[PPT] - Many-Sorted First-Order Model Theory Lecture 2 12 th June, 2020 1 / PowerPoint Presentation

SLIDE 1

Many-Sorted First-Order Model Theory

Lecture 2 12th June, 2020

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SLIDE 2

Conservative signature morphisms

Definition 1 (Conservative signature morphism)

A signature morphism χ: Σ → Σ′ is conservative iff each Σ-model has a χ-expansion.

Exercise 1

◮ A signature morphism χ: Σ → Σ′ is conservative iff it is injective. ◮ For any conservative signature morphism χ: Σ → Σ and any sets of Σ-sentences Γ1 and Γ2 we have Γ1 | =Σ Γ2 iff χ(Γ1) | =Σ′ χ(Γ2). ◮ Show that χ(Γ1) | =Σ′ χ(Γ2) implies Γ1 | =Σ Γ2 doesn’t hold if χ is not conservative.

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SLIDE 3

Term models

Definition 2 (Non-void signatures)

A signature Σ is non-void if all sorts are inhabited by some terms, i.e. TΣ,s = ∅ for all s ∈ S.

Fact 3 (Term models)

If Σ is non-void then the set of Σ-terms TΣ can be regarded as a Σ-model: ◮ the carrier set for each sort s ∈ S is TΣ,s (which is not empty); ◮ for each σ : s1 . . . sn → s ∈ F, the function σTΣ : TΣ,s1 × · · · × TΣ,sn → TΣ,s is defined by σTΣ(t1, . . . , tn) = σ(t1, . . . , tn) for all i ∈ {1, . . . , n} and ti ∈ TΣ,si; ◮ TΣ interprets each relation symbol as the empty set, i.e. πTΣ = ∅ for all (π : w) ∈ P.

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SLIDE 4

More on term models

Theorem 4 (Initiality)

Given any non-void signature Σ and any Σ-model A, there is a unique homomorphism h: TΣ → A defined by the interpretation of each term into the model A, i.e. h(t) = tA for all sorts s ∈ S and terms t ∈ TΣ,s.

Proof.

Assume there exists another homomorphism g : TΣ → A. We prove that g(t) = h(t) for all sorts s ∈ S and terms t ∈ TΣ,s. We proceed by induction on the structure of terms: ◮ h(σ(t1, . . . , tn)) = σA(h(t1), . . . , h(tn)) IH = σA(g(t1), . . . , g(tn)) = g(σ(t1, . . . , tn)).

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SLIDE 5

More on term models

Notation

Let Σ be a signature and X a set of vari- ables for Σ such that Σ[X] is non-void. We let TΣ(X) denote TΣ[X] ↾Σ, the free model of terms with variables from X. Σ[X] TΣ[X]

↾ιX

Σ
ιX
TΣ(X) = TΣ[X] ↾Σ

Exercise 2 (Freeness)

Assume ◮ a signature Σ and a set of variables X for Σ such that Σ[X] is non-void, and ◮ a Σ-model A and an evaluation v : X → |A|. TΣ(X)

v#

A

X

u
v
Then there exists a unique homomorphism v# : TΣ(X) → A such that

v = u; v#, where TΣ(X) = TΣ[X] ↾Σ and u : X ֒ → TΣ(X).

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SLIDE 6

Substructures

Definition 5 (Substructure/submodel)

Let Σ be a signature and A and B two Σ-models. A is a substructure of B, in symbols A ⊆ B, iff there exists an inclusion homomorphism h: A → B.

Exercise 3

A ⊆ B iff

1. |A| ⊆ |B|,
2. σB(a) ∈ As for all (σ : w → s) ∈ F and a ∈ Aw,
3. πA ⊆ πB.

Exercise 4

Let Σ be a signature and h: A → B a Σ-homomorphism. Then h(A) is a submodel of B. Note that πh(A) = {h(a) | a ∈ πA}.

