March 1, 2012

MATH43032/63032 Part 2

Modal Logic

1

A Motivating Example

Consider the following example of ‘follows’:

It is not possible that Bill and Monica are both telling the truthPossibly Monica is telling the truth∴ It is not necessarily the case that Bill is telling the truth

As with the previous sections on monotonic and nonmonotonic reasoning we want to formallycapture in what sense this conclusion ‘follow’ from the premises, why we are justified in writing∴ .

Clearly in this example the modalities (as they are called) ‘possible’ and ‘necessary’ matter.It is true that if we just ignored them then the conclusion would follow from the premises bythe propositional (= sentential) calculus, however in doing this we would have changed themeaning of the statements: Saying ‘Possibly Monica is telling the truth’ is not the same assaying ‘Monica is telling the truth’.

Nevertheless we might hope to somehow express them in terms of the connectives ∧,∨,¬,→which we already have in the Sentential Calculus, SC. However that hope is easy dashed becauseclearly the possibility of Monica telling the truth isn’t completely determined by whether ornot she actually is telling the truth. In other words the truth value of Possibly Monica istelling the truth is not a function of the the truth value of Monica is telling the truth, that is‘possibility’ is not Truth Functional. However the connectives of SC are truth functional, i.e.the truth values of θ ∧ φ, θ ∨ φ, θ → φ,¬θ are completely determined by the truth values of θand φ individually, and the same goes for any compound connective built up from these. Sothere is no way of understanding examples such as the above purely within SC, we need toenlarge our language, proof theory and semantics.

Syntax of Modal Logic

A language L for modal logic is just the same as for SC, hence a non-empty set of propositionalvariables.From L we define the set SML of modal sentences of L using the connectives ∧, ∨, →, ¬ andthe modality 2 (read as ‘necessarily’ or ‘box’) and parentheses,(,) as follows:

M1 L ⊆ SML

M2 If θ, φ ∈ SML then ¬θ, (θ ∧ φ), (θ ∨ φ), (θ → φ), 2θ ∈ SML

M3 θ ∈ SML just if this can be derived on the basis of M1,M2.

As usual we define the length of θ ∈ SML, |θ|, as the number of symbols in θ (excludingcommas) and can use proof by induction on the length of θ to show that some property holdsfor all θ ∈ SML.

Notice that SL ⊆ SML. Again we have unique readability since the case of 2θ is the same asfor ¬θ in SC.

Abbreviation For θ ∈ SML we write

3θ = ¬2¬θ (read ”possibly θ”)

Unlike the propositional calculus, for Modal Logic(s) it seems most natural to consider analyz-ing the ”follows” relation initially by developing a proof theory and leave the semantics until

2

later. [In Modal Logic(s) it is not usual to look at ‘monotone modal consequence relations’but instead to look directly at the smallest such relation(s), i.e. the ‘proves’ relation. This isOK here because the form of the rules/postulates as so called positive Horn Clauses meansthat such relations are closed under intersections and hence there is a least one, unlike thesituation for Rational Consequence Relations. Furthermore we could give a characterizationof the ‘modal monotone consequence relations’ just as we did for the non-modal SC so in thissense all the action is really taking place with the smallest such relation, i.e. ‘proves’.]

Proof systems of modal logic

Unlike with the Propositional Calculus, many systems of Modal Logic are studied. All havethe REF-axiom1 θ | θ for θ ∈ SML, and the rules of SC applied to sentences and subsets ofSML, viz: For Γ ⊆ SML finite etc.

And right:Γ | θ ∆ | φΓ,∆ | θ ∧ φ AND

And left:Γ, θ, φ |ψΓ, θ ∧ φ |ψ ANL

Or right:Γ | θ

Γ | φ ∨ θ andΓ | θ

Γ | θ ∨ φ ORR

Disjunction, Or left:Γ, θ |ψ ∆, φ |ψΓ,∆, θ ∨ φ |ψ DIS

Implies right:Γ, θ | φ

Γ | θ → φIMR

Modus Ponens:Γ | θ ∆ | θ → φ

Γ,∆ | φ MP

Not in:Γ, θ | φ ∆, θ | ¬φ

Γ,∆ | ¬θ NIN

Not not out:Γ | ¬¬θΓ | θ NNO

Monotonicity, or Weakening:Γ | θ

Γ,∆ | θ MON

So our modal logics extend SC, and consequently we also have the derived rule AO (And out).All our systems of modal logic we consider also have the necessity rule,

| θ|2θ NEC

1The left hand side here should really be the set θ rather than just the sentence θ but it is a commonabbreviation to drop the braces in such cases. Similarly we may write Γ, θ |φ for Γ ∪ θ |φ.

3

Notice that the left hand sides here are empty.

After this the different systems are characterized by the remaining axioms where:

K : 2(θ → φ) | (2θ → 2φ)D : K + 2θ | 3θ T : K + 2θ | θB : K + θ | 23θ S4 : T + 2θ | 22θS5 : T + 3θ | 23θ

Definition. A proof (in K, T , . . . ) is a finite sequence of sequents

Γ1| θ1, . . . ,Γm| θm,

where θi ∈ SML and Γi ⊆ SML are finite, such that for each i = 1, . . . , m, either Γi| θi is aninstance of an axiom (of K, T ,. . . ) or for some j1, . . . , js < i

Γj1| θj1 . . . Γjs| θjsΓi| θi

is an instance of one of the rules NEC or AND-MON from SC.

The natural number m here is called the length of the proof.

Definition. Let J be one of the systems K,B, . . . , S5 (a convention we shall adopt through-out). For θ ∈ SML and Γ ⊆ SML, possibly infinite, we define

Γ ⊢J θ ⇐⇒ there is a proof Γ1| θ1, . . . ,Γm| θm in J suchthat Γm ⊆ Γ and θm = θ.

In this case Γ1| θ1, . . . ,Γm| θm is called a proof 2 of θ from Γ in J .

Example.A formal proof of θ ⊢T

3θ:

1. 2¬θ | ¬θ by T

2. θ | θ by REF

3. θ,2¬θ | ¬θ by MON applied to 1.

4. θ,2¬θ | θ by MON applied to 2.

5. θ | ¬2¬θ by NIN (Not in) applied to 3. and 4.

Another exampleA formal proof of 2(θ ∧ φ) ⊢K

2θ:

1. θ | θ by REF

2. θ, φ | θ by MON from 1

3. θ ∧ φ | θ by ANL from 2

4. | (θ ∧ φ) → θ by IMR from 3

5. |2((θ ∧ φ) → θ) by NEC from 4

6. 2((θ ∧ φ) → θ) |2(θ ∧ φ) → 2θ by the K axiom

7. |2((θ ∧ φ) → θ) → (2(θ ∧ φ) → 2θ) by IMR from 6

2We insert the adjective ‘formal’ when we want to distinguish between a proof in this sense and an ‘informalproof’ such as we would give to justify a theorem.

4

8. |2(θ ∧ φ) → 2θ by MP from 5,7

9. 2(θ ∧ φ)|2(θ ∧ φ) → 2θ by MON from 8

10. 2(θ ∧ φ) |2(θ ∧ φ) by REF

11. 2(θ ∧ φ) |2θ by MP from 9, 10

Proposition 1 Let J be one of the systems K,B, . . . , S5. Let Γ ⊆ SML and θ ∈ SML.

(i) if θ ∈ Γ, then Γ ⊢J θ.

(ii) if φ ∈ SML and φ|θ is an axiom of the system J with φ ∈ Γ, then Γ ⊢J θ.

(iii) ifΓ1 | θ1, Γ2 | θ2, . . . , Γk | θk

Γ | θis an instance of a rule of proof, except that possibly the Γi,Γ are allowed to be infinite,and Γi ⊢J θi for i = 1, 2, . . . , k then Γ ⊢J θ.

Proof (i) follows from REF : θ | θ is a proof of θ from Γ.

(ii). If φ | θ is an axiom of the system J and φ ∈ Γ then, by definition, φ | θ is a proof of θ fromφ in J .

(iii). We do the case where the rule is AND. So suppose Γ ⊢J θ and Γ ⊢J φ, say Λ1 | θ1, . . . ,Λm | θmis a proof of Γ ⊢J θ and ∆1 | φ1, . . . ,∆h | φh is a proof of Γ ⊢J φ. So θm = θ, φh = φ and Λm,∆h

are finite subsets of Γ. Then it is easy to check that

Λ1 | θ1, . . . ,Λm | θm,Λm ∪∆h | θm,∆1 | φ1, . . . ,∆h | φh,Λm ∪∆h | φh,Λm ∪∆h | θm ∧ φh

is a proof (the justifications for the new steps being MON,MON, AND) ofΓ ⊢J θ ∧ φ since θm ∧ φh = θ ∧ φ and Λm ∪∆h is a finite subset of Γ.

