2. A Motivating Example
Consider the following example of ‘follows’:
It is not possible that Bill and Monica are both telling the truth
Possibly 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 formally
capture 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 by
the propositional (= sentential) calculus, however in doing this we would have changed the
meaning of the statements: Saying ‘Possibly Monica is telling the truth’ is not the same as
saying ‘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 because
clearly the possibility of Monica telling the truth isn’t completely determined by whether or
not she actually is telling the truth. In other words the truth value of Possibly Monica is
telling 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 θ
and φ individually, and the same goes for any compound connective built up from these. So
there is no way of understanding examples such as the above purely within SC, we need to
enlarge 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 propositional
variables.
From L we define the set SML of modal sentences of L using the connectives ∧, ∨, →, ¬ and
the 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 (excluding
commas) and can use proof by induction on the length of θ to show that some property holds
for all θ ∈ SML.
Notice that SL ⊆ SML. Again we have unique readability since the case of 2θ is the same as
for ¬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
3. 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 is
OK here because the form of the rules/postulates as so called positive Horn Clauses means
that such relations are closed under intersections and hence there is a least one, unlike the
situation for Rational Consequence Relations. Furthermore we could give a characterization
of the ‘modal monotone consequence relations’ just as we did for the non-modal SC so in this
sense 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 have
the REF-axiom1 θ | θ for θ ∈ SML, and the rules of SC applied to sentences and subsets of
SML, viz: For Γ ⊆ SML finite etc.
Γ|θ ∆|φ
And right: AND
Γ, ∆ | θ ∧ φ
Γ, θ, φ | ψ
And left: ANL
Γ, θ ∧ φ | ψ
Γ|θ Γ|θ
Or right: and ORR
Γ|φ∨θ Γ|θ ∨φ
Γ, θ | ψ ∆, φ | ψ
Disjunction, Or left: DIS
Γ, ∆, θ ∨ φ | ψ
Γ, θ | φ
Implies right: IMR
Γ|θ → φ
Γ|θ ∆|θ → φ
Modus Ponens: MP
Γ, ∆ | φ
Γ, θ | φ ∆, θ | ¬φ
Not in: NIN
Γ, ∆ | ¬θ
Γ | ¬¬θ
Not not out: NNO
Γ|θ
Γ|θ
Monotonicity, or Weakening: MON
Γ, ∆ | θ
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
NEC
| 2θ
1
The left hand side here should really be the set {θ} rather than just the sentence θ but it is a common
abbreviation to drop the braces in such cases. Similarly we may write Γ, θ | φ for Γ ∪ {θ} | φ.
3
4. 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 | θ
B: K + θ | 23θ S4 : T + 2θ | 22θ
S5 : 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 an
instance of an axiom (of K, T ,. . . ) or for some j1 , . . . , js < i
Γj1 | θj1 . . . Γjs | θjs
Γi | 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 such
that 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 example
A 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
2
We insert the adjective ‘formal’ when we want to distinguish between a proof in this sense and an ‘informal
proof’ such as we would give to justify a theorem.
4
5. 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) if
Γ1 | θ1 , Γ2 | θ2 , . . . , Γk | θk
Γ|θ
is 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 from
φ 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 | θm
is 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) of
Γ ⊢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 θφ be the result of simultaneously replacing each occurrence of pi in θ by φi for
1 ≤ i ≤ n. Then
(i) θφ ∈ SML
(ii) If Γ1 | θ1 , . . . , Γm | θm is a proof in J with Γi ⊆ SML(n) , θi ∈ SML(n) , then Γφ | θ1 , . . . , Γφ | θm
1
φ
m
φ
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 for
K, B, . . . , S5 , we may directly lift proofs and derivability results from SC to these systems. For
example, 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
5
6. In any of the systems J if ⊢J (θ → φ) then ⊢J (2θ → 2φ) and ⊢J (3θ → 3φ). More generally
for 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 these
notes. Of course Propositions 1 and 2 are useful here.] As usual once we have the Completeness
Theorems we will have much more straightforward ways to show these, and similar. On the
other hand we actually need some of these to prove the Completeness Theorems in the first
place!