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SLIDE 7

Congruence

Definition 6 (Congruence)

Let Σ be a signature and A be a Σ-model. A congruence ≡= {≡s}s∈S on A is

1. an equivalence on |A|, i.e. an S-sorted relation ≡s⊆ As × As for all s ∈ S satisfying the

following properties: ◮ (Reflexivity) a ≡s a for all s ∈ S and a ∈ As ◮ (Symmetry) a1 ≡s a2 a2 ≡s a1 for all s ∈ S and a1, a2 ∈ As ◮ (Transitivity) a1 ≡s a2 a2 ≡s a3 a1 ≡s a3 for all s ∈ S and a1, a2, a3 ∈ As

2. compatible with the function symbols in Σ

◮ (Congruence) a1 ≡s1 a′

1 . . . an ≡sn a′ n

σA(a1, . . . , an) ≡s σA(a′

1, . . . , a′ n)

for all (σ : s1 . . . sn → s) ∈ F and ai, a′

i ∈ Asi for all i ∈ {1, . . . , n}.

We will drop the subscript s from ≡s whenever it is clear from the context.

Convention

Let ≡ be a congruence on a model A. If (a) w = s1 . . . sn ∈ S∗ (b) a = a1 . . . an ∈ Aw, and (c) a′ = a′

1 . . . a′ n ∈ Aw then a ≡w a′ iff ai ≡si a′ i for all i ∈ {1, . . . , n}. 7 / 22

SLIDE 8

Examples of congruences

Example 7 (NAT)

spec NAT is sort Nat .

p

0 : -> Nat .

p

s_ : Nat -> Nat . end We define the congruence ≡ on N as follows: ◮ n1 ≡ n2 iff (n1 mod 2) = (n2 mod 2) for all n1, n2 ∈ N.

Example 8 (INT)

spec INT is sort Int .

p 0 : -> Int .
p s_ : Int -> Int .
p p_ : Int -> Int .

end Let ≡ be the congruence on T(ΣINT) generated by the following two sets of pairs

f terms:

◮ {p s t ≡ t | t ∈ T(ΣINT)}, and ◮ {s p t ≡ t | t ∈ T(ΣINT)}.

Exercise 5

Prove that ≡ defined on N above is a congruence.

Exercise 6

Prove that the intersection of two congruences is again a congruence.

Exercise 7

Let E be a set of equations over a non-void signature Σ. Prove that ≡E := {(t1, t2) | E | = t1 = t2} is the least congr. on TΣ generated by {(t1, t2) | (t1 = t2) ∈ E}.

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SLIDE 9

Kernel

Lemma 9

Let Σ be a signature and h : A → B a Σ-homomorphism. We define the congruence ker(h) = {ker(h)s}s∈S on A as follows: ◮ ker(h)s = {(a, b) | hs(a) = hs(b)} for all s ∈ S The relation ker(h) is a congruence on A.

Proof.

◮ (Reflexivity): Obviously, h(a) = h(a), which implies (a, a) ∈ ker(h) ◮ (Symmetry): We assume (a, b) ∈ ker(h) and we show that (b, a) ∈ ker(h). We have: a ker(h)b iff h(a) = h(b) iff h(b) = h(a) iff b ker(h) a. ◮ (Transitivity): We assume (a, b) ker(h) and (b, c) ∈ ker(h), and we show that (a, c) ∈ ker(h). Since (a, b) ∈ ker(h) and (b, c) ∈ ker(h), we have h(a) = h(b) and h(b) = h(c). We obtain h(a) = h(c). Hence, (a, c) ∈ ker(h). ◮ (Congruence): Let (σ : w → s) ∈ F. We assume that (a, b) ∈ ker(h)w and we prove that (σA(a), σA(b)) ∈ ker(h)s. Since (a, b) ∈ ker(h)w, we have hw(a) = hw(b). It follows that hs(σA(a)) = σB(hw(a)) = σB(hw(b)) = hs(σA(b)). Hence, (σA(a), σA(b)) ∈ ker(h)s.

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SLIDE 10

Quotient

Notation

Let ≡ be a congruence on a Σ-model A. Let s ∈ S and a ∈ As. The class of a is a := {a′ ∈ As | a ≡ a′} sometimes denoted also by a/≡.

Fact 10

Note that a ⊆ As for all s ∈ S and a ∈ As. This means that ≡ determines a partition of the universe |A|. As

a

Example 11

Consider the congruence defined in Example 7. We have: ◮ 0 = {0, 2, 4, 6, . . . } ◮ 1 = {1, 3, 5, 7, . . . }

Example 12

Consider the congruence defined in Example 8. We have: ◮ 0 = {0, p s 0, s p 0, p s p s 0, . . . } ◮ p 0 = {p 0, p s p 0, s p p 0, p s p s p 0 . . . }

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SLIDE 11

Convention

If w = s1 . . . sn and a = (a1, . . . , an) ∈ Aw then we let a denote the tuple ( a1, . . . , an).