The other rules are similar. [See also the appended Example Sheet.]

Notation. We write L(n) for the finite language p1, . . . , pn.

Proposition 2 Let J be one of the systems K,B, . . . , S5, let φ1, . . . , φn ∈ SML and let θ ∈SML(n). Let θ

~φ be the result of simultaneously replacing each occurrence of pi in θ by φi for1 ≤ i ≤ n. Then

(i) θ~φ ∈ SML

(ii) If Γ1 | θ1, . . . ,Γm | θm is a proof in J with Γi ⊆ SML(n), θi ∈ SML(n), then Γ~φ1 | θ

~φ1 , . . . ,Γ

~φm | θ~φm

is a proof in J .

Proof (i) is a straightforward induction on |θ| for θ ∈ SML(n).(ii) is a straightforward induction on m.

A useful consequence this proposition is that since any SC-proof for L(n) is also a proof forK,B, . . . , S5, we may directly lift proofs and derivability results from SC to these systems. Forexample, since (p1 → p2) ↔ (¬p2 → ¬p1) is a tautology (in SC), we have ⊢K (φ1 → φ2) ↔(¬φ2 → ¬φ1) for all φ1, φ2 ∈ SML.

Some provability results worth noting

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In any of the systems J if ⊢J (θ → φ) then ⊢J (2θ → 2φ) and ⊢J (3θ → 3φ). More generallyfor 2, if θ1, θ2, . . . , θn ⊢J φ then 2θ1,2θ2, . . . ,2θn ⊢J

2φ, see 3 on the Example Sheet.

If θ ⊢J φ holds then so does θ ⊢J ′

φ for any extension J ′ of J .

By AO and AND,3

⊢J (θ ↔ φ) ⇐⇒ ⊢J (θ → φ) and ⊢J (φ→ θ) ⇐⇒ θ ⊢J φ and φ ⊢J θ

by MON, REF, MP. From this it follows that the relation ⊢J (θ ↔ φ) is an equivalence relation,called provable equivalence (in J).

⊢K : (k1) 2(θ ∧ φ) ↔ (2θ ∧ 2φ) (k2) 2θ ↔ ¬3¬θ(k3) 2¬θ ↔ ¬3θ (k4) ¬2θ ↔ 3¬θ

⊢T : (t0) 2θ → θ (t1) θ → 3θ

⊢S4 : (s1) 2θ ↔ 22θ (s2) 3θ ↔ 33θ

(s3) 23θ ↔ 2323θ (s4) 32θ ↔ 3232θ

⊢S5 : (σ0) 2θ ↔ 22θ (σ1) 3θ ↔ 23θ

(σ2) 2θ ↔ 32θ (σ3) 2(θ ∨2φ) ↔ (2θ ∨ 2φ)

(σ4) 2(θ ∨3φ) ↔ (2θ ∨3φ)

We leave these as a fun exercises for you to do. [See the Examples Sheet at the end of thesenotes. Of course Propositions 1 and 2 are useful here.] As usual once we have the CompletenessTheorems we will have much more straightforward ways to show these, and similar. On theother hand we actually need some of these to prove the Completeness Theorems in the firstplace!

Corollary 3 S5 extends S4 (in fact strictly).

Proof By (σ0) there is a proof in S5 whose last sequent is | (2θ → 22θ) and hence a proofPθ with last sequent 2θ |22θ. Hence if Q is a proof in S4 we can obtain a proof in S5 withthe same final sequent by replacing every occurrence of the S4 axiom, 2θ |22θ, in Q by Pθ.(We’ll see later that this extension is strict.)

Corollary 4 Let Mi be 2 or 3 for i = 1, 2, . . . , n, n ≥ 1. Then in S4 M1M2 . . .Mnθ isprovably equivalent to one of

2θ, 3θ, 23θ, 32θ, 232θ, 323θ.

Corollary 5 Let Mi be 2 or 3 for i = 1, 2, . . . , n, n ≥ 1. Then in S5, M1M2 . . .Mnθ isprovably equivalent to Mnθ (so one of 2θ, 3θ)

3As usual θ ↔ φ is an abbreviation for (θ → φ) ∧ (φ → θ).

6

Semantics for the modal logic K

Definition. A frame for a language L is a triple 〈W,E, V 〉, such that

1. 〈W,E〉 is a directed graph, i.e. W is a nonempty set and E ⊆W ×W .4

2. V = (Vi)i∈W where Vi : L→ 0, 1 (i ∈ W ).

The elements i ∈ W are called worlds.

Definition. Let 〈W,E, V 〉 be a frame for the language L. For i ∈ W , we define 〈W,E, V 〉, i |=θ (or just i |= θ if the frame is clear from the context) for θ ∈ SML by induction on |θ|:

if θ = p ∈ L, i |= θ ⇐⇒ Vi(p) = 1if θ = ¬φ, i |= θ ⇐⇒ i 6|= φif θ = (φ ∧ ψ), i |= θ ⇐⇒ i |= φ and i |= ψif θ = (φ ∨ ψ), i |= θ ⇐⇒ i |= φ or i |= ψif θ = (φ → ψ), i |= θ ⇐⇒ i 6|= φ or i |= ψif θ = 2φ, i |= θ ⇐⇒ ∀j ∈ W ((i, j) ∈ E ⇒ j |= φ)

By unique readability this definition is correct.

Remark. By definition we have i |= 2¬θ ⇐⇒ ∀j ∈ W ((i, j) ∈ E ⇒ j 6|= θ). Hence

i |= 3θ ⇐⇒ ∃j ∈ W ((i, j) ∈ E and j |= θ).

For example in the frame W = 0, 1, 2, E = (0, 0), (0, 1), (1, 1), (1, 2), (2, 2), L = p,V0(p) = V1(p) = 1, V2(p) = 0, diagrammatically:

0 1 2

pp ¬p0 2p

0 2 22p

0 2p ∧ ¬22p

1 2 2p

1 2 2¬p

2 2¬p

2 2 2p

Definition. For Γ ⊆ SML and θ ∈ SML we say that θ is a logical consequence of Γ in K,denoted

Γ |=K θ,

if for every frame 〈W,E, V 〉 and every i ∈ W we have5

i |= Γ ⇒ i |= θ.

4E is sometimes referred to as an accessibility relation. So if (i, j) ∈ E we say that j is accessible from i.5i Γ stands for i φ for all φ ∈ Γ.

7

Theorem 6 (Correctness Theorem for K) If Γ ⊆ SML and θ ∈ SML, then

Γ ⊢K θ ⇒ Γ |=K θ.

Proof Let Γ1|θ1, . . . ,Γm|θm be a proof of Γ ⊢K θ. Hence Γm ⊆ Γ and θm = θ and it is enoughto show by induction on k for k = 1, 2, . . . , m that Γk |=K θk.

Case 1. Γk|θk is an axiom of K.

If Γk|θk is an instance of REF, then clearly Γk |=K θk. Otherwise Γk|θk is an instance of K,i.e. of the form 2(ψ → φ) | (2ψ → 2φ). Let 〈W,E, V 〉 be a frame of L and let i ∈ W withi |= 2(ψ → φ). We have to show i |= (2ψ → 2φ), i.e. we may assume i |= 2ψ and we showi |= 2φ. Suppose j ∈ W with (i, j) ∈ E but j 6|= φ. Since i |= 2ψ we have j |= ψ, hencej 6|= (ψ → φ). Since (i, j) ∈ E, this contradicts our assumption i |= 2(ψ → φ).

Case 2. Γk|θk is obtained by applying a rule ofK to some Γr1 |θr1, . . . ,Γrs|θrs , where r1, . . . , rs <k.

By the induction hypothesis we know Γrn |=K θrn for n = 1, . . . , s.

We do each rule separately. Except for the necessity rule, all follow as in the propositionalcalculus, since the verification always remains within one world. It remains to show that |=K

respects the necessity rule|θ|2θ

for every θ ∈ SML. In other words we may assume |=K θ and we must show |=K2θ.

