has the nextcardinal number beyond that of the countable assemblage; the proofof this theorem would, therefore, form a new bridge between thecountable assemblage and the continuum.1

The “very plausible theorem” Hilbert recommends for attention is the Con-tinuum Hypothesis (CH) of Georg Cantor. the notion of “equivalence” invokedin its statement is what we have called “equipollence”: X is equipollent to Y ifand only if there is a bijection from X onto Y (we write X ∼ Y for this). Recallthat a set X is countable if and only if X ∼ ω or (∃n ∈ ω)X ∼ n). If we writeR for the continuum, that is, the set of real numbers, or speaking geometrically,the set of points on the line, we can then state the CH very simply as:

Conjecture 1 For every X ⊆ R, if X is uncountable, then X ∼ R.

We noted that this is a particularly direct formulation of the hypothesis,unencumbered by the need to explain the sequence of ℵ’s, or invoke the Axiomof Choice, as required by the alternative formulation, |R| = ℵ1, or, yet moreencumbered, 2ℵ0 = ℵ1.

We began by reminding ourselves that R is uncountable, via Cantor’s orig-inal proof.

Theorem 19 (Cantor) R is uncountable.

Proof: It suffices to show that if ζi, i ∈ ω is a sequence of points in the unitinterval [0, 1], then there is a ζ ∈ [0, 1] such that for every i ∈ ω, ζ 6= ζi. Weconstruct a descending chain of nonempty closed intervals [0, 1] ⊃ I0 ⊃ I1 ⊃ . . .to satisfy:

(∀k ∈ ω)ζk 6∈ Ik. (1)

1Hilbert, David, Mathematical Problems, Bulletin (New Series) of the American Mathe-matical Society,Volume 37, Number 4, Pages 407-436. Article electronically published on June26, 2000.


We begin by trisecting the unit interval into three closed subintervals J01 , J

02 , J

03

and we choose I0 to be the leftmost of these subintervals which omits ζ0 (itcan’t lie in all three - that’s why we trisected rather than bisected!). Similarly,at stage n + 1 of the construction, we may suppose that In has already beenconstructed and we trisect it into subintervals Jn1 , J

n2 , J

n3 and choose In+1 to

be the leftmost of these that omits ζn+1. Now consider the set I =⋂n∈ω In.

It is clear from our construction that the sequence we have defined satisfies 1,and hence that for every k ∈ ω, ζk 6∈ I. To complete the proof, it suffices toobserve that, by the Least Upper Bound Principle, among the most fundamentalproperties of R, I 6= ∅.

We discussed the intuitive basis for the Least Upper Bound Principle. ThePrinciple states that any subset of R that is bounded from above has a leastupper bound. In geometric terms, it guarantees that if line is partitioned into two pieces, one of them leftward closed and the other rightward closed, oneof the pieces must have an endpoint; if this weren’t so, the line would not becontinuos, that is, the offending partition would witness the existence of a gap.We also presented the (easy) argument that the L.U.B. Principle implies thatthe intersection of a nested sequence of nonempty closed intervals is nonempty.

To gain further insight into the CH, we began to discuss Cantor’s solutionof a special case. In order to explain this special case, we briefly discussed somenotions from point-set topology.

Definition 13 1. A set X ⊆ R is open if and only if for all ζ ∈ X, there isan open interval I such that ζ ∈ I ⊆ X.

2. A set X ⊆ R is closed if and only if R−X is open.

3. Let X ⊆ R. ζ is a limit point of X if and only if for every open intervalI, if ζ ∈ I, then I ∩X 6= ∅.

4. Let X ⊆ R. ζ is an isolated point of X if and only if ζ ∈ X and ζ is nota limit point of X.

5. Let X ⊆ R. X is perfect if and only if X is closed and contains no isolatedpoints.

6. Let X ⊆ R. D(X) (the Cantor-Bendixson derivative of X) is the set oflimit points of X.

7. A collection B of open subsets of R is called a basis for the topology on Rif and only if every open subset of R is a union of sets contained in B.

8. A topological space is separable if and only if it has a countable basis.

Next time, we will prove that

Theorem 20 (Cantor) If X ⊆ R is closed and uncountable, then X ∼ R.

To get in the spirit, attempt the following exercises.


Exercise 1 Let X ⊆ R. Show that X is closed if and only if D(X) ⊆ X.

