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The Logic of Logical Necessity

Andrew Bacon

Kit Fine

March 12, 2021

Abstract

Prior to Kripke’s seminal work on the semantics of modal logic, McKinsey oﬀered

an alternative interpretation of the necessity operator, inspired by the Bolzano-Tarski

notion of logical truth. According to this interpretation, ‘it is necessary that A’ is true

just in case every sentence with the same logical form as A is true. In our paper, we

investigate this interpretation of the modal operator, resolving some technical ques-

tions, and relating it to the logical interpretation of modality and some views in modal

metaphysics. In particular, we present an hitherto unpublished solution to problems

41 and 42 from Friedman’s 102 problems, which uses a diﬀerent method of proof from

the solution presented in the paper of Tadeusz Prucnal.

A common conception of a logical truth, often credited to Bolzano, is that of a sentence

true in virtue of its logical form alone. In a given interpreted language one might make this

precise by stipulating a sentence to be logically true if and only if the result of uniformly

substituting any of the non-logical constants with expressions of the same grammatical

category is true, and dually, logically consistent if and only if some substitution instance is

true. For instance, ‘if John is tall then John is tall’ is a logical truth, since the result of

substituting any name and predicate for ‘John’ and ‘is tall’ respectively results in a truth,

whereas ‘John is tall’ is not a logical truth because it is either already false or the result of

substituting the predicate ‘not tall’ for ‘tall’ in it is false.

This analysis makes salient a formal analogy between the notion of a substitution in

Bolzano’s deﬁnition of logical truth and logical consistency, and the notion of a possible

world in the analysis of necessity and possibility in a Kripke model. Indeed, prior to Kripke’s

work on the semantics of modal logic, McKinsey proposed a substitutional interpretation —

or more accurately, a constraint on an interpretation — of the modal operator exactly along

Bolzanean lines.1Given a language Lcontaining the usual truth functional connectives and

a unary connective 2, McKinsey laid out some conditions that the set of true sentences

of this language should satisfy, on the intended interpretation of 2. Apart from the usual

clauses for the truth functional compounds, McKinsey requires that for every sentence A:

We cross generations in paying tribute to Saul’s work on modal logic, one of us having learned of his

work at its outset and the other having learned of it after several decades of development. He has done

more than anyone to put the semantical study of modal logic on a ﬁrm technical footing, and our debt to

his pioneering work should be evident throughout the paper.

Thanks are due to Lloyd Humberstone and Peter Fritz for correspondence with AB on this material that

greatly beneﬁted the paper. Theorems 13, 24 and 34 are due to AB, and theorems 9, 26, 41 and section 6

are due to KF. The results in section 4.2 were joint. Other minor propositions and observations are due to

AB unless otherwise indicated.

1See McKinsey (1945).

1

The Substitutional Constraint 2Ais true if and only if iA is true for every substitution

iof the language.

McKinsey leaves it open what sort of language Lis, and the exact nature of the class of

substitutions, stipulating only that the sentences of the language be closed under the truth-

functional and modal connectives, and that the substitutions be closed under composition

and contain a trivial substitution. We will consider more concrete versions shortly.

Under an alternative approach, the logical truths are not as Bolzano characterized them,

but are nonetheless given by some theory ∆. Since ∆ is being informally understood as the

set of logical truths we should require that ∆ be a subset of the truths, and one could then

replace McKinsey’s constraint with:

The Metalogical Constraint 2Ais true if and only if A∈∆

constraining the interpretation of the modal operator 2to include in its extension propo-

sitions expressed by sentences in ∆ and to exclude propositions expressed by sentences not

in ∆. Provided the theory ∆ is closed under the rule of substitution, one direction of McK-

insey’s constraint is ensured, although not necessarily the other. In the case where Lis the

language of propositional modal logic, and ∆ a logic (which we may assume to be at least

closed under modus ponens and the rule of necessitation), we obtain an interpretation of

propositional modal logic hit upon independently by Meyer and Fine in the 70s.2

Whether or not these constraints are plausible will depend both on the interpretation of

the modal operator and on the range of interpretations of the non-logical constants of the

language in which the constraints are formulated.

