Canonical Model

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Part 2 overview

Welcome to Part 2!We will cover:

7. Logics, soundness, completeness. Canonical model for K.

8. Soundness and completeness for more frame classes.

9. Finite model property, via filtrations. Decidability.

10. Modal -calculus.

Thanks to Nick Bezhanishvili and Clemens Kupke (who taughtpart 2 in 2009 and 2010) for major contributions to the slides,

exercises, and solutions.

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7. Logics, soundness, completeness

Recall (definition 9) the notion of validity of a modal formula A ina class C of frames: A is valid in C if A is true at every world of

every model whose frame is in C.Each class C of frames gives a logic:

Log(C) = {A : A is a modal L-formula valid in C}.

Log(C) (the logic ofC) comprises the laws of chemistry ofC.

We want a more concrete characterisation of Log(C): C comes straight from the application, so is too close to it.

C is probably too big/abstract (to automate etc).

We want the underlying rules controlling the form of Log(C).We want to reason about formulas directly.

Reasoning via standard translation is possible especially withmodern first-order theorem provers but clumsy, unnatural, anddoesnt take advantage of modal nature of formulas.

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Hilbert systems

We will use Hilbert systems to capture the logics of classes offrames. A Hilbert system consists of

1. axioms

2. rules of inference.

The axioms are given truths. They should be valid in some givenclass C of frames.

The inference rules allow us to derive new truths from old. If theformulas above the line in a rule are already derived, the rule allowsus to derive the formula underneath the line. (See next slide.)

Rules should be chosen so as not to lead us from truth intofalsehood. Formulas derived from C-valid formulas should also be

valid in C.

Hilbert systems are simple, powerful, flexible, and in common use.(But not suited for practical reasoning.)

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The Hilbert system K

The most basic Hilbert system for modal logic is called K.Note: K depends on L (our set of propositional atoms).

Pick any two distinct p, q L.K-axioms All propositional tautologies (valid propositional formulas).

Eg., p q p, (p q) ((q r) (p r)), etc. Normality: (p q) (p q), for p, q as above.

K-rules (if youve got the top, you can derive the bottom). Here,A, B are any formulas and p L is any atom.

modus ponens (MP): A, A BB

Universal generalisation (UG): AA

(also called Necessitation)

Substitution (Sub): AA(B/p)

A(B/p) is the formula got by replacing all occurrences of theatom p in A by B. Eg. (p p q)(q/p) = q q q.

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Proofs in a Hilbert system

Definition 50Let H be a Hilbert system. We write H A, and say that A is atheorem of H, if there are formulas A1, . . . , An such that

An = A

each Ai (1 i n) is either an axiom of H or is obtained from some of A1, . . . , Ai1 by a rule of H.

The sequence A1, . . . , An is called a proof of A (in H), or aderivation of A (in H), of length n.

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Example of a proof

Example 51

Let A, B be any modal formulas. Then K (AB) A. For,

1. p q p axiom (tautology)

2. A B A Sub(1)

3. (A B A) UG(2)

4. (p q) (p q) axiom (normality)5. (A B A) ((A B) A) Sub(4)

6. (A B) A MP(3,5)

is a proof of(A B) A in K, of length 6.

In practice we often omit lines 1 and 4, and justify lines 2 and 5 byinstance of tautology, instance of normality.

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Useful properties ofK

Lemma 52 (essentially MP)

Let A, B be formulas. IfK A andK A B then K B.Proof.Suppose that X1, . . . , X n1, A is a proof of A (in K),and Y1, . . . , Y m1, A B is a proof of A B. Then by MP,X1, . . . , X n1, A , Y 1, . . . , Y m1, A B, B is a proof of B.

We will use this later, justified by MP.

Exercise 53If A B is an instance of a tautology, show thatK A B.

So as far as K is concerned, we can write (A B C) withoutworrying whether we mean ((AB)C) or (A (BC)), etc.

Exercise 54 (important)

Show that K A B (A B) for any formulas A, B.

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Logic generated by a Hilbert system

In this course, all Hilbert systems will have at least the axioms of K (they can have more axioms than K

even infinitely many more) exactly the inference rules of K: that is, MP, UG, and Sub.

Definition 55

1. The logic generated by a Hilbert system H is the set of alltheorems of H:

{A : A a modal formula, H A}.We denote this set by Thm(H), or simply by H.