Exercise 2 Show that R (with the topology described above) is separable.


13 Lecture 10.18

We went over the exercises suggested at the end of the preceding memoir.Here are some further exercises to prepare for the proof of Cantor’s resolution

of CH for closed sets.

Exercise 3 Show that if X ⊆ R is perfect and nonempty, then X ∼ R.

Exercise 4 Show that if X ⊆ R is closed, then there is a countable set Y anda perfect set Z such that X = Y ∪ Z.


14 Lecture 11.15

This memoir is as much a preview of coming lectures as it is an account of ourclass meeting of November 15.

We established the following corollary to the abstract completeness theorem.

Corollary 7 Suppose Σ is a semi-decidable set of sentences of an effectivelypresented first-order language. If Cn(Σ) is complete, then Cn(Σ) is decidable.

Proof: We first showed that if Σ is semi-decidable, then Cn(Σ) is semi-decidable. In order to do this, we introduced the notion of a computably enu-merable set (c.e. set). A set X ⊆ A∗ is c.e. if and only if X = ∅ or there isan algorithmically computable function F : N 7→ A∗ such that ran(F ) = X.We then proved that a set X is c.e. if and only if X is semi-decidable. Wefirst established the left to right implication. Suppose X is c.e. and let F be acomputable enumeration of X (it is clear that the empty set is decidable, hencesemi-decidable). The the following algorithm computes a semi-decision proce-dure for X: for input σ ∈ A∗, enumerate in order F (0), F (1), . . . and compareeach of these with σ. If and when σ appears in the list, output 1. It is clearthat this procedure outputs 1 if and only if σ ∈ X, that is, it is a semi-decisionprocedure for X.

For the implication from right to left, we need to examine the notion of asemi-decision procedure a bit more closely. We imagine the procedure executedby a machine M . It is supplied with an input σ ∈ A∗ (the set of strings oversome finite alphabet A). If within some finite number of computation steps t,M finishes its computation and outputs a 1, then σ is a member of the set WM

semi-decided by M ; otherwise, σ 6∈ WM . We write C(M,σ, t) to denote thatM with input σ halts within t steps of computation and outputs a 1. We maythink of all the pairs 〈σ, t〉 for σ ∈ A∗ and t ∈ N coded by natural numbers;let’s use p(〈σ, t〉) to denote the natural number code of the pair 〈σ, t〉.

Now, let X be a non-empty semi-decidable set and let σ0 be the least elementof X. We define an algorithmically computable function F : N 7→ A∗ as follows:

F (n) =

{σ if n = p(〈σ, t〉) ∧ C(M,σ, t)σ0 otherwise

Sinceσ ∈WM ⇐⇒ (∃t)C(M,σ, t),

it is clear that ran(F ) = WM .Now, returning to the main argument, we show that if Σ is semi-decidable,

then Cn(Σ) is semi-decidable. Let ϕ,ϕ1, . . . be a computable enumeration ofΣ, which exists since Σ is c.e. Now we describe an algorithmic semi-decisionprocedure for Cn(Σ). Given input ψ the procedure invokes the semi-decisionprocedure for validity (which exists by the abstract completeness theorem) andruns it for k-steps on each of the inputs ϕ1 → ψ, . . . , (ϕ1 ∧ . . . ∧ ϕk) → ψ. Ifany of these computations terminates with output 1, our procedure outputs 1.


It follows at once from the abstract completeness theorem and the compactnesstheorem that this is a correct semi-decision procedure for Cn(Σ).

To complete the proof of the corollary, it now suffices to show that if a set ofsentences Γ is semi-decidable and complete, then Γ is decidable. We describe analgorithmic procedure to decide membership in Γ. With input ϕ the procedureinvokes the semi-decision procedure for Γ with both the input ϕ and the input¬ϕ. Since Γ is complete, one of these computations halts and outputs 1.

We presented a sketch of the proof of

Theorem 21 (Church’s Theorem) The set of valid sentences of an effec-tively presented first order language is not decidable.

Our proof of Church’s Theorem relied on the following development of fun-damental notions in the theory of computability.

