5.6 Soundness and Completeness
5.6.1 Completeness
Completeness. If M, m⊧C thenM, m⊢C
Our proof of completeness will be by the usual canonical model construction, modified for our language. In defining our canonical model, some care needs to be taken to ensure that the relation R is universal.33
To that end, assume thatM, m⊬C. ThenM,m, andC(C)is consistent. Hence there exists an M.C.S. Λ containing M,m, and C(C).
First, we first define our canonical model: Let MC= ⟨WC, DC, RC, OC, cC, vC⟩ where:
WC={Γ:Γ is a maximally consistent having exactly the same modal singular sen- tences as Λ.}.
DC={n: n occurs in some maximally consistent set Γ∈WC}.
ΓRCΘ iff 1) Γ and Θ are inWC and 2) for all terms singular termscand termsA
ifcLe A∈Γ then c e A∈Θ.
OC(Γ)={n: noccurs in Γ and Γ in WC}.
cC(n) =n.
v(Γ, A) = {n∶naA∈Γ}
At first glance, it may seem that our requirement that Γ is a maximally consistent set having exactly the same modal singular sentences as Λ is too weak a condition to ensure that Γ and Λ make all of the same modal formulae true. In fact it is not, as the following proof shows:
For all M.C.S. Γ, ∆ if Γ and ∆ contain exactly same modal singular sentences then for all modal formulae φ φ∈Γ if and only ifφ∈∆.
Take arbitrary Γ and ∆ and assume the antecedent. We only need to consider the 16 possible modal categorical formulae. For the cases of A∇i B recall thatA∇i B∈Γ if and only if ∃msuch thatmMa A∈Γ andm∇a B∈Γ by the M.C.S. Lemma 23. We then have
m Ma A ∈ Γ and m ∇a B ∈ Γ if and only if m Ma A ∈ ∆ and m ∇a B ∈ ∆ since Γ and ∆ contain exactly the same modal singular formulae. An analogous proof holds for the o
categorical formulae using M.C.S. lemma 24.
For the various universal propositions, we prove the cases forALa B and observe that the rest are similar. Assume thatA La B ∈Γ this holds if and only if C(ALa B) ∉Γ by M.C.S. 2, if and only ifAMo B∉Γ by MCS Lemma 24, this holds if and only if there is somem such thatmMa A∈Γ and for allmifmMa A∈Γ thenmLa B∈Γ. However, since Γ and ∆ agree on all singular modal formulae, both of these properties are preserved in ∆. Hence there is some m such that m Ma A ∈∆ and for all m if mMa A∈ ∆ then
mLa B∈∆. Hence by M.C.S. Lemmas 24 and 2, it follows thatALa B∈∆.
With these definitions in place, it is routine to verify that we can define the canonical analogues and that they have the following properties:
i VC(Γ, A) =v(Γ, A) ∩O(Γ) = {n∶naA∈Γ}.
ii VC(Γ,¬A) =D∖ (v(Γ, A) ∩O(Γ)) = {n∶naA∉Γ}.
iii MC(Γ, A) = {n∶ There is some ∆ such that ΓR∆ and n a A∈∆}
= {n∶ There is some ∆ such thatn a A∈∆}
iv LC(Γ, A) = {n∶For all ∆ if ΓR∆ thennLa A∈∆} = {n∶For all ∆nLa A∈∆} In the proof of i., the first equivalence is definitional. The proof of ii follows by basic set theory. In the case of iii. the right to left inclusion is trivial. For the left to right direction, take an arbitrary term m∈ {n∶ There is some ∆ such thatn a A∈∆}. We need to show that ΓR∆ this follows immediately by our construction ofW. The reason for this is because of how we constructed WC. To see this, observe that, for any two M.C.S.s Γ and ∆, Γ and ∆ contain all of the same modalised singular formulae. (Recall that Γ and ∆ are M.C.S.s in WC). Now, observe that the contrapositive of the second condition onRC and the maximality of Γ and Θ this is equivalent to:
For all singular terms c and terms AifcaA∈Θ thencMa A∈Γ.
Now, clearly since all the M.C.S. that are in WC agree on their singular modal for- mulae, the consequent of the hypothetical is satisfied. This form of the R condition is very useful.
Hence the consequent of the contrapositive of R is satisfied. The case of iv. is similar.