Let 〈W,E, V 〉 be a frame of L and i ∈ W . Since |=K θ we know j |= θ for every j ∈ W .Therefore in particular j |= θ for every j ∈ W with (i, j) ∈ W . This shows i |= 2θ as desired.

The Correctness Theorem is very useful because it enables us to show results of the formΓ 0K φ by showing Γ 2K φ, and this simply requires us to exhibit a suitable frame 〈W,E, V 〉and i ∈ W such that i Γ but i 2 φ.

For example in the frame W = 0, 1, 2, E = (0, 0), (0, 1), (1, 1), (1, 2), (2, 2), L = p,V0(p) = V1(p) = 1, V2(p) = 0 given earlier we have 0 2p but 0 2 22p so 2p 0K

22p, andhence K must be strictly weaker that S4.

Definition Γ ⊆ SML is J-inconsistent if Γ ⊢J θ ∧ ¬θ for some θ ∈ SML,

– equivalently Γ ⊢J θ and Γ ⊢J ¬θ for some θ ∈ SML,

– equivalently Γ ⊢J φ for any φ ∈ SML.

Γ is J-consistent if not J-inconsistent.6

As a corollary to Theorem 6 we have:

Corollary 7 K is consistent, i.e. we cannot have ⊢K φ and ⊢K ¬φ for any sentence φ.Equivalently ∅ is consistent in K.

Proof If not we would have i φ and i ¬φ for any world i in any frame for K. But clearly(by unique readability) this is impossible.

6We shall see later that this notion is independent of L.

8

Completeness for the modal logic K

Our aim in this section is to show the Completeness Theorem for K, that is that ⊢K andK are actually the same thing. The structure of the proof is similar to that for SC (or thePredicate Calculus), as you will see.

Lemma 8 (i) If ∆ ⊆ SML is consistent in K and θ ∈ SML, then (at least) one of ∆∪θor ∆ ∪ ¬θ is consistent in K.

(ii) Suppose Γ0 ⊆ Γ1 ⊆ Γ2 ⊆ . . . are K-consistent subsets of SML. Then⋃

∞

n=0 Γn is K-consistent. More generally the union of a set of K-consistent subsets of SML, which istotally ordered by inclusion, is K-consistent.

Proof The proof is the same as for SC:

(i) If ∆∪θ is K-inconsistent then ∆, θ ⊢K φ and ∆, θ ⊢K ¬φ for (any) φ so ∆ ⊢K ¬θ by NIN.Hence if ∆∪¬θ is also K-inconsistent, ∆ ⊢K ¬¬θ so ∆ is also K-inconsistent, contradiction.

(ii) Suppose on the contrary ∆1 | φ1, . . . ,∆m | φm was a proof of θ ∧ ¬θ from⋃

∞

n=0 Γn. Then∆m ⊆ ⋃

∞

n=0 Γn so because ∆m is finite it must actually be a subset of some Γj . To see thissuppose ∆m = ψ1, . . . , ψr . Each ψi ∈ ⋃

∞

n=0 Γn so for each i there is some ni such thatψi ∈ Γni

. But then since the Γn are increasing each ψi for i = 1, 2, . . . , r is in Γn where n isthe maximum of the ni, so ∆1 | φ1, . . . ,∆m | φm is a proof of θ ∧¬θ from Γn, contradicting theK-consistency of Γn.

Definition. A subset Ω of SML is called maximal K-consistent if Ω is K-consistent and everyΩ′ ⊆ SML properly containing Ω is K-inconsistent

Lemma 9 Let Ω ⊆ SML be maximal K-consistent. Then for θ, φ ∈ SML

1. Ω ⊢K θ ⇐⇒ θ ∈ Ω.

2. θ 6∈ Ω ⇐⇒ ¬θ ∈ Ω.

3. θ ∧ φ ∈ Ω ⇐⇒ θ ∈ Ω and φ ∈ Ω.

4. θ ∨ φ ∈ Ω ⇐⇒ θ ∈ Ω or φ ∈ Ω.

5. (θ → φ) ∈ Ω ⇐⇒ θ /∈ Ω or φ ∈ Ω.

Proof (1) If θ /∈ Ω then by maximal consistency Ω∪θmust be inconsistent, so Ω, θ ⊢K φ∧¬φ(any) φ. But then Ω ⊢K θ → (φ ∧ ¬φ) so if Ω ⊢K θ them Ω ⊢K φ ∧ ¬φ and Ω itself would beK-inconsistent, contradiction. Hence Ω 0K θ.

Conversely if θ ∈ Ω then Ω ⊢K θ by Proposition 1(i).

(2) If θ 6∈ Ω then Ω ∪ θ must be K-inconsistent so Ω, θ ⊢K φ and Ω, θ ⊢K ¬φ. By NINΩ ⊢K ¬θ so by (1) ¬θ ∈ Ω.

Conversely, if θ ∈ Ω we cannot also have ¬θ ∈ Ω, otherwise Ω ⊢K θ and Ω ⊢K ¬θ and Ω wouldbe K-inconsistent.

(3) This follows from (1) since Ω ⊢K θ ∧ φ iff Ω ⊢K θ and Ω ⊢K φ.

(4) Right to left follows from (1) since Ω ⊢K φ implies Ω ⊢K θ ∨ φ etc. In the other directionsuppose θ /∈ Ω and φ /∈ Ω. Then by (2) and ¬θ,¬φ ∈ Ω and by (1) and AND Ω ⊢K ¬θ ∧ ¬φ.Hence Ω ⊢K ¬(θ ∨ φ), since ⊢K (¬θ ∧ ¬φ) ↔ ¬(θ ∨ φ), so θ ∨ φ /∈ Ω by (1) and (2).

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(5) Since ⊢K (θ → φ) ↔ (¬θ ∨ φ), Ω ⊢K (θ → φ) iff Ω ⊢K ¬θ ∨ φ and so this case follows from(4) with the help of (1) and (2).

Lemma 10 Every K-consistent subset of SML is contained in a maximal K-consistent subsetof SML.

Proof Assume for the moment that L is countable. Then we can enumerate the sentences ofSML as θ1, θ2, θ3, . . . (exercise!). Let ∆ be a K-consistent subset of SML and set Γ0 = ∆,

Γn+1 =

Γn ∪ θn+1 if this is K−consistentΓn otherwise.

By induction on n each Γn is consistent so⋃

∞

n=0 Γn is consistent and extends ∆ = Γ0. Itis also maximal consistent for if θ (= θn+1 say) could be added to this set and still preserveconsistency then it would already have been added in to Γn so would already be included in⋃

∞

n=0 Γn.

In the case L is uncountable we need to assume a well ordering of L (as follows from the Axiomof Choice). This in turn gives a well ordering of SML and the proof then goes exactly as aboveusing transfinite induction on this ordering (and taking unions at limit ordinals).

Definition. A subset Γ of SML is said to be satisfiable (in K) or K-satisfiable, if there is aframe 〈W,E, V 〉 and some i ∈ V such that i |= Γ.

Note that this notion is independent of L.

The Completeness Theorem for K will follow as an easy corollary of our next theorem.

Theorem 11 Γ ⊆ SML is K-satisfiable if and only if Γ is K-consistent.

Proof First suppose Γ is satisfiable in K. Let 〈W,E, V 〉 be a frame and let i ∈ V such thati |= Γ. Suppose Γ is not consistent in K, hence Γ ⊢K θ and Γ ⊢K ¬θ for some θ ∈ SML. Bythe Correctness Theorem for K we have Γ |=K θ and Γ |=K ¬θ. Since i |= Γ we get i |= θ andi |= ¬θ, a contradiction.

Conversely suppose Γ is consistent in K. LetW be the set of all maximal K-consistent subsetsof SML. Define7

E = (Ω,Λ) ∈ W ×W | Λ ⊇ θ |2θ ∈ Ω

and for each Ω ∈ W define a map VΩ : L −→ 0, 1 by

VΩ(p) = 1 ⇐⇒ p ∈ Ω.

Then with V = (VΩ)Ω∈W , 〈W,E, V 〉 is a frame and we claim that

(†) Ω |= θ ⇐⇒ θ ∈ Ω

for all θ ∈ SML. We prove this by induction on |θ|. If θ = p ∈ L, the claim holds true bydefinition.

7It is useful to notice that the requirement that Λ ⊇ θ |2θ ∈ Ω is equivalent to Ω ⊇ 3θ | θ ∈ Λ .