Under what interpretations of the modal operator could these schemas be plausible? One

possibility, following Quine, is to treat the modal operator as ‘crypto-quotational’, so that a

sentence embedded under a modal operator should be understood as residing inside invisible

quotation marks, and the modality ascribes some metalinguistic property to the embedded

sentence. But although we motivated the constraints by analogy with the metalinguistic

notion of logical truth, one doesn’t have to identify the modality under either constraint

with a metalinguistic property. One could instead posit a genuine propositional operator

that yields an interpretation of 2under which the substitutional or metalogical conditions

are satisﬁed with respect to a suitable language L. The schemas then impose a substantive

constraint on a propositional operator by articulating a sense in which the postulated notion

of logical necessity stands to the world as logical truth stands to language, without thereby

identifying the two.

The constraints are not plausible on arbitrary interpretations of the non-logical con-

stants either. For instance, if the language in question contained predicates ‘bachelor’ and

‘married’ with their customary meanings then either constraint will imply, given reasonable

side conditions, that it is possible that there are married bachelors.3While this result might

be acceptable when it is interpreted as a metalinguistic logical consistency claim in disguise,

2Speciﬁcally, the metalogical constraint is satisﬁed by any metavaluation (see Meyer (1971), and section

5.2 below): a valuation mapping sentences of propositional modal logic to 1 or 0, satisfying the conditions

that v(A∧B) = min(v(A), v(B)), v(¬A) = 1 −v(A), and v(2A) = 1 iﬀ A∈∆. If, additionally, v(A) = 1

whenever A∈∆ for any metavaluation v, ∆ is called coherent. Fine applies the metalogical interpretation

of 2to obtain simple proofs of the disjunction problem in modal and intuitionistic logics. For instance, any

coherent modal logic, ∆, will have the disjunction property, for if 2A1∨... ∨2An∈∆, then v(2A1∨... ∨

2An) = 1 is true in any metavaluation vbased on ∆, and so Ai∈∆ for some i.

3We can assume ∆ is a logical theory, so it will not prove speciﬁc relationships concerning the non-

logical constants, and we also assume the substitutions are rich enough we may substitute for ‘married’ and

‘bachelor’ predicates, like ‘married’ and ‘not a bachelor’, that are coinstantiated.

2

it is arguably not so on the worldly interpretation proposed above. For plausibly, to be a

bachelor just is to be a man who is not married, so, by Leibniz’s law, the alleged possibility

would imply that it is possible that there are married men who are not married, which is

certainly not true on any candidate interpretation of the modal operator.(Such an appeal to

Leibniz’s law would not be legitimate on a crypto-quotational reading of the modal opera-

tor as logical consistency, since Leibniz’s law does not permit the substitution of identicals

within quotation marks.)

One can avoid these untoward results by working in what Russell calls a logically per-

fect language: a language where the non-logical constants do not denote logically complex

properties and no two non-logical constants codenote.4The former requirement rules out

predicates like ‘bachelor’ that denote conjunctions of simpler properties, and the latter pairs

of synonymous predicates like ‘lawyer’ and ‘attorney’. The schemas then capture a sort of

Humean vision in which the logically simple, or fundamental, properties and relations can,

in the proposed sense of ‘can’, stand in any consistent logical relationship to one another.

Lastly, McKinsey left it open what sort of language could be plugged into his constraint.

It could simply be the language of propositional modal logic — and this will be the option

we will explore in this paper — although it could be a more expressive language, such as a

ﬁrst-order or higher-order language. If it is a higher-order language there is the possibility

of a Tarskian analysis of logical truth. A sentence A(c1...cn) in non-logical constants c1...cn

is a logical truth, according to Tarski, when it is true under every possible interpretation of

c1...cn. We can achieve generality over the interpretations c1...cnin a higher-order language

by simply quantifying into the positions they occupy, leaving the logical truth of A(c1...cn)

amounting to the truth of ∀x1...xn.A(x1...xn), where each xiis a variable of the same type

as ci. Thus, for instance, the logical truth of ‘If John is tall then John is tall’ amounts

to the mere truth of ‘for any individual xand property Yif xis Ythen xis Y’. In this

case, we can fully internalize what in the ﬁrst two constraints was stated in metalinguistic

terms, since ∀x1...xn.A(x1...xn) is a sentence of the object language. So another constraint,

in the same spirit as the substitutional and metalogical constraints, replaces the Bolzanoian

conception of logical truth with the Tarskian, yielding:

The Tarskian Constraint 2A(c1...cn)↔ ∀x1...xn.A(x1...xn)