Semantic clauses for the various formulae are analogous to the ones given before and can be seen below.
As an easy corollary of this and the MCS Lemma, observe that we have the following:
1. if ΓRCΘ andAiB∈Θ then AMi B∈Γ 2. if ΓRCΘ andALe B∈Γ then A e B∈Θ
Second, observe that we an alternative way to demonstrate that ΓRCΓ. Take arbi- trary cand A. From cMe A∈Γ it follows thatc e A∈Γ by the rule reflexivity.
Before proving each categorical case, the following lemmas will be very useful in what follows.
Existence Lemma
Let Γ be a Maximally Consistent Set inWC. Then we claim that:
AMi B∈Γ if and only if there exists M.C.Ss ∆ and Ξ such that, for some m, 1. mMa A∈Γ andmMa B∈Γ
2. m a A∈∆ and m a B∈Ξ 3. ΓRC∆ and ΓRCΞ, and so
4. MC(Γ, A) ≠ ∅and MC(Γ, B) ≠ ∅. Left to Right:
Assume that AMi B∈Γ. Then by the MCS Lemma part 21 for some m,mMa A∈Γ and
mMa B∈Γ, proving (1).
We construct ∆ and Ξ as follows: let δ = {bLe A∈Γ : for every singular term b and categorical termA} ∪ {bLa A∈Γ : for every singular term b and categorical term A}.
By constructionδ⊆Γ and soδ is consistent. We claim thatδ∪{maA}andδ∪{maB}
are both consistent. Assume not. We prove the case forδ∪{maA}and observe that the other case is analogous. First, suppose δ∪ {maA} is inconsistent: it follows by C-Intro that δ ⊢ meA. As δ is a set of modalised singular formulae, it follows by Necessity Introduction that δ ⊢ mLe A. Since δ ⊆Γ it follows that Γ ⊢m Le A contradicting the consistency of Γ.
Let ∆ be an MCS extending δ∪ {maA} and let Ξ be an MCS extending δ∪ {maB}. First, observe that ∆ ∈ WC and Ξ ∈ WC. This follows since Γ ∈ WC and ∆ and Ξ agree with Γ on all modalised singular formulae. Second, observe that m a A∈∆ and
m a B∈Ξ, proving (2).
Now recall that ΓRCΘ iff for all singular terms c and terms A if c Le A ∈ Γ then
c e A∈Θ. Take an arbitrary singular term g and term A such thatgLe A∈Γ. It then follows that g Le A∈ δ and hence by construction that gLe A∈∆ and g Le A∈ Ξ. That
g e A∈∆ andg e A∈Ξ follows by the ruleReflexivity. Hence ΓRC∆ and ΓRCΞ, proving (3).
Proof of 4: This follows from 2 and 3.
Right to Left
Immediate from 1. and the MCS Lemma part 21.
In what follows we will mostly be using Existence Lemma 4. The previous result, togeather with the results:
1. if ΓRCΘ andAiB∈Θ then AMi B∈Γ 2. if ΓRCΘ andALe B∈Γ then A e B∈Θ
forms the basis of what is normally called the Existence Lemma in standard treatments of modal logic. e.g. [13, p.200]. As will become clear, our existence lemma together with these two observations will have a very similar function.
To establish some other important properties, the following lemmas will be used ex- tensively:
MC-Lemma
Let Γ be an M.C.S. Then for all singular termscand terms, A:
c∈MC(Γ, A) if and only ifcMa A∈Γ.
Proof:
Left to Right direction:
Assume that c ∈ MC(Γ, A). Then by iii on 138 there is some M.C.S. ∆ such that 1) ΓR∆ and 2) caA∈∆. Recall that the contrapositive of 1) states that:
ifc a A∈∆ thencMa A∈Γ. This combined with 2) entails that cMa A∈Γ as desired.
Right to Left direction:
Assume thatcMa A∈Γ. We construct an M.C.S. ∆ such thatcaA∈∆ and ΓR∆. Let δ = {d Le A ∶ d e AL ∈ Γ}. We claim that δ∪ {caA} is consistent. Assume it is not. Then δ ⊢ ceA by C-Introduction. Since δ is modalised, it follows by Necessity Introduction thatδ⊢cLe A. Now, since δ⊆Γ it follows thatcLe A∈Γ, contradicting the consistency of Γ. So δ∪ {caA} is consistent. Let ∆ be an M.C.S. based on δ∪ {caA}. Clearly caA ∈∆, per construction. Likewise, ΓR∆, since per construction ∆ contains every instance ofdLe A for any dand A that occur in Γ.