10

Using Lemma 9 and the induction hypothesis that the result holds for all φ with 1 ≤ |φ| < |θ|we have for each Ω ∈ W :

if θ = ¬φ, Ω |= θ ⇐⇒ Ω 6|= φ ⇐⇒ φ 6∈ Ω ⇐⇒ θ ∈ Ωif θ = (φ ∧ ψ), Ω |= θ ⇐⇒ Ω |= φ and Ω |= ψ ⇐⇒ φ ∈ Ω and ψ ∈ Ω

⇐⇒ (φ ∧ ψ) ∈ Ωif θ = (φ ∨ ψ), Ω |= θ ⇐⇒ Ω |= φ or Ω |= ψ ⇐⇒ φ ∈ Ω or ψ ∈ Ω

⇐⇒ (φ ∨ ψ) ∈ Ωif θ = (φ→ ψ), Ω |= θ ⇐⇒ Ω 6|= φ or Ω |= ψ ⇐⇒ φ 6∈ Ω or ψ ∈ Ω

⇐⇒ (φ→ ψ) ∈ Ω

It remains to show (†) in the case θ = 2φ. If θ ∈ Ω, then by definition of E, whenever(Ω,Λ) ∈ E we have φ ∈ Λ, i.e. Λ |= φ by induction. This proves Ω |= θ.

Conversely suppose θ 6∈ Ω. We must find some Λ ∈ W with (Ω,Λ) ∈ E such that Λ 6|= φ,i.e. φ 6∈ Λ by induction. By Lemma 10 and the definition of E, it is enough to show that¬φ ∪ ψ ∈ SML | 2ψ ∈ Ω is K-consistent.

Otherwise, there are ψ1, . . . , ψk, η ∈ SML with 2ψi ∈ Ω, and ¬φ, ψ1, . . . , ψk ⊢K η ∧ ¬η.By AO, NIN and NNO we get ψ1, . . . , ψk ⊢K φ and hence by Q3 on the Examples sheet2ψ1, . . . ,2ψk ⊢K

2φ. Since 2ψ1, . . . ,2ψk ∈ Ω, this gives Ω ⊢K2φ and hence 2φ = θ ∈ Ω,

the required contradiction. Hence (†) is proved.To complete the proof notice that since Γ is K-consistent, there is some Ω ∈ W containing Γ.By (†) we have Ω |= θ for all θ ∈ Γ, thus 〈W,E, V 〉,Ω |= Γ.

Corollary 12 (The Completeness Theorem for K) Let Γ ⊆ SML and θ ∈ SML. Then

Γ ⊢K θ ⇐⇒ Γ |=K θ.

Proof Γ ⊢K θ is equivalent to Γ∪¬θ beingK-inconsistent (exercise!). Γ |=K θ is equivalentto Γ∪ ¬θ being not satisfiable (another exercise!). Hence the corollary is a reformulation ofTheorem 11.

Of course since proofs are finite we also get the

Theorem 13 (Compactness Theorem for K) Γ ⊆ SML is K-satisfiable if and only ifevery finite subset of Γ is K-satisfiable.

A nice property for a modal logic to possess is that it has the finite model property:

Definition J has the finite model property if every finite J-satisfiable subset Γ of SML, for Lfinite, is satisfiable by a finite frame (for J), that is a frame with a finite set of worlds.

Just for interest we mention that:

Theorem 14 K has the finite model property.

Semantics for other modal logics

By putting further restriction on the directed graph 〈W,E〉, we obtain frames suitable forsemantics of the other modal systems.

Definition. Let 〈W,E, V 〉 be a frame. We says that 〈W,E, V 〉 is a

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• D-frame, if E is serial, i.e. ∀i ∈ W ∃j ∈ W (i, j) ∈ E.

• T-frame, if E is reflexive, i.e. ∀i ∈ W (i, i) ∈ E.

• B-frame, if E is symmetric, i.e. ∀i, j ∈ W [(i, j) ∈ E ⇒ (j, i) ∈ E].

• S4-frame, if E is reflexive and transitive, i.e. ∀i, j, k ∈ W [(i, j), (j, k) ∈ E ⇒ (i, k) ∈ E].

• S5-frame, if E is an equivalence relation, i.e. E is reflexive, symmetric and transitive.

As usual let J be one of the systems K,B,D, T, S4 or S5. For Γ ⊆ SML and θ ∈ SML wedefine

Γ |=J θ ⇐⇒ for every J-frame 〈W,E, V 〉 and each i ∈ W , if i |= Γ then i |= θ.

We say that Γ ⊆ SML is satisfiable in J , or J-satisfiable, if there is a J-frame and a world iof that frame satisfying Γ, i.e. i |= Γ.

Theorem 15 For J = K, T,B,D, S4, S5 and Γ ⊆ SML, Γ is J-satisfiable iff Γ is J-consistent.

Proof In each case the proof is just a refinement of the proof of Theorem 11. We illustratethis with some key steps:

⇐ J = T : Suppose Γ is satisfied at i ∈ W in the reflexive frame 〈W,E, V 〉 but Γ ⊢T θ ∧ ¬θ.Then there is a proof Γ1 | θ1,Γ2 | θ2, . . . ,Γn | θn in T of θ∧¬θ from Γ (so Γn ⊆ Γ, θn = (θ∧¬θ)).We show by induction on m for 1 ≤ m ≤ n that Γm T θm.

This divides into cases according to the justification for Γm | θm being in the proof. The casesfor the rules and axioms of K are exactly as before so all we need to check is the T axiom, i.e.we need to show 2θ T θ. So let 〈W ′, E ′, V ′〉 be any reflexive frame and j ∈ W ′. If it is notthe case that j |= 2θ ⇒ j |= θ then j 2θ and j 2 θ. But then since (j, j) ∈ E (because E isreflexive) we obtain also j θ from j 2 θ, contradiction! Hence j |= 2θ ⇒ j |= θ and 2θ T θas required.

Having shown Γn T θn we see that since i |= Γn (because i |= Γ and Γn ⊆ Γ), i |= θn, i.e.i |= θ ∧ ¬θ, so i |= θ and i |= ¬θ which is impossible. Hence Γ must be T -consistent.

The proofs for the other values of J are exactly similar, we just have to check in each case thatthe ‘new’ J axioms are valid in the J-frames.

⇒ For each value of J the construction of the frame with worlds the maximal consistent subsetsof SML (so satisfying (1)-(5) of Lemma 9) is just as before but with J replacing K, so forexample,

(Ψ,Λ) ∈ E ⇐⇒ Λ ⊇ η |2η ∈ Ψ.The only additional feature is that we must show that the graph 〈W,E〉 we get out has theappropriate structure (reflexive, serial, etc) for J . We show this for a couple of values of J :

Case 1 J = T . We must show the graph is reflexive. Let Ω ∈ W . Then since ⊢T2θ → θ,

2θ ∈ Ω ⇒ Ω ⊢T θ ⇒ θ ∈ Ω by (1). Therefore Ω ⊇ θ |2θ ∈ Ω so (Ω,Ω) ∈ E, as required.

Case 2 J = D. We must show that 〈W,E〉 is serial. Let Ω ∈ W . It is enough to showthat η |2η ∈ Ω is D-consistent since then it can be extended to a maximal D-consistentsubset Λ of SML and we shall have (Ω,Λ) ∈ E. So assume on the contrary that this set isnot D-consistent, say η1, . . . , ηj ∈ η |2η ∈ Ω and η1, . . . , ηj ⊢D φ and η1, . . . , ηj ⊢D ¬φ forsome φ. Then by 3 on the Examples Sheet, 2η1, . . . ,2ηj ⊢D

2φ and 2η1, . . . ,2ηj ⊢D2¬φ.

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Hence Ω ⊢D2φ and Ω ⊢D

2¬φ since 2η1, . . . ,2ηj ∈ Ω. Using now the D axiom (plus IMR,MP etc) we obtain from Ω ⊢D

2φ that Ω ⊢D ¬2¬φ, which with Ω ⊢D2¬φ shows Ω to be

D-consistent, contradiction.

Case 3 J = S5. We must show that E is an equivalence relation on W . Since S5 extends TE is certainly reflexive so it is enough to show that if (Ω,Λ), (Ω,Ψ) ∈ E then (Λ,Ψ) ∈ E sincethis implies both transitivity and symmetry. So suppose Ω,Λ,Ψ ∈ W and

Λ ⊇ η |2η ∈ Ω , Ψ ⊇ η |2η ∈ Ω .