And this can likewise be thought of as articulating a Humean metaphysics of freely recom-

binable fundamental entities.5

The paper is organized as follows. In section 1, McKinsey’s substitutional constraint

on the interpretation of necessity is made precise within the context of propositional modal

logic, and associated notions of valuation and validity are deﬁned. In section 2 we raise

the question of when an assignment of truth values to the sentence letters can be extended

uniquely to a valuation of the modal language satisfying the substitutional constraint. We

show that unique extensions exist for any class of non-modal substitutions, and for the full

substitution class we present a previous unpublished proof of the existence of an extension,

leaving the uniqueness as a conjecture. Sections 3 and 4 concern the logic of logical necessity.

4Russell writes: ‘In a logically perfect language the words in a proposition would correspond one by

one with the components of the corresponding fact, with the exception of such words as “or”, “not”, “if”,

“then”, which have a diﬀerent function. In a logically perfect language, there will be one word and no more

for every simple object, and everything that is not simple will be expressed by a combination of words, by a

combination derived, of course, from the words for the simple things that enter in, one word for each simple

component.’ Russell (1940), p25

5This principle is the main subject of Bacon (2020).

3

We consider the validity of various principles of modal logic with respect to various substitu-

tion classes, such as the McKinsey and Grzegorczyk axioms, and settle some questions raised

in Humberstone (2016). Section 5 treats the metalogical and Tarskian constraints, and the

paper concludes with some remarks on the substitutional approach to modal predicate logic.

1 Substitutional interpretations of 2

1.1 Preliminaries

McKinsey does not specify the language he is working in or provide a concrete account of

what he means by a substitution. Rather, he proceeds by imposing some abstract constraints

on the language and the substitutions. He assumes that the language is closed under the

formation of truth-functional compounds and a necessity operator. For simplicity, we will

assume that the language Lis closed at least under conjunction and negation and the

formation of necessity sentences:

L1. If A, B ∈ L then (A∧B)∈ L

L2. If A∈ L then ¬A∈ L

L3. If A∈ L then 2A∈ L

We will regiment McKinsey’s account of substitution in terms of a set, S, of abstract sub-

stitutions and an action µ:S× L → L where µ(i, A) — written as iA — informally

represents the result of applying the substitution i∈Sto a sentence Aof Lto produce

another sentence of L. We shall say that a pair (S, µ) is a substitution class if and only

if conditions Commutativity, Identity and Composition, below, are satisﬁed. We will often

suppress mention of the action µ, and refer to a substitution class solely by its associated

set of substitutions provided no ambiguity arises.

Commutativity The action of each substitution i∈Son Lcommutes with the logical

connectives, ∧,¬,2:

1. i(A∧B)=(iA ∧iB)

2. i¬A=¬iA

3. i2A=2iA

Identity There is a substitution i∈Ssuch that iA =Afor every sentence Aof L.

Composition For any two substitutions, i, j ∈Sthere is as substitution k∈Ssuch that

kA =i(jA) for every sentence Aof L

As previously mentioned, these constraints may be satisﬁed by some very expressive lan-

guages, including ﬁrst and higher-order languages, however we will mostly restrict our atten-

tion to propositional languages. Going forward, we will use Lto refer to the propositional

modal language with letters p0, p1, p2, ... and connectives ∧,¬,2. For technical reasons, it

will sometimes be convenient to suppose that the language contains primitive 0-ary con-

stants >and ⊥, in which case the Commutativity condition must be extended to include

the equations i>=>and i⊥=⊥for any i∈S. (Note that these equations might not hold

if ⊥, for instance, was treated as a deﬁned connective, such as (p0∧ ¬p0).)

4

Occasionally we will need to talk about other languages with further connectives, or fewer

connectives. For any subset {C1...Cn} ⊆ {∧,∨,¬,→,>,⊥,3,2}we will write L(C1...Cn)

to represent the sublanguage in the connectives C1...Cn: the smallest set containing the

sentence letters, and containing CmA1...Akwhenever it contains A1...Ak,kthe arity of Cm

and 1 ≤m≤n.L() thus refers to the set of sentence letters. A substitution class for a

language with some of these logical connectives is deﬁned as above except that we require the

substitutions commute with any additional logical connectives, as with i(A∨B) = (iA ∨iB)

or i3A=3iA. We say that a language is complete iﬀ it is truth-functionally complete and

contains either 2or 3.