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 proof
Pθ with last sequent 2θ | 22θ. Hence if Q is a proof in S4 we can obtain a proof in S5 with
the 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 M1 M2 . . . Mn θ is
provably 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 , M1 M2 . . . Mn θ is
provably equivalent to Mn θ (so one of 2θ, 3θ)
3
As usual θ ↔ φ is an abbreviation for (θ → φ) ∧ (φ → θ).
6
7. 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) = 1
if θ = ¬φ, i |= θ ⇐⇒ i |= φ
if θ = (φ ∧ ψ), i |= θ ⇐⇒ i |= φ and i |= ψ
if θ = (φ ∨ ψ), i |= θ ⇐⇒ i |= φ or i |= ψ
if θ = (φ → ψ), i |= θ ⇐⇒ i |= φ or i |= ψ
if 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 |= 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
p p ¬p
0 2p 1 2p 2 2¬p
0 22p 1 2¬p 2 2p
0 2p ∧ ¬22p
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.
4
E is sometimes referred to as an accessibility relation. So if (i, j) ∈ E we say that j is accessible from i.
5
i Γ stands for i φ for all φ ∈ Γ.
7
8. 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 enough
to 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 with
i |= 2(ψ → φ). We have to show i |= (2ψ → 2φ), i.e. we may assume i |= 2ψ and we show
i |= 2φ. Suppose j ∈ W with (i, j) ∈ E but j |= φ. Since i |= 2ψ we have j |= ψ, hence
j |= (ψ → φ). Since (i, j) ∈ E, this contradicts our assumption i |= 2(ψ → φ).
Case 2. Γk |θk is obtained by applying a rule of K 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 propositional
calculus, 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 |=K 2θ.
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 form
Γ K φ by showing Γ K φ, and this simply requires us to exhibit a suitable frame W, E, V
and i ∈ W such that i Γ but i 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 22p so 2p K 22p, and
hence 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.
6
We shall see later that this notion is independent of L.
8
9. Completeness for the modal logic K
Our aim in this section is to show the Completeness Theorem for K, that is that ⊢K and
K
are actually the same thing. The structure of the proof is similar to that for SC (or the
Predicate Calculus), as you will see.
Lemma 8 (i) If ∆ ⊆ SML is consistent in K and θ ∈ SML, then (at least) one of ∆ ∪ {θ}
or ∆ ∪ {¬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 is
totally 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 ⊆ ∞ Γn so because ∆m is finite it must actually be a subset of some Γj . To see this
n=0
suppose ∆m = { ψ1 , . . . , ψr }. Each ψi ∈ ∞ Γn so for each i there is some ni such that
n=0
ψi ∈ Γni . But then since the Γn are increasing each ψi for i = 1, 2, . . . , r is in Γn where n is
the maximum of the ni , so ∆1 | φ1 , . . . , ∆m | φm is a proof of θ ∧ ¬θ from Γn , contradicting the
K-consistency of Γn .
Definition. A subset Ω of SML is called maximal K-consistent if Ω is K-consistent and every
Ω′ ⊆ 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 ∈ 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∪{θ} must 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 be
K-inconsistent, contradiction. Hence Ω K θ.
Conversely if θ ∈ Ω then Ω ⊢K θ by Proposition 1(i).
(2) If θ ∈ Ω then Ω ∪ {θ} must be K-inconsistent so Ω, θ ⊢K φ and Ω, θ ⊢K ¬φ. By NIN
Ω ⊢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 would
be 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 direction
suppose θ ∈ Ω and φ ∈ Ω. Then by (2) and ¬θ, ¬φ ∈ Ω and by (1) and AND Ω ⊢K ¬θ ∧ ¬φ.
/ /
Hence Ω ⊢ ¬(θ ∨ φ), since ⊢K (¬θ ∧ ¬φ) ↔ ¬(θ ∨ φ), so θ ∨ φ ∈ Ω by (1) and (2).
K
/
9
10. (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 subset
of SML.
Proof Assume for the moment that L is countable. Then we can enumerate the sentences of
SML as θ1 , θ2 , θ3 , . . . (exercise!). Let ∆ be a K-consistent subset of SML and set Γ0 = ∆,
Γn ∪ {θn+1 } if this is K−consistent
Γn+1 =
Γn otherwise.