As an easy corollary of this construction, observe that if Γ∈WC then ∆∈WC. This follows since ∆ contains every instance ofdMa A for any dand A that occur in Γ, as we noted above.
LC−Lemma
Let Γ be an M.C.S. Then for all singular termscand terms, A:
c∈LC(Γ, A) if and only if cLa A∈Γ.
Proof:
Left to Right direction:
Assume thatc∈LC(Γ, A). Assume thatcLa A∉Γ. Then cMe A∈Γ, since Γ is an M.C.S. We claim that if this were the case then there exists and M.C.S. ∆ such thatc e A∈∆ and ΓR∆, contradicting our assumption that c∈LC(Γ, A).
We construct ∆ as follows: Let δ = {ALe B ∶ALe B∈Γ}. We claim that δ∪ {ceA} is consistent. This follows by an argument analogous to the one used in theMC only using the negative singular form ofNecessity Introduction. As before ∆ is an M.C.S. based on
Right to Left direction:
Assume that cLa A∈Γ. Now, assume that there is some ∆ such that ΓR∆ andceA∈∆ i.e. c ∉ LC(w, A). Then c a A ∈ ∆ since c La A ∈ Γ and ΓR∆. This contradicts the consistency of Γ. Hence c∈LC(w, A).
With these two lemmas in place, we now introduce the following notational shorthand: Similarly to our semantics, we define the following negative operations:
MC(Γ,¬A) = {c∶ΓR∆ and ceA∈∆}
LC(Γ,¬A) = {c∶ if ΓR∆ thenceA∈∆}
From theMC and theLC Lemmas we have the following easy corollaries: For all MCS Γ
1. c∈MC(Γ,¬A) if and only if cMe A∈Γ 2. c∈LC(Γ,¬A) if and only ifcLe A∈Γ
To prove 1. observe thatc∈MC(Π,¬A) if and only ifc∉LC(Π, A). HencecLa A∉Π. Since Π is an M.C.S. it follows thatcMe A∈Π
The proof of 2. is similar.
In what follows we will refer to 1. as the ¯MC Lemma and 2. as the ¯LC Lemma.
We now prove the main lemma for our completeness proof:
Truth Lemma:
For all formulae φ, and all MCS Π∈WC we have:
φ∈Π if and only if MC,Π⊧φ
As each categorical and singular proposition has a contradictory, by MCS Lemma 1. we will only need to consider the left to right direction. The right to left direction will follow from the observation that if φ∉Π then C(φ) ∈Π (by the MCS Lemma) and so
MC,Π⊧ C(φ)(which will follow from the left to right direction) which entailsMC,Π⊭φ
since Π is an MCS.
For singular propositions we have the following cases to consider: 1. if caA∈Π thenc∈VC(Π, A). 2. if ceA∈Π thenc∉VC(Π, A). 3. if cLa A∈Π thenc∈LC(Π, A). 4. if cMe A∈Π thenc∉LC(Π, A). 5. if cMa A∈Π thenc∈MC(Π, A). 6. if cLe A∈Π thenc∉MC(Π, A).
7. ifcQa A∈Π thenc∈M(w, A) ∩M(w,¬A). 8. ifcQe A∈Π thenc∈M(w, A) ∩M(w,¬A). 9. ifc ¯ Q a A∈Π thenc∉M(w, A) ∩M(w,¬A). 10. ifcQa A¯ ∈Π thenc∉M(w, A) ∩M(w,¬A).
The cases of 1 and 2 are trivial and follow by construction. 3-6 follow from the
MC Lemma and the LC Lemma. 7-10 are also easy consequences of the MC and ¯MC
Lemmas.
For categorical formulae we have 16 cases to consider.
Assertoric cases:
Assume thatAaB ∈Π. First observe thatVC(Π, A)is non-empty. To see this observe that since Π is a maximally consistent set, it follows by the MCS Lemma thatAiB∈Π and so there is some m such that maA ∈ Π. Hence VC(Π, A) is non-empty. Next, observe that VC(Π, A) ⊆ VC(Π, B). To see this, assume not. Then there is some n
such that naA∈ Π and neB ∈ Π. Then it follows by MCS Lemma 22 that Π ⊢ AoB
and soAoB∈Π, contradicting the consistency of Π. Hence VC(Π, A) ⊆VC(Π, B) and
VC(Π, A) is non-empty. i.e. MC,Π⊧AaB.