Definition 13 (Quotient substructures)

Let ≡ be a congruence on a Σ-model A. The quotient structure of A modulo ≡ is the Σ-structure A (also denoted A/≡) defined below: ◮ As = { a | a ∈ As} for all sorts s ∈ S, ◮ for all function symbols (σ : w → s) ∈ F, the function σ

A :

Aw → As is defined by σ

A(

a) = σA(a) for all a ∈ Aw; ◮ for all relation symbols (π : w) ∈ P, the relation π

A is defined by π A = {

a | a′ ∈ πA for some a′ ∈ a} .

Lemma 14

A is well-defined.

Proof.

σ

A :

Aw → As is a function: if a = b then a ≡ b, which implies σA(a) ≡ σA(b), and we get σ

A(

a) = σA(a) = σA(b) = σ

A(

b).

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SLIDE 12

First isomorphism theorem

Fact 15

The quotient map q : A → A defined by q(a) = a for all s ∈ S and a ∈ As is a homomorphism.

Theorem 16 (First isomorphism theorem)

Let h: A → B be a Σ-homomorphism and let ≡ be a congruence on A such that ≡⊆ ker(h). There exists a unique homomorphism g : A → B such that q; g = h, where q is the quotient homomorphism q : A → A.

A

∃!g

B

A

q

h
Proof.

We define g : A → B by g( a) = h(a) for all s ∈ S and a ∈ As. We show that g is well-defined: ◮ if a = a′ then a ≡ a′; since ≡⊆ ker(h), h(a) = h(a′); ◮ since h is compatible with the function and relation symbols, g is compatible with the function and relation symbols too. For the uniqueness part, let f : A → B such that q; f = q; g. For all s ∈ S and a ∈ As, f ( a) = f (q(a)) = g(q(a)) = g( a).

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SLIDE 13

Second isomorphism theorem/homomorphic image theorem

Theorem 17 (Second isomorphism theorem)

Let h : A → B be a Σ-homomorphism. Then A/ker(h) ∼ = im (h).

A

∃!g

im (h)

A

q

h
Proof.

By a slightly abuse of notation we denote by h the co-restriction of h : A → B to im (h). Let q : A → A by the quotient map from Fact 15. We apply Theorem 16 with ≡= ker(h). There exists a unique homomorphism g : A → im (h) such that q; g = h. Clearly g is surjective. We need to show that g is injective: Assume that g( a) = g( a′). We have h(a) = g( a) = g( a′) = h(a′). Since ≡= ker(h), we get a = a′.

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SLIDE 14

Basic set of sentences

Definition 18 (Basic set of sentences)

A set of Σ-sentences E is basic if there exists a Σ-model AE such that for all Σ-models A, A | = E iff there exists a homomorphism AE → A. We say that AE is a basic model of E. If in addition the homomorphism AE → A is unique then the set E is called epi-basic.

Theorem 19

Any set of atomic Σ-sentences E is basic. If Σ is non-void then E is epi-basic.

Proof.

Firstly, we prove that E is epi basic assuming that Σ is non-void. Let ≡E := {(t1, t2) | E | = t1 = t2} be the congruence on TΣ. The basic model AE is obtained from TΣ/≡E by interpreting each (π : w) ∈ P as follows: πAE := { t | E | = π(t)}. We show that A | = E iff there exists a unique arrow AE → A: “⇒” By Theorem 4, there exists a unique arrow h: TΣ → A. We show that ≡E ⊆ ker h: t1 ≡E t2 iff E | = t1 = t2; since A | = E, A | = t1 = t2; we have tA

1 = tA 2 iff h(t1) = h(t2) iff

(t1, t2) ∈ ker(h). By Theorem 16, there exists a unique arrow g′ : TΣ/≡E → A such that q; g′ = h. Since A | = E, there exists a unique arrow g : AE → A such that q; g = h, defined by g( t) = g′( t) for all t ∈ TΣ. TΣ/≡E

∃!g′

A

TΣ

q

h
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SLIDE 15

Proof of Theorem 19.