Example 1 (The full monoid of substitutions of propositional modal logic).Let Sbe the set

of concrete substitutions of L, i.e. functions i:L() → L from sentence letters to arbitrary

sentences of L.µ(i, A)may be deﬁned as the result of uniformly substituting i(pk)for pkin

A, for each k∈N.

Commutivity and L1-L3 are clearly satisﬁed. Identity is witnessed by the element that

maps each letter to itself, and composition by the element of Sthat maps each letter pkto

µ(i, j(pk)).

Note, however, that a set satisfying McKinsey’s requirements may not be isomorphic

to the full monoid of substitutions of a language. There might be distinct substitutions

i, j ∈Ssuch that iA =jA for every sentence A∈ L (there could, for instance, be two

identity elements satisfying Identity). Say that i∼jiﬀ iA =jA for every A∈ L. If one

quotients a set Ssatisfying Commutativity, Identity and Composition by this equivalence

relation one gets another set S/ ∼which satisﬁes Commutativity, Identity and Composition

under the action deﬁned by setting µ∼([i]∼, A) = µ(i, A). Indeed, under this quotienting

operation Sforms a monoid: the unit ιmay be deﬁned as the equivalence class [i]∼of

any element of Ssatisfying Identity, and [i]◦[j] may deﬁned to be the equivalence class

of any ksatisfying Composition. For most purposes we can simply treat Sas a set of

concrete substitutions of the language containing the identity substitution and closed under

composition of substitutions. Even if we impose these additional conditions, one still does

not get the usual property that any function from the non-logical constants (the sentence

letters in this case) to arbitrary expressions (sentences in this case) extends to a substitution

in S. This property is distinctive to the full substitution class alone.

Example 2 (Substitutions within a sublanguage).Write S(C1...Cn)for the substitution

class deﬁned by the set of functions

i:L() → L(C1...Cn)

When L0is a language containing C1...Cn(so that L(C1...Cn)⊆ L0), the action of S(C1...Cn),

µ(i, A), may be deﬁned in the usual way (as in example 1), thereby satisfying Commutativity.

Since L(C1...Cn)contains each sentence letter, S(C1...Cn)contains the identity sub-

stitution and satisﬁes Identity. Since L(C1...Cn)is itself a language, it is closed under

substitutions of letters by sentences in L(C1...Cn). So Composition is also satisﬁed.

We will also investigate a class of substitutions introduced in Humberstone (2016), in

relation to McKinsey’s theory of necessity:

Example 3 (Humberstone substitutions).A Humberstone substitution is a function i:

L() → L(>⊥)such that

5

i(pk)is either pk,>or ⊥.

These substitutions act on the language with primitive >and ⊥connectives (i.e. L(¬ ∧

2>⊥)). µ(i, A)is deﬁned in the usual way, so that McKinsey’s conditions are satisﬁed. We

call the set of Humberstone substitutions H.

A positive formula is one whose letters all occur positively, where the letters occurring

positively and negatively in a formula are deﬁned by a simultaneous recursion as follows:

P(pk) = {pk},N(pk) = ∅,

P(A∧B) = P(A)∪P(B), N(A∧B) = N(A)∪N(B),

P(¬A) = N(A), N(¬A) = P(A)

P(2A) = P(A), N(2A) = N(A).

Example 4 (Positive substitutions).A positive substitution is a substitution that maps

letters to positive formulas, and act on formulas in the usual way. We will call the class of

positive substitutions P.

Lastly, we consider a special sort of substitution that will play a role later in our discus-

sion of Carnap’s theory of logical necessity.

Example 5 (Carnapian substitutions).A Carnapian substitution is a function ion sentence

letters such that

i(pk)is the result of preﬁxing some number (possibly zero) of ¬sign to pk.

and µ(i, A)is deﬁned as before. We call the set of Carnapian substitutions K.

1.2 S-valuations, Pre-validity and Validity

We introduce the notion of an S-valuation: a valuation, taking sentences of the language L

to truth values, that satisﬁes McKinsey’s constraints.