2. A modal logic is a set of the form Thm(H), for some Hilbertsystem H. (Strictly, it is called a normal modal logic.)

The definition is entirely syntactic.Aim: find classes C of frames and Hilbert systems H such thatLog(C) = Thm(H). Then we get a syntactic handle on validity.Many people find this very impressive.

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Soundness and completeness for K

We begin with the simplest case:

C is the class of all frames.So Log(C) is the set of all valid formulas.

H = K, the basic Hilbert system.

We are going to show that Log(C) = Thm(K) that is, the logicof the class of all frames is just K. And here it is:

Theorem 56A formula is a theorem of K iffit is valid.

is called soundness.Proofs are sound: they do not prove false things.

is called completeness.The proof system can prove all valid formulas.

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Soundness ofK

Soundness is easy. Its really just an exercise!

Proposition 57 (soundness for K)

Any theorem of K is valid.

Proof. We prove a bit more. Let F= (W, R) be any frame.Claim 1. Any axiom of K is valid in F.Proof of claim.

All propositional tautologies are plainly valid in F.Proposition 8 showed that (p q) (p q) is valid in any

frame.So all K-axioms are valid in F. Claim

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Proof of soundness ctd.

Claim 2. The rules of K preserve validity over F.Proof of claim. Suppose that A and A B are valid in F.

Let p be any atom and C any formula.We show that B, A, and A(C/p) are valid in F.

So let M = (W,R,h) be any model on F. Let t W be arbitrary.

As A and A B are valid in F, we have M, t |= A andM, t |= A B. So plainly, M, t |= B.

If u W and R(t, u), then M, u |= A (as A is valid in F).But u was arbitrary, so M, t |= A.

Let N = (W,R,g), whereg(p) = {w W : M, w |= C}g(q) = h(q) for each atom q= p.

We made p true in N just where C is true in M.So (exercise) M, w |= A(C/p) iffN, w |= A for all w W.But A is valid in F. So N, t |= A. Hence M, t |= A(C/p).

Claim

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Proof of soundness ctd.

An easy induction on the length of a proof now shows that everyK-theorem is valid in F.Try it yourself; details in theorem 75 later.

Completeness is harder. . .

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Why is completeness harder?

We want to prove that if a given formula A0 is valid then K A0.

How? Two possible ways:

1. Assume A0 is valid. Show K A0 by constructing a proof ofA0 in K. (How?!)

2. Assume K A0. Show A0 is not valid, by constructing amodel in which A0 is true at some world.

Either way, we have to construct something. This is always tricky.

It turns out that (2) is viable. (Easier to build infinite things!)But what will the model be made of? What will its worlds be?Answer: syntactic objects namely, sets of formulas!

But first we try to give the idea, by having a look at a model.

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Descriptions of worlds in a model

Let M = (W,R,h) be any model.For each t W, let t be the set of all formulas true at t in M:

t = {A : M, t |= A}.

t is a full description of t in modal logic. ( delta.)

We try to represent t by this sett.(It doesnt matter what t actually is, after all.)

So we form a new set of worlds: W

d

= {t : t W}.This Wd is the set of all full descriptions of worlds in M.

It is made of sets of formulas: syntactic!u

t

d

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Can we make a sensible model on Wd? one suggested by the original modelM, but made without peeking at M?

First, try to define an assignment hd into Wd. Let p be an atom.

The descriptiont of some t knows whether p is true at t:M, t |= p p t.

u

t

d

Here, we know M, t |= p and M, t |= q, because p,q t.So why not let

hd(p) = { Wd : p }, for each atom p L.

Then p is true at iff its in !

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Accessibility relation on Wd?

In M, we know that if R(t, u) and M, t |= A then M, u |= A.

That is, if R(t, u

)and

A

t then A

u.u

tA

B

AB

d

So R(t, u) leaves traces in t,u!

So why not try: for , Wd ( gamma), letRd(,) iff for every A, ifA then A .

This all makes sense, thinking of the as full descriptions ofworlds. Anyway, we have defined a model Md = (Wd, Rd, hd).

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Md is well-behaved

It turns out that the full descriptions making up Md describethemselves in Md (so (Md)d = Md):

Lemma 58For every formula A and Wd, we haveMd, |= A iffA .

Proof.An easy induction, using that for all , Wd and formulas A, B:

1.