Let 2φ ∈ Λ. It is enough to show that φ ∈ Ψ. But if it isn’t then 2φ /∈ Ω so by (2) ¬2φ ∈ Ω.∴ by (k4) 3¬φ ∈ Ω so 23¬φ ∈ Ω by using the S5 axiom and (1). This means that 3¬φ ∈ Λ,so by (k4) ¬2φ ∈ Λ, which contradicts the consistency of Λ and the result follows.

Exactly as in the case of K we now have T,B,D, S4, S5 are consistent and:

Corollary 16 (Completeness Theorem for J) For Γ ⊆ SML, θ ∈ SML and J one ofK, T,D,B, S4, S5,

Γ ⊢J θ ⇐⇒ Γ J θ.

Corollary 17 (Compactness Theorem for J) Γ ⊆ SML is J-satisfiable iff every finitesubset of Γ is satisfiable.

For interest we mention that we also have:

Theorem 18 (Finite Model Property for J) K, T,D,B, S4, S5 each satisfy the finite modelproperty.

Again Corollary 16 in the forward direction is useful in showing certain non-provability results,all one needs to do is construct a suitable frame of the appropriate type. We give someexamples:

Example 1 S4 properly extends T by the previous example of a reflexive frame W = 0, 1, 2,E = (0, 0), (0, 1), (1, 1), (1, 2), (2, 2), V0(p) = V1(p) = 1, V2(p) = 0 since 0 2p, 0 2 22p so0T (2p→ 22p) whilst ⊢S4 (2p→ 22p).

Example 2 S4 does not extend B (in terms of what is derivable in it) since ⊢B (p → 23p)but ¬(p→ 23p) is consistent with S4 since it holds at vertex 0 in the reflexive and transitiveframe W = 0, 1, E = (0, 0), (0, 1), (1, 1), V0(p) = 1, V1(p) = 0.

Also B does not extend S4 since ⊢S4 (2p→ 22p) whilst ¬(2p→ 22p) holds at the vertex 0in the symmetric frame W = 0, 1, E = (0, 1), (1, 0), V0(p) = 0, V1(p) = 1.

Notice that since θ ⊢S5 23θ and 2θ ⊢S5 22θ the system S5 properly extends both B and S4.[Of course now we have the Completeness Theorem for S5 it is easy to derive these last two,by simply considering equivalence frames notice that they hold with S5 in place of ⊢S5 .]

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Example 3 In the following reflexive, serial and symmetric graph with n+ 2 vertices

0pppppp ¬p

at world 0 we have 0 2np ∧ ¬2mp for every m > n, where 2

np is p preceded by n copies of2. This shows that all the modalities 2n are inequivalent in T,D,B.

Other Interpretations

Although historically 2 was introduced to represent necessity, where the logic S5 seems to bestcapture its properties, there are now many other interpretations studied for which the sameor similar axioms to T, S4, S5 etc. are appropriate. For example treating the propositionalvariables as sentences of some language for the Predicate Calculus and interpreting 2p tomean p is provable in some axiom system, and similarly for sentences built up from these p.

Some other interpretations of 2θ are

θ will be true at some time in the future8 θ is obligatory

θ is against the law θ is believed

θ is known9

and for each of these various axioms, frames etc. are judged to be appropriate.

Logics are also studied which use multiple modalities, a currently particularly popular instancebeing in the analysis ofmulti-agent systems. One example here is where we have agents 1, . . . , nand 2iθ stands for ‘agent i knows θ’. In addition to the usual axioms and rules of SC thefollowing axioms and rules seem appropriate here:

For i = 1, . . . , n axioms 2i(θ → φ) |2iθ → 2iφ and rules

2iθ | θ NEC| θ

|2iθ2iθ |2i2iθ

¬2iθ |2i¬2iθ i.e. S5 for each agent

A suitable frame for this calculus is of the form 〈W,E1, E2, . . . , , Ek, V 〉 where each 〈W,Ei, V 〉is a frame for S5 (i.e. Ei is an equivalence relation on W ). So now given maximal consistentΩ,Λ we put

(Ω,Λ) ∈ Ei ⇐⇒ Λ ⊇ φ |2iφ ∈ Ω .

14

Example Sheet on Modal Logic

1. Show that| ¬θ| ¬3θ

is a derived rule of K.

2. Show that for any of the systems J if ⊢J (θ → φ) then ⊢J (2θ → 2φ) and ⊢J (3θ → 3φ).

3. Show that if θ1, θ2, . . . , θn ⊢J φ then 2θ1,2θ2, . . . ,2θn ⊢J2φ

4. Show Proposition 1(iii) in the case the rule is IMR i.e. show that if Γ, θ ⊢J φ thenΓ ⊢J θ → φ.

5. Show that the following are provable in K

(k1) 2(θ ∧ φ) ↔ (2θ ∧2φ) (k2) 2θ ↔ ¬3¬θ(k3) 2¬θ ↔ ¬3θ (k4) ¬2θ ↔ 3¬θ

6. Show that the following are provable in S4

(s1) 2θ ↔ 22θ (s2) 3θ ↔ 33θ

(s3) 23θ ↔ 2323θ (s4) 32θ ↔ 3232θ

[Hint: For (s3) use (t1) with 23θ in place of θ.]

7. Show that the following are provable in S5

(σ0) 2θ ↔ 22θ (σ1) 3θ ↔ 23θ

(σ2) 2θ ↔ 32θ (σ3) 2(θ ∨ 2φ) ↔ (2θ ∨2φ)

(σ4) 2(θ ∨3φ) ↔ (2θ ∨3φ)

8. Analyze the reasoning used in the original Bill-Monica example.

9. Show that provability (in J) is independent of the particular overlying language chosen byshowing that if

Γ1 | θ1,Γ2 | θ2, . . . ,Γn | θnis a proof in J and Γn ⊆ SML, θn ∈ SML then there is a proof

∆1 | φ1,∆2 | φ2, . . . ,∆k | φk

in J such that ∆k | φk = Γn | θn and each ∆i ⊆ SML, and each φi ∈ SML. [In other words ifΓ ⊢J θ then there is a proof of this which never mentions any proposition variable not alreadymentioned in Γ or θ.]

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10. Let W = a, b, c, , E = (a, b), (b, b), (a, c), (c, a) , L = p, Va(p) = Vb(p) = 1, Vc(p) =0. Which of the following hold at a in the frame 〈W,E, V 〉?

(i) 2p, (ii) 3p , (iii) 22p, (iv) 323¬p, (v) 2(p → 2p)

11. Show that none of the following are provable in K (i.e. from the empty set):

(i) p→ 3p),

(ii) 2(p ∨ q) → (2p ∨ 2q),

(iii) 22p→ 2p,

(iv) 3p→ p,

(v) 3p→ 33p.

12. By showing that K θ and using the Completeness Theorem for K show that ⊢K θ foreach of the following sentences θ:

(i) 3(θ ∧ φ) → (3θ ∧3φ),

(ii) 2(θ → φ) → (3θ → 3φ),

(iii) (2θ ∧3φ) → 3(θ ∧ φ).

13. By using the Completeness Theorems for D and B show that ⊢D ¬(2p ∧ 2¬p) but0B ¬(2p ∧2¬p).Find a sentence θ such that ⊢B θ but 0D θ. [So this question shows that B * D and D * B.]

14. Let 〈W,E, V 〉 be a frame for K, let w ∈ W , w′ /∈ W and let 〈W ′, E ′, V ′〉 be a framewith W ′ = W ∪ w′ E = E ∪ (w′, w) and Vu = V ′

u for u ∈ W . Show that for θ ∈ SML andu ∈ W,

〈W,E, V 〉, u θ ⇐⇒ 〈W,E, V 〉, u θ.

Hence show that if ⊢K2θ then ⊢K θ.

Let K be the result of adding the rule|2θ| θ

to K. Show that

⊢K θ ⇐⇒ ⊢K θ.

[Notice that this does not imply that K = T !]

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15. Let K4 be K augmented with the axiom schema 2θ |22θ. Write Γ K4 θ if for alltransitive frames 〈W,E, V 〉 and i ∈ W , if i Γ then i θ. Outline the key new steps (i.e.beyond those used in proving the corresponding result for K) of a proof that

Γ K4 θ ⇐⇒ Γ ⊢K4 θ.

16. Show that there is a not-transitive frame 〈W,E, V 〉 such that for any i ∈ W and θ ∈ SML,

i |= 2θ → 22θ.