Deﬁnition 1. Suppose that L,Sand µsatisfy L1-L3, Commutativity, Identity and Com-

position. An S-valuation is a function v:L→{0,1}such that

v(A∧B) = min(v(A), v(B))

v(¬A)=1−v(A)

v(2A) = 1 if and only if, for every i∈S,v(iA) = 1

By a truth-value assignment we mean a function deﬁned on the sentence letters, v−:L() →

{0,1}. A valuation vextends a truth-value assignment, v−, iﬀ vL() =v−.

Formally speaking we identify truth and falsity with the numbers 1 and 0, but we shall

sometimes talk of a sentence being true or false in a valuation rather than receiving the

value 1 or 0.

In cases where some of >,⊥,∨or 3are also taken as primitive, the notion of valuation

should respect the natural clauses for those connectives:

v(A∨B) = max(v(A), v(B))

6

v(>)=1

v(⊥)=0

v(3A) = 1 if and only if, for some i∈S,v(iA) = 1.

We introduce two senses in which a formula can be valid relative to a substitution class.

The ﬁrst is not so straightforwardly related to the Bolzano-Tarski conception of validity,

since the set of valid sentences in this sense need not be closed under the rule of substitution:

Deﬁnition 2 (Pre-validity).A sentence Ais pre-valid relative to the substitution class S

if and only if v(A)=1for every S-valuation v.

Consider any substitution class Swhich contains a substitution isuch that i(p0) = >

(such as S(C1...Cn) for any set of connectives C1...Cncontaining >). For any S-valuation v,

v(3p0) = 1 because v(ip0) = v(>) = 1. So 3p0is pre-valid relative to the class S. However

the substitution instance 3⊥is not pre-valid; indeed it is false in every S-valuation.

Validity proper thus cannot be understood as the result of closing the pre-validities under

the rule of substitution. Rather we consider a sentence valid relative to a substitution class

Sonly if all of its substitution instances are pre-valid (including, possibly, substitutions

outside S)

Deﬁnition 3 (Validity).A sentence Ais valid relative to the substitution class Sif and

only if iA is pre-valid relative to the substitution class S, for every substitution iin the full

substitution class S(¬ ∧ 2).

If Sis a substitution class, then we will write L(S)for the set of sentences valid with

respect to S.

Let us remark brieﬂy on the choice of primitives. It is easily veriﬁed that standard

deﬁnitions of arbitrary truth-functional compounds in terms of ¬and ∧receive the same

truth value in any valuation. Similarly 3may be deﬁned, as usual, in terms of ¬and 2.

In most contexts it is more economic to work in the language L, whose primitives consist

of only ∧,¬and 2. But in some cases, the presence of the other primitives makes the

deﬁnitions simpler: for instance if we identiﬁed >and ⊥with particular deﬁnitions, such

as ¬(p0∧ ¬p0) and p0∧ ¬p0, then our oﬃcial deﬁnition of S(>⊥) would not satisfy the

commutativity equations i⊥=⊥and i>>.

The diﬀerences between the logics of diﬀerent substitution classes will be the subject of

the next few sections. However, we are already in a position to see how diﬀerent substitution

classes may behave diﬀerently at the level of pre-validity. We have, for instance, the following

contrast between full and non-modal substitutions:

Proposition 1. The sentence 3(3p∧3¬p∧2(p→2p)) (for pa sentence letter) is pre-valid

over S(¬ ∧ 2), while its negation is pre-valid over S(¬∧).

Proof. Let vbe any S(¬ ∧ 2) valuation. First note that v(2>) = 1 since v(i>) = v(>)=1

for every i∈S(¬ ∧ 2). Similarly for any sentence letter q,v(¬2q) = 1: v(2q) = 0 since

v(r∧ ¬r) = 0 and for i∈S(¬ ∧ 2), iq =r∧ ¬r.

We shall show that upon substituting 2qfor pin (3p∧3¬p∧2(p→2p)) we get

something true in v, and so 2qwitnesses the truth of 3(3p∧3¬p∧2(p→2p)). (i)

v(32q) = 1 since 2qhas a true-in-vsubstitution, namely v(2>) = 1. (ii) v(3¬2q) = 1

since ¬2qis already true in v. (iii) v(2(2q→22q)) = 1. Take any i∈S(¬ ∧ 2). If

7

v(2iq) = 1 then v(jiq) = 1 for every j. This means v(kliq) = 1 for any kand l, and so

v(22iq) = 1. Thus v(2iq →22iq) = 1 for any i∈S(¬ ∧ 2), and so v(2(2q→22q) = 1.