“⇐” Assuming a homomorphism g : AE → A we prove A | = E. By Theorem 4, h = q; g and h(t) = tA for all t ∈ |TΣ|.

1. Let t1 = t2 ∈ E. We have t1 ≡E t2, which implies

g( t1) = g( t2). It follows that tA

1 = h(t1) =

g(q(t1)) = g( t1) = g( t2) = g(q(t1)) = h(t2) = tA

2 . Hence, A |

= t1 = t2.

2. Let π(t) ∈ E. By the definition of AE,

t ∈ πAE . Since g is a homomorphism, g( t) = g(q(t)) = h(t) = tA ∈ πA. Hence, A | = π(t). AE

g

A

TΣ

q

h
Secondly, we show that each set of atomic sentences E over any signature Σ if basic.

Let C be the S-sorted set which consists of a new constant for each sort s ∈ S that in not inhabited by the Σ-terms. It follows that Σ[C] is non-void. Let Aχ(E) be the basic model of χ(E). We show that Aχ(E) ↾Σ= AE is a basic model of E, i.e. for all Σ-models A we have: A | =Σ E iff there exists a Σ-homomorphism AE → A. “⇒” Assume that A | = E. There exists an ιC-expansion B of

A. By the satisfaction condition B |

= E. Since E is epi- basic, there exists a unique homomorphism h′ : Aχ(E) →

B. Hence, we get h = h′ ↾Σ : AE → A.

Σ[C] Aχ(E)

↾Σ

h′

B

↾Σ

Σ
ιC
AE

h

A

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SLIDE 16

Proof of Theorem 19.

“⇐” Assume a homomorphism h: AE → A and show that A | = E. Note that Aχ(E) = (AE, f : C → |AE|). Let B = (A, (f ; h): C → |A|) be the ιC- expansion of A, which interprets each constant c ∈ C as h(cAχ(E)). Let h′ : Aχ(E) → B be the ιC- expansion of h such that h′(a) = h(a) for all s ∈ S and a ∈ Aχ(E),s. Since χ(E) is epi-basic B | = E. By the satisfaction condition A | = E. AE

h

A

C

f

f ;h
Exercise 8

Any set of existentially quantified atomic sentences ∃X · ρ (ρ is an equation or a relation) is basic but not necessarily epi-basic. Hint: Firstly, prove that any sentence of the form ∃X · E, where E is a finite set of atomic sentences, is basic.

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SLIDE 17

Substitutions

Definition 20 (Substitution)

Let Σ be a signature and X, Y two sets of variables for Σ. A Σ-substitution between Σ[X] and Σ[Y ] is a function θ : X → TΣ[Y ]. (z, s, Σ[X])

(z, s, Σ[Y ])

Sen(Σ[X])

θ

Sen(Σ[Y ])

A substitution θ: X → TΣ[Y ] determines

1. a sentence translation θSen : Sen(Σ[X]) → Sen(Σ[Y ]) which is

◮ the identity on the symbols in Σ, and ◮ maps every variable x ∈ X to θ(x) ∈ TΣ[Y ];

2. a model reduct ↾θ: |Mod(Σ[Y ])| → |Mod(Σ[X])| defined by

◮ (B↾θ)s = Bs for all sorts s ∈ S, ◮ σ(B

↾θ) = σB for all function symbols σ ∈ F,

◮ π(B

↾θ) = πB for all relation symbols σ ∈ F,

◮ x(B

↾θ) = θ(x)B for all x ∈ X.

for all models B ∈ |Mod(Σ[Y ])|. We drop the superscript Sen from θSen when there is no danger of confusion.

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SLIDE 18

Theorem 21 (Satisfaction condition for substitutions)

Let θ : X → TΣ[Y ] be a substitution. Then for all Σ[Y ]-models B and all Σ[X]-sentences γ, B | =Σ[Y ] θSen(γ) iff B↾θ| =Σ[X] γ. Similarly to the case of satisfaction condition for signature morphisms, we need two lemmas in order to prove Theorem 21.

Lemma 22

For all substitutions θ: X → TΣ[Y ], all Σ[X]-terms t and all Σ[Y ]-models B, we have tB

↾θ = θ(t)B.

Proof.