We introduced the notion of a Turing machine, an abstract computationaldevice. A Turing machine has a reading head, an infinite two way tape ruledinto cells, and a control which may be in one of a fixed finite number of states.Each cell of the tape is either blank or has a 0 or a 1 written on it. A machineM is specified by its program which consists of a finite set of instructions. Aninstruction I is a quadruple 〈q, s, a, q′〉 where q and q′ are control states, s isa tape symbol (either B, for blank, or 1), a is an action (B for print a blank,0 for print a 0, 1 for print a 1, R for move the tape one cell to the right, orL for move the tape one cell to the left). If M ’s control is in state q and itsreading head is positioned over a cell containing the symbol s then M executesthe instruction I by performing action a and transitioning to control state q′.We suppose that M is deterministic, that is, for all q and s there is at mostone instruction whose first and second coordinates are q, s. One of M ’s controlstates q0 is singled out as its start state. The (partial) function FM : N 7→ Ncomputed by a Turing machine M is defined as follows. FM (n) = k if and onlyif M , when started with its reading head over the leftmost nonblank square ofa tape containing the numeral for n in binary and its control in the start state,performs a computation which halts with the binary numeral for k on its tape(a machine halts if its control is in state q and its reading head is positionedover a cell containing the symbol s and it has no instruction whose first andsecond coordinates are q, s). We may define the (partial) binary, ternary, ...,function computed by a Turing machine in a similar manner. For example, thefollowing Turing machine computes the constant function with value 1.

〈q0, 0, 1, q3〉〈q0, 1, L, q1〉〈q1, 0, B, q2〉〈q1, 1, B, q2〉〈q1, B,B, q3〉〈q2, B, L, q1〉

We note that there are countably many Turing machines, hence, they canbe enumerated as Me, e ∈ N . We write F ke for the partial k-ary function com-puted by Me. The enumeration can be performed in such a way as to verify


the following fundamental result, which may be considered as the foundationof computer science, since it, together with Turing’s Thesis, discussed below,essentially established the possibility of a general purpose programmable com-puter.

Theorem 22 (Universal Turing Machine Theorem) There is a u such thatfor every e and every n

F 2u(e, n) = Fe(n).

Turing’s Thesis: If F is an algorithmically computable function, then forsome e, F = Fe.

The best evidence for Turing’s thesis comes via Turing’s analysis of compu-tation which gave rise to the notion of Turing machine and which provides astrong argument for the stronger claim (sometimes called Turing’s super-thesis)that the execution of any algorithm can be adequately captured by the compu-tation of some Turing machine, that is, Turing machines adequately simulateall algorithms themselves, not just their outputs. Turing’s thesis underlies allclaims that given mathematical problems are effectively undecidable, since themathematical argument for all such claims involves establishing that there is noTuring computable decision procedure for the problem. Turing’s thesis can alsobe invoked as ground for claims that various intuitively computable functionsare Turing computable, and it will be implicitly deployed in this way through-out the following developments. (Indeed, we appealed to Turing’s Thesis in ourintuitive argument for the Universal Turing Machine Theorem in class.)

Given sets X and Y we are interested in comparing the computational diffi-culty of deciding membership in X versus Y . One such comparison is providedby the notion of effective many-one reducibility. X is effectively many-one re-ducible to Y if and only if there is a Turing computable F such that for everya,

a ∈ X ⇐⇒ F (a) ∈ Y.

It follows at once that if X is reducible to Y and Y is decidable (resp., semi-decidable), then X is decidable (resp., semi-decidable). This provides a usefultechnique to establish that a set is not decidable, namely, show that an unde-cidable set is reducible to it.

We now sketch an argument for Church’s Theorem. We first show that thereis a semi-decidable set which is not decidable. First note that a set is decidableif and only if both it and its complement are semi-decidable. So, it suffices toexhibit a semi-decidable set whose complement is not semi-decidable. Recall thecomputation relation C(M,σ, t) defined above. We modify this to allow inputof a Turing machine index e instead of Me and a natural number n in place ofσ and we write We for the semi-decidable set of natural numbers accepted byMe. Observe that for every e and n,

n ∈We ⇐⇒ (∃t)C(e, n, t).

Now, let K = {e | e 6∈ We}. It is easy to see that K is semi-decidable, butthat its complement is not. To conclude the proof of Church’s Theorem, it now


suffices to show that every semi-decidable set is many-one reducible to the setof valid sentences of first-order logic. Here we invoke the arithmetization of thetheory of Turing machines, that is, there is a finite set of sentences {ρ1, . . . , ρk}and a formula χ(e, n, t) in the language of first-order arithmetic such that forall e and n

(∃t)C(e, n, t) ⇐⇒ ∅ |= (ρ1 ∧ . . . ∧ ρk)→ (∃t)χ(e, n, t).