Now let vbe any S(¬∧)-valuation, take any i∈S(¬∧), and let A:= ip. By deﬁnition A

is non-modal. Suppose for reductio that v(3A∧3¬A∧2(A→2A)) = 1. Since v(3A)=1

and Ais non-modal, there is some valuation of propositional logic, u, on the non-modal

language that makes Atrue. Now construct a substitution jthat maps each letter pin A

to a literal that is true in v, if u(p) = 1, and a literal that is false in vif u(p) = 0, and in

such a way that we never pick two literals containing the same letter. Since for each letter

pmoccuring in Awe have v(jpm) = u(pm), we have that v(A(jp1...jpn)) = u(A(p1...pn)

(writing Aas a function of its sentence letters), which means v(jA) = u(A) = 1. Moreover,

since v(3¬A) = 1, Ais not a tautology, and so jA is not either (from the way it was

constructed). So v(2jA) = 0 (since if Bis non-modal, v(2B) = 1 only if Bis a tautology).

So we have v(j A) = 1 and v(2jA) = 0 so v(jA →2jA) = 0 and so v(2(A→2A)) = 0.

This contradicts the assumption that v(3A∧3¬A∧2(A→2A)) = 1.

Thus v(3ip ∧3¬ip ∧2(ip →2ip)) = 0 for every non-modal substitution i; and so

v(3(3p∧3¬p∧2(p→2p))) = 0 for every S(¬∧)-valuation v.

This proposition technically doesn’t rule out the pre-validities of S(¬∧) and S(¬ ∧ 2)

coinciding by being the inconsistent logic. We will eventually rule out this possibility by

showing the existence S(¬∧) and S(¬ ∧ 2)-valuations. It’s also worth noting, however, that

despite the stark contrast between the pre-validities of these two classes, for all we know

the logics of these two substitution classes are identical.

1.3 Relationship to Kripke models

We brieﬂy remark on the relationship between the foregoing substitutional interpretation

of propositional modal logic and the more familiar interpretation due to Kripke (1959).

Although the clauses for substitutional necessity are on the surface quite diﬀerent from the

clauses for necessity in a Kripke model, any S-valuation may be reconceived as a Kripke

model.

Deﬁnition 4. For any given substitution class, S, the Kripke frame associated with S,

(WS, RS), is deﬁned as follows:

WS:= S

RS:= {(i, j ◦i)|i, j ∈S}, that is, RSik if and only if there is some j∈Ssuch that

k=j◦i.

Given an S-valuation, v, the Kripke model associated with vis obtained by deﬁning a

valuation V:WS× L() → {0,1}on (WS, RS)as follows:6

V(i, pk) = v(ipk).

Vmay then be extended to arbitrary formulas in the usual manner.

In what follows we reserve the lower case letters vand ufor S-valuations and use the

uppercase letters Vand Ufor valuations in a Kripke model.

Proposition 2. For any i∈Sand formula A,V(i, A) = v(iA).

6See Drake (1962).

8

Proof. By induction. The clauses for the sentence letters and truth functional connectives

are trivial. For the modal clause, we reason as follows.

V(i, 2A) = 1 if and only if, for every j∈S,V(j◦i, A) = 1. By the inductive hypothesis,

this holds iﬀ v(jiA) = 1 for every j∈S, which holds iﬀ v(i2A) = 1, as required.

We will also frequently appeal to the notion of p-morphism of frames:

Deﬁnition 5 (p-morphism).Ap-morphism from a Kripke frame (W, R)to another Kripke

frame (W0, R0)is a function f:W→W0such that:

(i) For any x, y ∈W, if Rxy then R0f(x)f(y).

(ii) For any x∈Wand y0∈W0, if R0f(x)y0then there exists a y∈Wsuch that Rxy and

f(y) = y0.

If fis a p-morphism between the frames of the models (W, R, V ) and (W0, R0, V 0) and

fpreserves the truth values of some letters, p1...pn, — i.e. V(x, pk) = V0(f(x), pk) for

k= 1...n and any x∈W— then it may be shown that for any sentence Acontaining

only letters from p1...pn,V(x, A) = V0(f(x), a). If fpreserves the truth values of all the

letters, then fis sometimes a p-morphism between the models (W, R, V ) and (W0, R0, V 0).