[Compare this with the result in 15.]

17. Let φ ∈ SML and let 〈W,E, V 〉 be a frame with W = N′ E = (n,m) |n < m ∈ N .Show that for any i ∈ N,

i 32φ → (2(2φ → φ) → 2φ).

Show that if we replace N here by the set of rational numbers Q then this schema no longerholds.

18. Let K3 be K augmented with the axiom schema

θ |333θ.

By considering a suitable set of frames or otherwise show that p 0K3 33p.

19. Let L = p and let 〈W,E, V 〉 be the frame W = N,E = (n,m) |n ≤ m+ 1 , V0(p) = 0, vn(p) = 1 for n > 0. Show that in this frame

(i) 2 2p ∧ ¬22p.(ii) For any θ ∈ SML and n ∈ N, if n 2 2θ → 22θ then

n 2 2(22θ → 222θ).

Let J be the system T augmented with the axiom schema

|2(22θ → 222θ) → (2θ → 22θ).

Show that ¬(2p → 22p) is satisfiable in some frame for J (i.e. a frame in which the sentences

2(22θ → 222θ) → (2θ → 22θ) and 2θ → θ

are true at all worlds in this frame) but is not satisfied in any finite frame for J .[Bit harder than the average question. Compare this with the situation for K, T,B, S4, S5

where the finite model property does hold.]

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MATH43032/63032 Second Take Home Test

Solutions to be handed in to the Reception in the ATB by 4.00pm Thursday 22nd March.Solutions handed in after that time will automatically have 20% of the marks deducted forevery day late.

1. Let F be the modal logic K augmented with the additional axiom schema

| 2θ ∨2¬θ

for θ ∈ SML.

Give proofs of the key new features (new in the sense that they do not already essentiallyappear in the corresponding proof for K) in the proof of the following completeness theoremfor F :

For Γ ⊆ SML, θ ∈ SML,

Γ ⊢F θ ⇐⇒ for all frames < W,E, V > such that

∀j, k, l ∈ W, (< l, j >,< l, k >∈ E ⇒ j = k) and

for all i ∈ W, if i |= Γ then i |= θ.

[8 marks]

Hence show that 22p 0F2p. [2 marks]

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Solutions to Modal Logic Examples

1.1. 2¬θ |2¬θ, REF

2. ¬2¬θ | ¬2¬θ, REF

3. 2¬θ,¬2¬θ |2¬θ, MON 1

4. 2¬θ,¬2¬θ | ¬2¬θ, MON 2

5. 2¬θ, | ¬¬2¬θ, NIN 3,4

6. |2¬θ → ¬3θ, IMR 5Hence if we have proof of | ¬θ then NEC gives |2¬θ and by MP with the above proof weobtain | ¬3θ.

2. Assume ⊢J θ → φ. By Proposition 1(iii) (which we will henceforth use without mention)and NEC ⊢J

2(θ → φ). Using the K axiom 2(θ → φ) ⊢J2θ → 2φ and IMR ⊢J

2(θ → φ) →(2θ → 2φ) so ⊢J (2θ → 2φ by MP. Similarly for θ, φ transposed so with AND ⊢J

2θ ↔ 2φ.

Also if ⊢J θ → φ then since⊢J (θ → φ) → (¬φ → ¬θ),

(by Proposition 2 since ⊢SC (p1 → p2) → (¬p2 → ¬p1)), ⊢J ¬φ → ¬θ. By the first part⊢J

2¬φ → 2¬θ and by the above trick again ⊢J ¬2¬θ → ¬2¬φ, as required.

3. From θ1, . . . , θn ⊢J φ IMR repeatedly gives

⊢J (θ1 → (θ2 → . . . (θn → φ) . . .)).

By NEC⊢J

2(θ1 → (θ2 → . . . (θn → φ) . . .)).

By the K axiom

⊢J2(θ1 → (θ2 → . . . (θn → φ) . . .)) → (2θ1 → 2(θ2 → (θ3 → . . . (θn → φ) . . .))

so by MP,⊢J (2θ1 → 2(θ2 → (θ3 → . . . (θn → φ) . . .)).

MON, REF and MP now give

2θ1 ⊢J2(θ2 → (θ3 → . . . (θn → φ) . . .).

The result follows by repeating this step for each of θ2, θ3, . . . , θn.

4. Suppose that Γ1 | θ1, . . . ,Γm | θm is a proof of Γ, θ ⊢J φ, so θm = φ and Γm ⊆ Γ∪ θ. Then

Γ1 | theta1, . . . ,Γm | θm,Γ ∪ θ | θm,Γm ∪ θ − θ | θ → φ

is also a proof (the last two sequents being justified by MON and IMR) and si furthermore aproof of Γ ⊢J θ → φ since θ → θm = θ → φ and

Γm ∪ θ − θ ⊆ Γ− θ ⊆ Γ.

5. (k1) It is enough to show 2(θ ∧ φ) ⊢K2θ ∧2φ and 2θ ∧2φ ⊢K

2(θ ∧ φ). By the notes2(θ ∧ φ) ⊢K

2θ and similarly 2(θ ∧ φ) ⊢K2φ so the first conclusion follows by AND. For the

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second θ ⊢K φ→ (θ∧φ) (using Proposition 2 on p1 SC p2 → (p1∧p2)) so 2θ ⊢K

2(φ → (θ∧φ))by 2 above. Using K, REF and MP, we now get 2θ ⊢K

2φ → 2(θ ∧ φ). REF, MON and MPthis gives 2θ,2φ ⊢K

2(θ ∧ φ) and ANL completes the job.

(k2) Since θ ⊢K ¬¬θ (because p ⊢SC ¬¬p), 2θ ⊢K2¬¬θ (by 2 above). ∴ 2θ,3¬θ ⊢K

2¬¬θand 2θ,3¬θ ⊢K ¬2¬¬θ by REF and MON (notice ¬2¬¬θ = 3¬θ) so by NIN 2θ ⊢K ¬3¬θ.In the other direction, since ¬¬θ ⊢K θ, 2¬¬θ ⊢K

2θ. Also ¬¬2¬¬θ(= ¬3¬θ) ⊢K2¬¬θ so

¬3¬θ ⊢K2θ by REF, MON, IMR, MP.

(k3) By (k2) with ¬θ in place of θ, ⊢K2¬θ ↔ ¬3¬¬θ. Using the facts that ⊢K θ ↔ ¬¬θ,

⊢K3θ ↔ 3¬¬θ, the required ⊢K

2¬θ ↔ ¬3θ follows from the fact that ‘provable equivalence(in K)’ is an equivalence relation.

From (k2) ⊢K2θ ↔ ¬3¬θ and with Proposition 2 and ⊢SC (p1 ↔ p2) → (¬p1 ↔ ¬p2) etc.

we obtain ⊢K ¬2θ ↔ 3¬θ.

6. (s1) 2θ ⊢S4 22θ from the S4 axiom and 22θ ⊢S4 2θ from the T axiom which gives it.

(s2) ⊢S4 2 6= θ ↔ 22¬θ by (s1) so ⊢S4 ¬2¬θ ↔ ¬22¬θ. By (k4), ⊢S4 ¬22¬θ ↔ 3¬2¬θand the result follows by provable equivalence (in S4).

(s3) By (t1) ⊢S4 23θ → 323θ. By 4 above ⊢S4 223θ → 2323θ and using the S4

axiom, ⊢S4 23θ → 223θ, we obtain ⊢S4 23θ → 2323θ. For the converse, by (t1),⊢S4 2¬θ → 32¬θ. By 2 and S4, ⊢S4 2¬θ → 232¬θ. ∴ ⊢S4 ¬232¬θ → ¬2¬θ, so by 2⊢S4 2¬232¬θ → 2¬2¬θ, i.e. ⊢S4 2323θ → 23θ, as required.

(s4) This follows directly from (s3) by replacing θ by ¬θ and putting ¬ in front of bothequivalents.

7. (σ0) ⊢S5 22θ → 2θ is as in the proof of (s1). For the converse, from S5, 3¬θ ⊢S5 23¬θ,so ¬23¬θ ⊢S5 ¬3¬θ , giving by (k2) and 2 that 32θ ⊢S5 2θ∗. By (t1), 2θ ⊢S5 232θ sousing S5, 2θ ⊢S5 232θ. But by ∗ with 4, 232θ ⊢S5 22θ, so 2θ ⊢S5 22θ as required.