(Here e and n are the formal numerals for e and n.) This concludes the proofsketch. In order to fill out the details of this sketch, and to prove Godel’s FirstIncompleteness Theorem, we need to introduce an alternative mathematicalcharacterization of effective computability, namely, numeralwise representabil-ity.

Numeralwise Representable Functions and Relations We begin by defin-ing the class of numeralwise representable functions and relations. Let Γ be asatisfiable set of sentences in a first-order language L which includes a sequenceof countably many distinct closed terms n, n ∈ ω. We call {n | n ∈ ω} thenumerals of L. We say a relation R ⊆ ωk is numeralwise represented by theformula θ(x1, . . . , xk) in Γ if and only if for all n1, . . . , nk ∈ ω,

if 〈n1, . . . , nk〉 ∈ R, then Γ |= θ(n1, . . . , nk, )

andif 〈n1, . . . , nk〉 6∈ R, then Γ |= ¬θ(n1, . . . , nk)

A relation is numeralwise representable in Γ is there is a formula that numeral-wise represents it, and a function is numeralwise representable if and only if itsgraph is.

We first showed that if Γ is a finite set of sentences (in an effectively pre-sented language) and R is numeralwise representable in Γ, then R is decidable,that is, the characteristic function of R is mechanically computable. This isa direct corollary of the Completeness Theorem, which implies that the set ofconsequences of a finite set of sentences in an effectively presented first-orderlanguage is semi-decidable. So suppose that R ⊆ ωk is numeralwise representedby the formula θ(x1, . . . , xk) in Γ. In order to decide whether 〈n1, . . . , nk〉 ∈ Rwe run the semi-decision procedure for the set of consequences of Γ simulta-neously on θ(n1, . . . , nk) and on ¬θ(n1, . . . , nk). One of these computationsterminates in a yes answer, and thereby yields a decision on the membership of〈n1, . . . , nk〉 in R.

We will focus our attention of the relations and functions numeralwise repre-sentable in a particular finitely axiomatized theory, Robinson Arithmetic (RA).The axioms of RA are the (universal closures) of the following formulas.

1. ¬Sx < x

2. Sx = Sy → x = y


3. ¬x < 0

4. x < Sy ↔ (x < y ∨ x = y)

5. x < y ∨ x = y ∨ y < x

6. x+ 0 = x

7. x+ Sy = S(x+ y)

8. x · 0 = 0

9. x · Sy = x · y + x

10. xE0 = 1

11. xE(Sy) = (xEy) · x

The structure N = 〈ω,<, 0, S,+, ·, E〉, where < is the usual order on ω andthe constant and function symbols (E for exponentiation) are interpreted asusual, satisfies RA. The numerals n in the language of RA are the sequenceof closed terms Sn0. Recall that if t is a variable-free term, then tN is thedenotation of t in N. We began our study of numeralwise representability in RAby showing that

Lemma 6 For all variable-free terms t and all n ∈ ω, RA |= t = n if and onlyif tN = n.

Definition 14 The formula θ((x1, . . . , xk) is numeralwise determined by RA ifand only if for all n1, . . . , nk ∈ ω,

RA |= θ(n1, . . . , nk, )

orRA |= ¬θ(n1, . . . , nk, )

The next lemma will be useful in establishing the numeralwise representabilityof a variety of relations in RA.

Lemma 7 The relation R ⊆ ωk is numeralwise represented by the formulaθ(x1, . . . , xk) in RA if and only if θ is numeralwise determined by RA and θdefines R in N, that is, θ[N] = R.

We suggested that it would be useful to think about proving the followinglemma in advance of our next class.

Lemma 8 Every atomic formula is numeralwise determined by RA. If θ andχ are numeralwise determined by RA, then so are all boolean combinations of θand χ. If θ(y, x, x1, . . . , xk) is numeralwise determined by RA, then so are

(∃y)(y < x ∧ θ(y, x, x1, . . . , xk))

and(∀y)(y < x→ θ(y, x, x1, . . . , xk)).

Documents

1 Lecture 08 - University of Pennsylvaniaweinstei/006Memoirs.pdf · PHIL 006/506, Fall, 2016 Scott Weinstein 1 1 Lecture 08.30 We began with a survey of some of the highlights of