2. A iff A /

3. A and B iff A B

4. ifA and Rd(,) then A

5. ifA / then there is with Rd(,) and A / .

These properties are easily seen to be true, if you remember thatevery Wd is of the form t for some t W. See lemma 70for a similar proof done in full.

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The real story

This is all very well, but we needed a model M to start with.Were given only that K A0, and we need to make a modelsatisfying A0.

Still, this Md gives us a clue: why not build a modelMK (say)whose worlds are sets of formulas!

Define hd, Rd as above.

Establish the 5 properties in lemma 58 (dry but necessary).Then we can prove MK, |= A iff A , for all A and allsets included in MK.

Show A0 is contained in some set in MK.

Then A0 will be true at ! We will have our model.

But which sets of formulas do we include in MK, for this to work?Answer: maximal consistent sets of formulas.

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Consistency preliminaries

Let (Gamma) be a set of formulas (maybe infinite!).

The following definitions are good for any Hilbert system H.Definition 59If A is a formula, we write H A if there are formulasB1, . . . , Bn (for some n) such that H B1 . . . Bn A.

Possibly, n = 0. So ifH A then H A for any .

Idea: H A says that A follows (is provable) from formulas in .These formulas need not be valid.

Example 60

{p, p q} K q, because (p (p q)) q is a propositional

tautology, so K (p (p q)) q.Of course, p, p q are not valid!

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Consistency

Definition 61 (consistency)

is said to be H-consistent if

H .

is said to be consistent (with no H stated) if it is K-consistent.

Exercise 62Show that a set of modal formulas is H-consistent iffH (A1 . . . An) whenever A1, . . . , An .

This may be a nicer characterisation of consistency.

Example 63 {A,B,(A B)} is (K-)inconsistent, because(p q (p q)) is a tautology, so by (Sub),K (A B (A B)).

{p,q, r} is consistent, because otherwise, K (p q r),so by proposition 57, (p q r) is valid which it isnt.

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Maximal consistent sets

Definition 64

A set of formulas is said to be maximal H-consistent (anH-MCS) if it is H-consistent but no larger set is H-consistent.A maximal consistent set (MCS) is a maximal K-consistent set.

A clue to why MCSs are interesting is the following:

Exercise 65LetM = (W,R,h) be any Kripke model. Let t W. Show that

t = {A : M, t |= A}

is a maximal consistent set.

So our full description model Md

was made of MCSs!This suggests we should choose maximal consistent sets to put inour model.

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Canonical model

Definition 66 (Canonical model for K)

1. Let WK = {all maximal (K-)consistent sets}.

2. For , WK, define RK(,) if for every formula A,ifA then A .

3. For each atom p L, definehK(p) = { WK : p }.

Let MK be the Kripke model (WK, RK, hK).We call MK the canonical model for K.We call FK = (WK, RK) the canonical frame for K.

In the following, we show that the canonical model works.It gets a bit technical. But essentially we are establishing that

1. the five properties of lemma 58 hold2. A0 is a member of some MCS

Look out for these points as we go along.

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Extending a consistent setUseful lemma

Lemma 67If is consistent and A is a formula, then at least one of {A}, {A} is also consistent.

Proof.Otherwise, there are B1, . . . , Bn, C1, . . . , C m with

K A B1 . . . Bn B, say

and K A C1 . . . Cm C, say

.

Now K (A B ) [(A C ) (B C )] we can derive it from a tautology by Sub.

As K AB , MP gives K (AC ) (B C ).

But K A C as well. MP now gives K B C .This means that is inconsistent contradiction.

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Which formulas does a MCS have in it?

Lemma 68 (handy test)

If is a MCS, then for any formula A we have K A iffA .

Proof.If A then as K A A, we obviously have K A.

Conversely, suppose K A.So K B1 . . . Bn A, for some B1, . . . , Bn .

Claim. {A} is inconsistent.Proof of claim. We have B1, . . . , Bn,A {A}.Now the following is an instance of a tautology, so a theorem of K:

(B1 . . . Bn A) (B1 . . . Bn A ).By MP, K B1 . . . Bn A . Claim

By lemma 67, {A} must be consistent. And {A} .But is maximal consistent. So {A} = . So A .

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Some properties of MCSs

Lemma 69

Let

be a MCS and A,B,B1, . . . , Bn be formulas (for somen 1). Then:

1. A, B iffA B .

2. A iffA / .

3. IfB1, . . . ,Bn then (B1 . . . Bn) as well.

Proof.1. If A B , then as K (A B A), we have K A. Bylemma 68, A . Similarly, B .