∞
By induction on n each Γn is consistent so n=0 Γn is consistent and extends ∆ = Γ0 . It
is also maximal consistent for if θ (= θn+1 say) could be added to this set and still preserve
consistency 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 Axiom
of Choice). This in turn gives a well ordering of SML and the proof then goes exactly as above
using 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 a
frame 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 that
i |= Γ. Suppose Γ is not consistent in K, hence Γ ⊢K θ and Γ ⊢K ¬θ for some θ ∈ SML. By
the Correctness Theorem for K we have Γ |=K θ and Γ |=K ¬θ. Since i |= Γ we get i |= θ and
i |= ¬θ, a contradiction.
Conversely suppose Γ is consistent in K. Let W be the set of all maximal K-consistent subsets
of 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 by
definition.
7
It is useful to notice that the requirement that Λ ⊇ { θ | 2θ ∈ Ω} is equivalent to Ω ⊇ { 3θ | θ ∈ Λ }.
10
11. 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 |= φ ⇐⇒ φ ∈ Ω ⇐⇒ θ ∈ Ω
if θ = (φ ∧ ψ), Ω |= θ ⇐⇒ Ω |= φ and Ω |= ψ ⇐⇒ φ ∈ Ω and ψ ∈ Ω
⇐⇒ (φ ∧ ψ) ∈ Ω
if θ = (φ ∨ ψ), Ω |= θ ⇐⇒ Ω |= φ or Ω |= ψ ⇐⇒ φ ∈ Ω or ψ ∈ Ω
⇐⇒ (φ ∨ ψ) ∈ Ω
if θ = (φ → ψ), Ω |= θ ⇐⇒ Ω |= φ or Ω |= ψ ⇐⇒ φ ∈ 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 ∈ Ω. We must find some Λ ∈ W with (Ω, Λ) ∈ E such that Λ |= φ,
i.e. φ ∈ 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 sheet
2ψ1 , . . . , 2ψk ⊢K 2φ. Since 2ψ1 , . . . , 2ψk ∈ Ω, this gives Ω ⊢K 2φ 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} being K-inconsistent (exercise!). Γ |=K θ is equivalent
to Γ ∪ {¬θ} being not satisfiable (another exercise!). Hence the corollary is a reformulation of
Theorem 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 if
every 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 L
finite, 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 for
semantics of the other modal systems.
Definition. Let W, E, V be a frame. We says that W, E, V is a
11
12. • 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 we
define
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 i
of 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 illustrate
this 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 cases
for 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 not
the case that j |= 2θ ⇒ j |= θ then j 2θ and j θ. But then since (j, j) ∈ E (because E is
reflexive) we obtain also j θ from j θ, contradiction! Hence j |= 2θ ⇒ j |= θ and 2θ T θ
as 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 that
the ‘new’ J axioms are valid in the J-frames.
⇒ For each value of J the construction of the frame with worlds the maximal consistent subsets
of SML (so satisfying (1)-(5) of Lemma 9) is just as before but with J replacing K, so for
example,
(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 the
appropriate 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 ⊢T 2θ → 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 show
that { η | 2η ∈ Ω } is D-consistent since then it can be extended to a maximal D-consistent
subset Λ of SML and we shall have (Ω, Λ) ∈ E. So assume on the contrary that this set is
not D-consistent, say η1 , . . . , ηj ∈ { η | 2η ∈ Ω } and η1 , . . . , ηj ⊢D φ and η1 , . . . , ηj ⊢D ¬φ for
some φ. Then by 3 on the Examples Sheet, 2η1 , . . . , 2ηj ⊢D 2φ and 2η1 , . . . , 2ηj ⊢D 2¬φ.
12
13. Hence Ω ⊢D 2φ 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 ⊢D 2¬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 T
E is certainly reflexive so it is enough to show that if (Ω, Λ), (Ω, Ψ) ∈ E then (Λ, Ψ) ∈ E since
this 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 of
K, T, D, B, S4 , S5 ,
Γ ⊢J θ ⇐⇒ Γ J θ.