Given the above correspondence between valuations and Kripke models we can talk about

p-morphisms between substitution classes and Kripke frames.

Deﬁnition 6 (p-morphism between substitution classes and Kripke frames).Ap-morphism

from a substitution class Sto a Kripke frame (W0, R0)is a p-morphism from the frame

(WS, RS)to (W0, R0).

2 The Existence and Uniqueness of S-valuations

If there were no valuations for a given substitution class, S, the concepts of pre-validity

and validity would be trivial and uninteresting. However, given a substitution class S, the

existence of an S-valuation is not always obvious. Consider, for instance, a valuation for

the full substitution class mapping mapping letters to arbitrary sentences of L. While the

truth values of conjunctions and negations are determined by the truth values of sentences of

lower complexity (the conjuncts or the negatum), the truth value of modal formulas is not.

The truth value of 2Ais determined by the truth values of iA for all possible substitutions

of letters within A, including substitutions of higher complexity: imight map some of the

letters appearing in Ato Aitself, yielding a potentially vicious circularity.7

In many cases the circularity is not vicious. Let’s say that a substitution iis non-modal

iﬀ ipkis a formula of the propositional calculus for each k. The following proposition

shows that whenever Sis a class of non-modal substitutions, S-valuations exist and can be

constructed in a familiar inductive manner. Indeed, any assignment to the sentence letters

extends to a unique S-valuation on L.

Proposition 3. Suppose that Sis a class of non-modal substitutions and that v−:L() →

{0,1}is a truth-value assignment on the sentence letters of L. Then there exists a unique

S-valuation extending v−.

7See the related remarks in Kripke (1976), p332 concerning the substitutional interpretation of the

quantiﬁers.

9

Here and elsewhere we say that a valuation vextends a truth-value assignment v−when

vL()=v−.

Proof. vmay be deﬁned inductively on the modal degree of the the formula. In particu-

lar, suppose that v(A) has been deﬁned for every formula of modal degree n, and we are

attempting to evaluate v(2A) and v(3A) of modal degree n+ 1. Since each i∈Sis non-

modal, iA will have modal degree n, and so v(2A) and v(3A) may and must be deﬁned as

mini∈Sv(iA) and maxi∈Sv(iA) respectively.

When Scontains substitutions with modal formulas in their range, the constraints of

deﬁnition 1 can no longer be met through a straightforward inductive construction. Nonethe-

less, we might conjecture that an analogue of proposition 3 holds for the full substitution

class:

Conjecture 4 (The Uniqueness Conjecture).Suppose that Sis the class of all substitutions

on L, and suppose that v−:L() → {0,1}is a truth-value assignment on the sentence letters.

Then there exists a unique S-valuation vextending v−.

Indeed this conjecture is a slight variant of a conjecture made by Harvey Friedman in

Friedman (1975).8The existence portion of the conjecture was settled independently by Kit

Fine and Tadeusz Prucnal.9Prucnal’s method is rather indirect, and goes via a solution

to a related problem concerning intuitionistic logic. In section 2.1 we present Fine’s direct

proof of the existence of a valuation of the full substitution class. The uniqueness part of

the conjecture is still open.

For some classes of modal substitutions we can prove the existence of an S-valuation

through a slightly diﬀerent inductive construction. To illustrate, consider the ‘positive

substitutions’ from example 4, that map formulas to positive formulas (in which letters only

appear under an even number of negations). Let Pbe the class of all positive substitutions

on L.

Proposition 5. There exist P-valuations.

Proof. Let X0be the set of positive formulas, and Xn+1 the truth functional combinations

of formulas from {A, 2A|A∈Xn}. We deﬁne a valuation inductively as follows. Each

formula of X0is treated as true. For any positive substitution iand sentence A∈Xn, it

is readily shown that iA is also in Xn. Thus assuming the valuation is deﬁned on Xnwe

can extend the valuation (uniquely) to formulas of the form 2Afor A∈Xnand thus to

arbitrary formulas in Xn+1.

2.1 The existence of valuations of the full substitution class

Next we settle the harder question of whether there are any valuations for the full sub-

stitution class. We will show that every truth-value assignment to the sentence letters,

v−:L() → {0,1}, can be extended to a valuation of the full substitution class.

Let us write P0(X) for P(X)\ ∅.