If A, B then as K A B (A B) (!), we have K A B, so by lemma 68, A B .

2. Suppose A,A . Now K A A . So K .That is, is inconsistent. So at most one of A, A is in .

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Lemma ctd.

By lemma 67, {A} or {A} is consistent.By maximality, = {A} or = {A}.So A or A .

Weve shown that exactly one of A, A is in .This is equivalent to what is required.

3. We show B1, . . . ,Bn (B1 . . . Bn) ,by induction on n. The case n = 1 is trivial.

Let n 2. Inductively assume the statement for n 1.We prove it for n.

So assume that B1, . . . ,Bn .By exercise 54, K B1 B2 (B1 B2).So K (B1 B2).By lemma 68, (B1 B2) .

By inductive hypothesis, (B1 . . . Bn) .

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Truth lemma for canonical model

Lemma 70 (truth lemma)

For every formula A and every MCS WK, we have:

MK, |= A iffA .

Proof. By induction on A.

For , we have MK, |= and (exercise) .

If A is an atom p, then MK, |= p iff hK(p) iff p .Inductively, assume the lemma for A, B. Then:

MK, |= A iffMK, |= A,iff A / (by the inductive hypothesis),iffA (by lemma 69(2), as is maximal consistent).

MK, |= (A B) iffMK, |= A and MK, |= B,iff A and B (by the inductive hypothesis),iff A B (by lemma 69(1)).

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Truth lemma: the case A

Now the big case: A. Inductively assume the lemma for A.

Suppose A .

Take any WK with RK(,).By definition of RK, we have A .So (by inductive hypothesis) MK, |= A.As was arbitrary, by semantics of we get MK, |= A.

Conversely, assume A / . We want to find a MCS containing

A and with RK(,). That is, should contain

0 = {A} {B : B }.

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Truth lemma: the case A ctd.

Claim. 0 = {A} {B : B } is consistent.Proof of claim. Assume not. So there are B1, . . . ,Bn such that K A B1 . . . Bn .

Now K (A i Bi ) (

i Bi A) (instance of taut).

By MP, we get K B1 . . . Bn A.Crucially: by UG, we then get K (B1 . . . Bn A).Using normality and MP we get K (B1 . . . Bn) A.

Now B1, . . . ,Bn . By lemma 69(3), (B1 . . . Bn) .

So K A. By lemma 68, A . Contradiction. Claim

By proposition 71 (next slide), there is a MCS with 0.

For every B, we have B B 0 . So RK(,).

Also, A 0 . By lemma 69(2), A / .So by the induction hypothesis, MK, |= A.

It follows by semantics of that MK |= A, as required.

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Lindenbaums lemma

Proposition 71 (Lindenbaum)

Any consistent set0 extends to (is contained in) a MCS.

Proof. Enumerate all modal formulas, as A0, A1, . . ..We define a chain 0 1 of consistent sets.0 is given, and is consistent. Inductively, define

n+1 = n {An}, if this is consistent,

n, otherwise.

Each n is consistent. Let =n n. Then:

is consistent too (for a proof of from would be a prooffrom some n).

is maximal consistent. For, let A be a formula with A / .We know A = An for some n. By construction, n {A} is

inconsistent (else n+1 = n {A} ). So {A} isinconsistent. So no set bigger than is consistent.

And contains 0.

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Completeness is provedWere done now! Relax.

Proposition 72 (completeness for K)

Any valid formula is a theorem of K.

Proof.Suppose that A0 is not a theorem of K.

Claim. {A0} is consistent.

Proof of claim. If not, then by exercise 62, K A0.But K A0 A0 (instance of tautology).By MP, K A0, contradicting our assumption. Claim

Extend {A0} to a MCS (proposition 71).By the truth lemma (lemma 70), MK, |= A for every A .

But A0 . So MK, |= A0.We found a model (MK) satisfying A0. So A0 is not valid.

Propositions 57 and 72 prove theorem 56.

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Origin of canonical modelphotos: Dana Scott, David Makinson

The canonical model construction was introduced (independently)by Lemmon & Scott (1966), Makinson (1966), Cresswell (1967),and Schutte (1968). It remains important today.

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Canonical Model

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