Corollary 17 (Compactness Theorem for J) Γ ⊆ SML is J-satisfiable iff every finite
subset 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 model
property.
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 some
examples:
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 22p so
T
(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 transitive
frame 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 0
in 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 .]
13
14. Example 3 In the following reflexive, serial and symmetric graph with n + 2 vertices
0
p p p p p p ¬p
at world 0 we have 0 2n p ∧ ¬2m p for every m > n, where 2n p is p preceded by n copies of
2. 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 best
capture its properties, there are now many other interpretations studied for which the same
or similar axioms to T, S4 , S5 etc. are appropriate. For example treating the propositional
variables as sentences of some language for the Predicate Calculus and interpreting 2p to
mean 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 instance
being in the analysis of multi-agent systems. One example here is where we have agents 1, . . . , n
and 2i θ stands for ‘agent i knows θ’. In addition to the usual axioms and rules of SC the
following axioms and rules seem appropriate here:
For i = 1, . . . , n axioms 2i (θ → φ) | 2i θ → 2i φ and rules
|θ
2i θ | θ NEC
| 2i θ
2i θ | 2i 2i θ
¬2i θ | 2i ¬2i θ 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 consistent
Ω, Λ we put
(Ω, Λ) ∈ Ei ⇐⇒ Λ ⊇ { φ | 2i φ ∈ Ω }.
14
15. 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 ⊢J 2φ
4. Show Proposition 1(iii) in the case the rule is IMR i.e. show that if Γ, θ ⊢J φ then
Γ ⊢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 by
showing that if
Γ1 | θ1 , Γ2 | θ2 , . . . , Γn | θn
is 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 if
Γ ⊢J θ then there is a proof of this which never mentions any proposition variable not already
mentioned in Γ or θ.]
15
16. 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 for
each 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) but
B
¬(2p ∧ 2¬p).
Find a sentence θ such that ⊢B θ but D
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 frame
/
with W ′ = W ∪ {w ′ } E = E ∪ {(w ′, w)} and Vu = Vu′ for u ∈ W . Show that for θ ∈ SML and
u ∈ W,
W, E, V , u θ ⇐⇒ W, E, V , u θ.
Hence show that if ⊢K 2θ 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 !]
16
17. 15. Let K4 be K augmented with the axiom schema 2θ | 22θ. Write Γ K4 θ if for all
transitive 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
K4
Γ 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 longer
holds.
18. Let K3 be K augmented with the axiom schema
θ | 333θ.
K3
By considering a suitable set of frames or otherwise show that p 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 2θ → 22θ then
n 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.]
17
18. 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 for
every 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 essentially
appear in the corresponding proof for K) in the proof of the following completeness theorem
for 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]
F
Hence show that 22p 2p. [2 marks]
18
19. 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 5
Hence if we have proof of | ¬θ then NEC gives | 2¬θ and by MP with the above proof we
obtain | ¬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) ⊢J 2θ → 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
⊢J 2(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 ⊢J 2(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 a
proof 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) ⊢K 2θ ∧ 2φ and 2θ ∧ 2φ ⊢K 2(θ ∧ φ). By the notes
2(θ ∧ φ) ⊢K 2θ and similarly 2(θ ∧ φ) ⊢K 2φ so the first conclusion follows by AND. For the
19
20. 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 MP
this gives 2θ, 2φ ⊢K 2(θ ∧ φ) and ANL completes the job.
(k2) Since θ ⊢K ¬¬θ (because p ⊢SC ¬¬p), 2θ ⊢K 2¬¬θ (by 2 above). ∴ 2θ, 3¬θ ⊢K 2¬¬θ
and 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) ⊢K 2¬¬θ so
¬3¬θ ⊢K 2θ by REF, MON, IMR, MP.
(k3) By (k2) with ¬θ in place of θ, ⊢K 2¬θ ↔ ¬3¬¬θ. Using the facts that ⊢K θ ↔ ¬¬θ,
⊢K 3θ ↔ 3¬¬θ, the required ⊢K 2¬θ ↔ ¬3θ follows from the fact that ‘provable equivalence
(in K)’ is an equivalence relation.