Assume that AiB ∈Π. We claim that VC(Π, A) ∩VC(Π, B) ≠ ∅. This holds if and only if there is somem such that maA∈Γ and maB∈Γ. This is immediate from MCS Lemma 20. FromVC(Π, A) ∩VC(Π, B) ≠ ∅it follows that MC,Π⊧AiB, as desired.
The proofs for eand oare corollaries.
Modal Propositions: 1. IfAMa B∈Π thenMC,Π⊧AMa B 2. IfAMe B∈Π thenMC,Π⊧AMe B 3. IfAMi B∈Π thenMC,Π⊧AMi B 4. IfAMo B∈Π thenMC,Π⊧AMo B 5. IfALa B∈Π thenMC,Π⊧ALa B 6. IfALe B∈Π thenMC,Π⊧ALe B 7. IfALi B∈Π thenMC,Π⊧ALi B 8. IfALo B∈Π thenMC,Π⊧ALo B
9. If AQa B∈Π thenMC,Π⊧AQa B 10. If AQe B∈Π thenMC,Π⊧AQe B 11. If AQi B∈Π thenMC,Π⊧AQi B 12. If AQo B∈Π thenMC,Π⊧AQo B 13. If AQa B¯ ∈Π thenMC,Π⊧AQa B¯ 14. If AQe B¯ ∈Π thenMC,Π⊧AQe B¯ 15. If A ¯ Q i B∈Π thenMC,Π⊧A ¯ Q i B 16. If A ¯ Q o B∈Π thenMC,Π⊧A ¯ Q o B Proof of 1:
Assume that A Ma B ∈ Π. We need to show two things. First, that MC(Π, A) is non-empty and second, thatMC(Π, A) ⊆MC(Π, B). Observe that bySubalternationM
we have AMi B∈Π. By the Existence Lemma it follows that MC(Π, A) ≠ ∅.
To prove the second, take an arbitrary termmand assume thatm∈MC(Π, A). Then by the MC Lemma it follows that mMa A∈Π However, it follows from AMa B∈Π and
mMa A∈Π that mMa B∈Π, using DDO. Hence, m∈MC(Π, B) by theMC Lemma, as required.
Proof of 2:
Assume thatAMe B∈Π. We need to show thatMC(Π, A)∩LC(Π, B) = ∅. So, assume that MC(Π, A) ∩LC(Π, B) ≠ ∅.
From our assumption it follows that there is some term n such that n∈MC(Π, A) ∩ LC(Π, B). Then by the MC Lemma it follows that n Ma A ∈ Π and the LC Lemma entails thatnLa B ∈Π. However, nMa A∈Π and our assumption that AMe B∈Π entails
nMe B∈Π by the DDN, contradicting the consistency of Π.
Proof of 3:
Assume thatAMi B∈Π. We need to show thatMC(Π, A)∩MC(Π, B) ≠ ∅. This follows immediately from the Existence Lemma and theMC Lemma.
Proof of 4:
Assume that A Mo B ∈ Π. We need to show that either MC(Π, A) is empty, or that
MC(Π, A) ⊈LC(Π, B).
Assume that neither of these are the case, then 1)MC(Π, A) ≠ ∅and 2)MC(Π, A) ⊆ LC(Π, B). From 1) it follows that there is some M.C.S. Λ and some term m such that
maA ∈ Λ and ΠRΛ. By an argument analogous to the ones we have used above, it follows thatmMa A∈Π and so Π⊧mMa A.
Now, assume thatnMa A∈Π for an arbitrary n. Then, by theMC Lemma it follows that n∈MC(Π, A). By 2) it follows thatn∈LC(Π, B). By the LC Lemma, it follows thatnLa B ∈Π. Hence by the MCS Lemma 26, it follows that ALa B ∈Π, contradicting the consistency of Π.
Necessary Propositions:
Proof of 5: Assume thatALa B∈Π. We need to show that 1)MC(Π, A)is non-empty and that 2)MC(Π, A) ⊆LC(Π, B).
To show 1), observe that ALa B ⊢A Mi B. It then follows by the Existence Lemma