(σ1) ⊢S5 23θ → 3θ follows from T whilst by S5 ⊢S5 3θ23θ.

(σ2) This follows easily from (σ1) by replacing θ by ¬θ and using (k2), (k4).

(σ3) Using Proposition 2 with p1 ∨ p2,¬p2 ⊢SC p1 we get θ ∨ 2φ,¬2φ ⊢S5 θ. By 3, 2(θ ∨2φ),2¬2φ ⊢S5 2θ, ∴ 2(θ ∨ 2φ) ⊢S5 2¬2φ → 2θ. By (σ1) and 2 ⊢S5 ¬2φ → 3¬φ,⊢S5 3¬φ → 23¬φ, ⊢S5 23¬φ → 2¬2φ giving 2(θ ∨ 2φ) ⊢S5 ¬2φ → 2θ and hence2(θ ∨ 2φ) ⊢S5 2θ ∨ 2φ by using Proposition 2 and ⊢SC (¬p1 → p2) ↔ (p2 ∨ p1).For the other direction, using Proposition 2, ⊢S5 θ → (θ ∨ 2φ), so ⊢S5 2(θ → (θ ∨ 2φ)) andwith K, 2θ ⊢S5 2(θ ∨2φ). Similarly 22φ ⊢S5 2(θ∨2φ), so 2φ ⊢S5 2(θ ∨2φ) by using (σ0),2 etc. ∴ by DIS and Proposition 2, 2θ ∨2φ ⊢S5 2(θ ∨2φ), as required.

(σ4) This is proved exactly like (σ3) but using (σ1) in place of (σ0).

8. Using b for ‘Bill is telling the truth’ and m for ‘Monica is telling the truth’ the validityof the argument

It is not possible that Bill and Monica are both telling the truthPossibly Monica is telling the truth∴ It is not necessarily the case that Bill is telling the truth

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amounts to the validity of ¬3(b ∧ m),3M ⊢J ¬2b. In fact this follows with J = K sinceusing Proposition ?? ⊢K ¬(b ∧m) → (b→ ¬m) so by NEC (and some sentential stuff)

⊢K2¬(b ∧m) → 2(b→ ¬m) (1)

Since ⊢K ¬3(b ∧m) ↔ 2¬(b ∧m) we get ¬3(b ∧m) ⊢K2¬(b ∧m) and using (1) we obtain

¬3(b ∧ m) ⊢K2(b → ¬m). Applying the K axiom (plus some usual sentential stuff) gets

us to ¬3(b ∧ m) ⊢K2b → 2¬m and hence to ¬3(b ∧ m) ⊢K ¬2¬m → ¬2b and finally to

¬3(b ∧m),¬2¬m ⊢K ¬2b which is what we want since ¬2¬m = 3m.

9. Suppose that Γ1 | θ1, . . . ,Γn | θn was a proof which mentioned some propositional variablep /∈ L, where Γn ⊆ SML and θn ∈ SML. Let q ∈ L and for a sentence φ let φ∗ be the resultof replacing p everywhere in φ by q. By induction on |φ| for φ ∈ SML it is easy to see that φ∗

is also a sentence. Furthermore Γ∗1 | θ∗1, . . . ,Γ∗

n | θ∗n is still a proof. The reason for this, as youcan easily convince yourself, is that if φ | φ is an instance of REF then so is φ∗ | φ∗ whilst if

Ω1 |ψ1, . . . ,Ωt |ψt

Ω |ψ

is an instance of a rule thenΩ∗

1 |ψ∗1, . . . ,Ω

∗t |ψ∗

t

Ω∗ |ψ∗

is an instance of that same rule. Also since Γn, θn don’t mention p, Γ∗n | θ∗n = Γn | θn. So

Γ∗1 | θ∗1, . . . ,Γ∗

n−1 | θ∗n−1, Γn | θn is still a proof with the same final sequent which mentions oneless propositional variable not in L than Γ1 | θ1, . . . ,Γn | θn. Repeating for each of the finitelymany (because the Γi are finite) mentioned propositional not in L enables us to remove themall and obtain a proof which remains entirely within the language L.

10. (i) No, because (a, c) ∈ E but c 2 p.

(ii) Yes, because (a, b) ∈ E and b p.(iii) Yes, because c, b 2p.(iv) Yes, because c ¬p, a 3¬p, c 23¬p, a 323¬p.(v) Yes, because c p→ 2p and b p→ 2p.

11. By the Correctness (or Completeness) Theorem forK it is enough to find a frame 〈W,E, V 〉and a world i ∈ W at which these sentences fail. Suitable frames are:

(i) W = i, E = ∅, Vi(p) = 1

(ii) W = i, j, k, E = (i, j), (i, k), Vi(p) = Vi(q) = 1′ Vj(p) = 1,Vj(q) = 0, Vk(p) = 0, Vk(q) = 1

(iii) W = i, j, E = (i, j), Vi(p) = 1, Vj(p) = 0

(iv) W = i, j, E = (i, j), Vi(p) = 0, Vj(p) = 1

(v) W = i, j, k, E = (i, j), (j, k), Vi(p) = Vj(p) = 1, Vk(p) = 0

12. Let 〈W,E, V 〉 be a frame and i ∈ W throughout.

(i) If i 2 3(θ∧ φ) then i 3(θ ∧φ) → (3θ∧3φ) as required. So suppose i 3(θ ∧φ). [We’dnormally take this previous step as read, indeed I’ll do so from now on.] Then j θ ∧ φ forsome (i, j) ∈ E. ∴ j θ and j φ so, since (i, j) ∈ E, i 3θ and i 3φ. ∴ i 3θ ∧ 3φso again i 3(θ ∧ φ) → (3θ ∧3φ). Since the frame and world i in the frame were arbitrary

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this shows that K3(θ ∧ φ) → (3θ ∧3φ) and by the Completeness Theorem for K this also

holds for ⊢K in place of K . [In the examples which follow we will miss out this common laststep of appealing to the Completeness Theorem etc..]

(ii) Suppose i 2(θ → φ) and i 3θ. Then j θ for some (i, j) ∈ E and also for this j,j θ → φ since i 2(θ → φ) and (i, j) ∈ E. ∴ j φ. ∴ i 3φ, again since (i, j) ∈ E. ∴

i 2(θ → φ) → (3θ → 3φ), as required.

(iii) Suppose i 2θ ∧3φ. Then i 2θ and i 3φ so j φ for some (i, j) ∈ E. ∴ also j θsince i 2θ, so j θ ∧ φ. ∴ i 3(θ ∧ φ). ∴ i (2θ ∧3φ) → 3(θ ∧ φ), as required.

13. Let 〈W,E, V 〉 be a serial frame and i ∈ W . By seriality(!) (i, j) ∈ E for some j ∈ W .Either j p or j ¬p. ∴ either j 2 ¬p or l 2 p so since (i, j) ∈ E, either i 2 2¬p or i 2 2p.∴ i 2 2p ∧ 2¬p. ∴ by the Completeness Thm for D, ⊢D ¬(2p ∧ 2¬p).But in the symmetric frame with W = i, E = ∅, Vi(p) = 1 (or 0). i 2p and i 2¬p soi 2p ∧ 2¬p. ∴ 2B ¬(2p ∧ 2¬p) so by the Completeness Thm for B, 0B ¬(2p ∧ 2¬p).For the converse it is clearly enough to show that 0D p → 23p since ⊢B p → 23p. So bythe Completeness Thm for D it is enough to exhibit a serial frame 〈W,E, V 〉 and i ∈ W suchthat i 2 p → 23p. In this case a suitable frame is W = 0, 1, E = (0, 1), (1, 1), V0(p) =1, V1(p) = 0.

14. The proof is by induction on |θ|. If θ = p then, since u ∈ W , Vu(p) = V ′u(p) so the result

is true. Assume now it holds for all |φ| < |θ|. If θ = ¬φ then

〈W,E, V 〉, u ¬φ ⇐⇒ 〈W,E, V 〉, u 2 φ,

⇐⇒ 〈W ′, E ′, V ′〉, u 2 φ by IH

⇐⇒ 〈W ′, E ′, V ′〉, u ¬φ, as required,

and similarly for the other connectives.