From (k2) ⊢K 2θ ↔ ¬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 = θ ↔ 22¬θ by (s1) so ⊢S4 ¬2¬θ ↔ ¬22¬θ. By (k4), ⊢S4 ¬22¬θ ↔ 3¬2¬θ
and 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 both
equivalents.
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 so
using 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 hence
2(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)) and
with 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 validity
of the argument
It is not possible that Bill and Monica are both telling the truth
Possibly Monica is telling the truth
∴ It is not necessarily the case that Bill is telling the truth
20
21. amounts to the validity of ¬3(b ∧ m), 3M ⊢J ¬2b. In fact this follows with J = K since
using Proposition ?? ⊢K ¬(b ∧ m) → (b → ¬m) so by NEC (and some sentential stuff)
⊢K 2¬(b ∧ m) → 2(b → ¬m) (1)
Since ⊢K ¬3(b ∧ m) ↔ 2¬(b ∧ m) we get ¬3(b ∧ m) ⊢K 2¬(b ∧ m) and using (1) we obtain
¬3(b ∧ m) ⊢K 2(b → ¬m). Applying the K axiom (plus some usual sentential stuff) gets
us to ¬3(b ∧ m) ⊢K 2b → 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 variable
p ∈ L, where Γn ⊆ SML and θn ∈ SML. Let q ∈ L and for a sentence φ let φ∗ be the result
/
of 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∗ | θ1 , . . . , Γ∗ | θn is still a proof. The reason for this, as you
1
∗
n
∗
can 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 then
Ω∗ | ψ1 , . . . , Ω∗ | ψt
1
∗
t
∗
Ω∗ | ψ ∗
is an instance of that same rule. Also since Γn , θn don’t mention p, Γ∗ | θn = Γn | θn . So
n
∗
Γ∗ | θ1 , . . . , Γn−1 | θn−1, Γn | θn is still a proof with the same final sequent which mentions one
1
∗ ∗ ∗
less propositional variable not in L than Γ1 | θ1 , . . . , Γn | θn . Repeating for each of the finitely
many (because the Γi are finite) mentioned propositional not in L enables us to remove them
all and obtain a proof which remains entirely within the language L.
10. (i) No, because (a, c) ∈ E but c 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 for K 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 3(θ ∧ φ) then i 3(θ ∧ φ) → (3θ ∧ 3φ) as required. So suppose i 3(θ ∧ φ). [We’d
normally take this previous step as read, indeed I’ll do so from now on.] Then j θ ∧ φ for
some (i, j) ∈ E. ∴ j θ and j φ so, since (i, j) ∈ E, i 3θ and i 3φ. ∴ i 3θ ∧ 3φ
so again i 3(θ ∧ φ) → (3θ ∧ 3φ). Since the frame and world i in the frame were arbitrary
21
22. this shows that K 3(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 last
step 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 θ
since 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 ¬p or l p so since (i, j) ∈ E, either i 2¬p or i 2p.
∴ i 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 so
i 2p ∧ 2¬p. ∴ B ¬(2p ∧ 2¬p) so by the Completeness Thm for B, B
¬(2p ∧ 2¬p).
For the converse it is clearly enough to show that D p → 23p since ⊢B p → 23p. So by
the Completeness Thm for D it is enough to exhibit a serial frame W, E, V and i ∈ W such
that i 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) = Vu′ (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 φ,
⇐⇒ W ′ , E ′ , V ′ , u 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 ′ = u ∈ W, (u, t) ∈ E ′ ⇐⇒ (u, t) ∈ E,
⇐⇒ ∀t, (u, t) ∈ E ′ ⇒ W ′, E ′ , V ′ , t φ
by IH since (u, t) ∈ E ′ ⇒ t ∈ W,
⇐⇒ W ′ , E ′ , V ′ , u 2φ, as required.
To show the second part suppose that ⊢K 2θ but K 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 direction
suppose that ⊢K θ, say that Γ1 | θ1 , . . . , Γn | θn was a proof in K of ⊢K θ. We show by induction
on m for m = 1, 2, . . . , n that Γm ⊢K θm (which gives the result). Suppose that Γj ⊢K θj for
j < m. If Γm | θm is justified in this prof by one of the axioms or rules of K then Γm ⊢K θm
follows 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.e
Γm ⊢K θm , as required.