We proceed by induction on the structure of terms: ◮ xB

↾θ = θ(x)B for all variables x ∈ X;

◮ σ(t1, . . . , tn)B

↾θ = σB ↾θ(tB ↾θ 1

, . . . , tB

↾θ n

)

IH

= σB(θ(t1)B, . . . , θ(tn)B) = θ(σ(t1, . . . , tn))B.

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SLIDE 19

A

↾ιZ

Σ[X, Z]

θ′

Σ[Y , Z ′]

∃!C

↾θ′

↾ιZ′
A↾ιZ = B↾θ

Σ[X]

ιZ
θ

Σ[Y ]

ιZ′

B

↾θ

Lemma 23

Let θ : X → TΣ[Y ] be a Σ-substitution and Z a set of variables for Σ[X]. For all Σ[X, Z]-models A and all Σ[Y ]-models B such that A↾ιZ = B↾θ there exists a unique Σ[Y , Z ′]-model C such that C↾θ′= A and C↾ιZ′ = B, where ◮ ιZ : Σ[X] ֒ → Σ[X, Z] is an inclusion, ◮ Z ′ = {(z, s, Σ[Y ]) | (z, s, Σ[X]) ∈ Z} is the translation of Z along θ, ◮ ιZ′ : Σ[Y ] ֒ → Σ[Y , Z ′] is an inclusion, ◮ θ′ : X ∪ Z → TΣ[Y ,Z] is def. by ◮ θ′(x, s, Σ) = θ(x, s, Σ) for all (x, s, Σ) ∈ X, ◮ θ(z, s, Σ[X]) = (z, s, Σ[Y ]) for all (z, s, Σ[X]) ∈ Z.

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SLIDE 20

A

↾ιZ

Σ[X, Z]

θ′

Σ[Y , Z ′]

∃!C

↾θ′

↾ιZ′
A↾ιZ = B↾θ

Σ[X]

ιZ
θ

Σ[Y ]

ιZ′

B

↾θ

Proof of Lemma 23.

We define C as follows: (1) C interprets all symbols in Σ[Y ] as B; (2) C interprets Z as A, that is (z, s, Σ[Y ])C = (z, s, Σ[X])A for all (z, s, Σ[Y ]) ∈ Z ′. By (1), C↾ιZ′ = B. Since θ′ extends θ, by (2), C↾θ′= A. For the uniqueness part, assume another Σ[Y , Z]-model D such that D↾θ′= A and D↾ιZ′ = B. (a) Since D↾Σ[Y ]= B = C↾Σ[Y ], that is D interprets all symbols in Σ[Y ] as B, and in particular, as the model C; (b) for all (z, s, Σ[Y ]) ∈ Z ′, since D↾θ′= A, we have (z, s, Σ[Y ])D = θ′(z, s, Σ[X])D = (z, s, Σ[X])A = θ′(z, s, Σ[X])C = (z, s, Σ[Y ])C. By (a) and (b), we get C = D.

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SLIDE 21

Reachable models

Definition 24 (Reachable models)

Let Σ be a signature and A be a Σ-model. A is reachable iff each element of |A| is the denotation of some term, i.e. the unique mapping f : TΣ → |A| defined by f (t) = tA for all t ∈ TΣ, is surjective.

Theorem 25

Let Σ be a signature, and X a set of variables for Σ. Then A a reachable model iff each expansion B of A to the signature Σ[X] (i.e. a valuation v : X → |A|) generates a substitution θ : X → TΣ such that A↾θ= B.

Proof.

“⇒” Let v : X → |A| be a valuation. Since f : TΣ → |A| defined by f (t) = tA for all t ∈ TΣ, is surjective, there exists θ: X → TΣ such that θ; f = v. We show that A ↾θ= (A, v) = B. It suffices to show that x(A

↾θ) = x(A,v) for all

x ∈ X. For all x ∈ X, we have x(A

↾θ) = θ(x)A = f (θ(x)) =

v(x) = x(A,v). TΣ

f

|A|

X

θ

v
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SLIDE 22

Proof of Theorem 25.

“⇐” We show that each element in |A| is the denotation

f some term.

Let s ∈ S and a ∈ As. Let x = (x, s, Σ) and B the ιx-expansion of A interpreting x as a, that is xB = a. By our assumptions, there exists a substitution θ: {x} → TΣ such that A ↾θ= B. Hence, θ(x)A = xA