Finally if θ = 2φ then

〈W,E, V 〉, u 2φ ⇐⇒ ∀t, (u, t) ∈ E ⇒ 〈W,E, V 〉, t φ⇐⇒ ∀t, (u, t) ∈ E ′ ⇒ 〈W,E, V 〉, t φ

since because w′ 6= u ∈ W, (u, t) ∈ E ′ ⇐⇒ (u, t) ∈ E,

⇐⇒ ∀t, (u, t) ∈ E ′ ⇒ 〈W ′, E ′, V ′〉, t φby IH since (u, t) ∈ E ′ ⇒ t ∈ W,

⇐⇒ 〈W ′, E ′, V ′〉, u 2φ, as required.

To show the second part suppose that ⊢K2θ but 0K θ. Then by the Completeness thm for

K there is a frame 〈W,E, V 〉 and w ∈ W such that w ¬θ. Let 〈W ′, E ′, V ′〉, w′ be as above.Since ⊢K

2θ, by Completeness 〈W ′, E ′, V ′〉, w′ 2θ so 〈W ′, E ′, V ′〉, w θ since (w′, w) ∈ E ′,contradicting the first part above. ∴⊢K θ.

For the last part it is clear that since K extends K if ⊢K θ then ⊢K θ. In the other directionsuppose that ⊢K θ, say that Γ1 | θ1, . . . ,Γn | θn was a proof in K of ⊢K θ.We show by inductionon m for m = 1, 2, . . . , n that Γm ⊢K θm (which gives the result). Suppose that Γj ⊢K θj forj < m. If Γm | θm is justified in this prof by one of the axioms or rules of K then Γm ⊢K θmfollows as usual by Proposition 1. Otherwise the rule must be the new one, Γm | θm must be| θm and some earlier Γj | θj must be |2θm. By IH ⊢K

2θm so by the second part ⊢K θm, i.eΓm ⊢K θm, as required.

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15. ⇐ Suppose that Γ1 | θ1, . . . ,Γn | θn was a proof of Γ ⊢K4 θ. We prove by induction onm = 1, 2, . . . , n that if 〈W,E, V 〉 is a transitive frame and i ∈ W and i Γm then i θm, i.e.Γm K4 θm. So assume that Γj K4 θj for j < m. If Γm | θm is justified in this proof eitherby being an axiom or rule of K then just the same argument as was used for K shows thatΓm K4 θm. Thus the only new case is when Γm | θm is an instance of the new axiom 2θ |22θ.In this case suppose that i 2θ. To show that i 22θ we must show j 2θ when (i, j) ∈ E.So let (j, r) ∈ E. By transitivity (i, r) ∈ E so r θ (since i 2θ). ∴ j 2θ and hencei 22θ. ∴ Γm K4 θm. This completes the induction. Finally, since Γ ⊇ Γn and θ = θn weobtain Γ K4 θ, as required.

⇒ To prove this direction we assume Γ 0K4 θ and construct a frame 〈W,E, V 〉 and i ∈ Wsuch that i Γ but i 2 θ. The construction follows exactly the course as in the notes for Kbut with ⊢K4 in place of ⊢K . The only new feature is that we must check that the resultingframe is transitive. This amounts to checking that if Ω,Λ,Ψ ∈ W are maximal K4-consistentsubsets of SML and (Ω,Λ), (Λ,Ψ) ∈ E, i.e.

η |2η ∈ Ω ⊆ Λ, η |2η ∈ Λ ⊆ Ψ, (2)

then (Ω,Ψ) ∈ E, i.e. η |2η ∈ Ω ⊆ Ψ. So assume (2) and let θ ∈ η |2η ∈ Ω . Then2θ ∈ Ω so Ω ⊢K4 θ. But also 2θ ⊢K4

22θ (using the new axiom) so Ω ⊢K422θ and by the

first of the proven properties of max.con. sets, 22θ ∈ Ω. θ 2θ ∈ Λ and θ ∈ Ψ by (2). Wehave shown that η |2η ∈ Ω ⊆ Ψ, as required.

16. Let 〈W,E, V 〉 be a frame with W = a, b, E = (a, b), (b, a) and Va = Vb. Clearlythis is not-transitive since (a, b), (b, a) ∈ E but (a, a) /∈ B. Clearly also, by symmetry, for anyφ ∈ SML, a φ ⇐⇒ b φ. ∴ if a 2φ then b 2φ so a 22φ. ∴ a 2φ → 22φ andsimilarly for b.

17. Suppose i ∈ N and i 32φ and i 2(2φ → φ). Then, from the first of these andthe definition of E, m 2φ for some i < m ∈ N, so n φ for all m < n ∈ N, whilst fromthe second ∀j > i, j 2φ → φ. Hence m 2φtoφ so m φ. ∴ k φ for all k ≥ m. ∴

m− 1 2φ. If i = m− 1 then i 2φ, showing in general, that

i 32φ → (2(2φ → φ) → 2φ).

On the other hand if i < m − 1 then repeating the argument again with m − 1 in place ofm shows that m − 2 2φ. Clearly, by the well foundedness of the usual ordering on N, wecan continue this descent until we do reach i, and hence i 2φ, giving the required results asabove.

This proof relied on the well foundedness of N. It fails with Q in place of N because, e.g. r ∈ Q |

√2 < q has no glb in Q. E.g. take Vq(p) = 0 if q <

√2, Vq(p) = 1 if

√2 < q and

φ = p, i = 0.

18. Let J be the set of serial frames 〈W,E, V 〉 such that if (i, j), (j, k) ∈ E then (k, i) ∈ E.Write Γ K3 θ if for all frames 〈W,E, V 〉 ∈ J and i ∈ W , if i Γ then i θ. Claim thatfor any 〈W,E, V 〉 ∈ J and i ∈ W , if i θ then i 333θ. Since, by seriality there aresome j, k ∈ W such that (i, j), (j, k) ∈ E, and hence by definition of J , (k, i) ∈ E. So fromi θ, k 3θ, j 33θ and i 333θ, as required. ∴ the schema θ |333θ is satisfied in allframes in J .

Now suppose that p ⊢ K333p, say Γ1 | θ1, . . . ,Γm | θm was a proof of this. We show byinduction on k = 1, 2, . . . , m that Γk K3 θk. If Γk | θk is an axiom of K or rule of K then the

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result follows as in the course notes. The only essentially new case is when Γk | θk is an instanceof the new axiom θ |333θ. But in this case the above argument has shown that θ K3 333θ.By induction then Γk K3 θm, so p K3 33p. But this is refuted at vertex 1 in the frame(which is in J ) with W = 1, 2, 3 , E = (1, 2), (2, 3), (3, 1) V1(p) = 1, V2(P ) = V3(p) = 0.∴ p 0K3 33p.

19. (i) 0 2 p so 1 2 2p (since (1, 0) ∈ E) and 2 2 22p (since (2, 1) ∈ E). Also

(2, n) ∈ E ⇒ n ∈ 1, 2, 3, . . . ⇒ n p

so 2 2p. ∴ 2 2p ∧ ¬22p, i.e. 2 2 2p→ 22p.

(ii) Suppose that n 2 2θ → 22θ, so n 2θ, n 2 22θ, but n 2(22θ → 222θ). Firstnotice that if n ≤ m then (n, i) ∈ E ⇒ (m, i) ∈ E so since n 2θ, alsom 2θ. ∴ n+1 22θsince —(n + 1, i) ∈ E ⇒ n ≤ i. Since n 2(22θ → 222θ), n + 1 22θ → 222θ son + 1 222θ. ∴ n 22θ, contradiction.

From (i),(ii) and the fact that the frame 〈W,E, V 〉 is reflexive (so 2θ → θ is true at all worlds)we see that this is a frame for J which satisfies ¬(2p → 22p) at some world. However¬(2p → 22p) cannot be satisfied in any finite frame 〈W ′, E ′, V ′〉 for J . Since suppose itwas, at world a1 ∈ W ′ say. By (ii) then a1 2 2(22p → 222p) so there must be a worlda2 ∈ W ′ such that (a1, a2) ∈ E and a2 2 22p → 222p. Applying (ii) again now at a2with 2p in place of p shows the existence of a world a3 ∈ W ′ such that (a2, a3) ∈ E anda3 2 222p → 2222p. Continuing in this way we obtain worlds a1, a2, a3, a4, . . . such thata1 2p, a1 2 2

2p, a2 22p, a2 2 n3p, a3 2

3p, a3 2 24p, a4 2

4p, . . . . But since the Taxiom is true at every world in W ′, if ai 2

i+1p then ai 2jp for all j ≤ i + 1. Hence the

a1, a2, a3, . . . must all be distinct, contradicting the finiteness of 〈W ′, E ′, V ′〉.

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