22
23. 15. ⇐ Suppose that Γ1 | θ1 , . . . , Γn | θn was a proof of Γ ⊢K4 θ. We prove by induction on
m = 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 either
by being an axiom or rule of K then just the same argument as was used for K shows that
Γm 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 hence
K4
i 22θ. ∴ Γm θm . This completes the induction. Finally, since Γ ⊇ Γn and θ = θn we
obtain Γ K4 θ, as required.
⇒ To prove this direction we assume Γ K4 θ and construct a frame W, E, V and i ∈ W
such that i Γ but i θ. The construction follows exactly the course as in the notes for K
but with ⊢K4 in place of ⊢K . The only new feature is that we must check that the resulting
frame is transitive. This amounts to checking that if Ω, Λ, Ψ ∈ W are maximal K4-consistent
subsets 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 }. Then
2θ ∈ Ω so Ω ⊢K4 θ. But also 2θ ⊢K4 22θ (using the new axiom) so Ω ⊢K4 22θ and by the
first of the proven properties of max.con. sets, 22θ ∈ Ω. θ 2θ ∈ Λ and θ ∈ Ψ by (2). We
have 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 . Clearly
this is not-transitive since (a, b), (b, a) ∈ E but (a, a) ∈ B. Clearly also, by symmetry, for any
/
φ ∈ SML, a φ ⇐⇒ b φ. ∴ if a 2φ then b 2φ so a 22φ. ∴ a 2φ → 22φ and
similarly for b.
17. Suppose i ∈ N and i 32φ and i 2(2φ → φ). Then, from the first of these and
the definition of E, m 2φ for some i < m ∈ N, so n φ for all m < n ∈ N, whilst from
the 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 of
m shows that m − 2 2φ. Clearly, by the well foundedness of the usual ordering on N, we
can continue this descent until we do reach i, and hence i 2φ, giving the required results as
above.
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 that
for any W, E, V ∈ J and i ∈ W , if i θ then i 333θ. Since, by seriality there are
some j, k ∈ W such that (i, j), (j, k) ∈ E, and hence by definition of J , (k, i) ∈ E. So from
i θ, k 3θ, j 33θ and i 333θ, as required. ∴ the schema θ | 333θ is satisfied in all
frames in J .
Now suppose that p ⊢ K3 33p, say Γ1 | θ1 , . . . , Γm | θm was a proof of this. We show by
induction on k = 1, 2, . . . , m that Γk K3 θk . If Γk | θk is an axiom of K or rule of K then the
23
24. result follows as in the course notes. The only essentially new case is when Γk | θk is an instance
of 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 K3 33p.
19. (i) 0 p so 1 2p (since (1, 0) ∈ E) and 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 2p → 22p.
(ii) Suppose that n 2θ → 22θ, so n 2θ, n 22θ, but n 2(22θ → 222θ). First
notice that if n ≤ m then (n, i) ∈ E ⇒ (m, i) ∈ E so since n 2θ, also m 2θ. ∴ n+1 22θ
since —(n + 1, i) ∈ E ⇒ n ≤ i. Since n 2(22θ → 222θ), n + 1 22θ → 222θ so
n + 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 it
was, at world a1 ∈ W ′ say. By (ii) then a1 2(22p → 222p) so there must be a world
a2 ∈ W ′ such that (a1 , a2 ) ∈ E and a2 22p → 222p. Applying (ii) again now at a2
with 2p in place of p shows the existence of a world a3 ∈ W ′ such that (a2 , a3 ) ∈ E and
a3 222p → 2222p. Continuing in this way we obtain worlds a1 , a2 , a3 , a4 , . . . such that
a1 2p, a1 22 p, a2 22 p, a2 n3 p, a3 23 p, a3 24 p, a4 24 p, . . . . But since the T
axiom is true at every world in W ′ , if ai 2i+1 p then ai 2j p for all j ≤ i + 1. Hence the
a1 , a2 , a3 , . . . must all be distinct, contradicting the finiteness of W ′, E ′ , V ′ .
24