8Friedman’s conjecture (see problem 41 of Friedman (1975)) was concerned with the existence and unique-

ness of a valuation on the full substitution class making all the sentence letters true. Moreover, his version

of the conjecture postulates the existence and non-uniqueness of such a valuation.

9See Prucnal (1979).

10

Deﬁnition 7 (Medvedev frame).A Medvedev frame is a frame of the form (P0(X),⊇)for

some ﬁnite non-empty set X.

We write Med for the logic of Medvedev frames.

Our strategy is: ﬁrst to construct a certain rooted Kripke model, with the property that

2Ais true at the root if and only if A∈Med; second, to show that 2Ais true at the root

of this model if and only if, for every substitution i,iA is true at the root. A valuation of

the full substitution class is then secured by identifying the truth value of a sentence with

its truth value at the root of the given Kripke model.

To this end, enumerate all of the consistent sentences of Med,C1, C2, C3, .... For each

Ci, pick a model Mi= (P0(Xi),⊇, Vi) of Ci. We may assume without loss of generality

that Xiand Xjare disjoint when i6=j. Say that propositional formulas A1...Anconstitute

apropositional partition iﬀ they are individually consistent, pairwise inconsistent and have

a tautologous disjunction. We choose our models so that, for each Xiwith Xi={a1, ..., an}

there is a partition of propositional formulas Ai

1...Ai

nsuch that V({am}, Am) = 1 for m=

1, ..., n.10 This is always possible since we are free to assign truth values to the sentence

letters not appearing in Cias we please in Mi, without disrupting the truth value of Ci,

and we can build Ai

1...Ai

nout of these fresh letters accordingly.

For each non-empty set Y⊂Xi={a1, ..., an}we may construct a formula Di

Ythat is

true at Yand only at Yin Mi:

Di

Y:= Vam∈Y32Am∧Vam6∈Y¬32Am

32Amis true at Yiﬀ Ysees a terminal world at which Amis true. Moreover, {am}is the

only terminal world where Amis true by deﬁnition, 32Amis true at Yiﬀ {am}is a subset

of Y. So DYcan only be true at a world Y0if Y=Y0:Y0sees are exactly {{am} | am∈Y},

and so the singleton subsets of Y0are identical to the singleton subsets of Y.

Observe that the formulas Di

Yare pairwise incompatible in propositional logic. (They

do not quite have a tautologous disjunction since we have excluded the case where Y=∅,

though their disjunction is a theorem of S4M and so will also belong to Med.)

Given a truth-value assignment to the sentence letters v−:L() → {0,1}, we will con-

struct a global model M= (W, ⊃, V ) that encompasses all of the individual models above

and, at its root, agrees with v−on all the sentence letters:

X:= Si∈ωXi

W:= {w⊆X|w=Xor wis ﬁnite and non-empty}

V(w, pj) = 1 iﬀ w⊆Xifor some iand Vi(w, pj) = 1 or w=Xand v−(pj) = 1. (And

V(w, pj) = 0 if wneither Xnor a subset of Xifor any i.)

We shall show that V(X, 2A) = 1 if and only if V(X, iA) = 1 for every substitution iof

the modal language. Since the clauses for the truth-functional connectives hold straightfor-

wardly, a valuation vof the full substitution class may therefore be constructed by setting

v(A) = V(X, A).

Given any component model Mi, the formulas Di

Yfor Y∈P0(Xi) determine a partition

on the set of worlds Wthat is indexed by the worlds Y∈P0(Xi) of the component model,

10A propositional partition of size nmay always be constructed. Pick sentence letters p1...pkwhere 2k≥n

and a surjection σ: 2k→n.Ammay then be deﬁned as Wσ(u)=m(Vu(r)=1 pr∧Vu(r)=0 ¬pr). A valuation

in which V({am}, Am) = 1 can be constructed by picking some representative element ufrom f−1(m) and

letting V({am}, pr) = u(r).

11

where the cell of the partition indexed by Yis the set of worlds at which Di

Yis true. While

we earlier observed that the Di

Ypartitioned the worlds of Miinto singletons, the cells of

the partition on Wwill generally be much larger.

Similarly, we may show that any proposition (i.e. set of worlds) from a given component

model Mi, will determine a union of cells in W, so that we can, in particular, project the

component model Miup into M. To this end, for each sentence letter p, we set:

Di

p:= WVi(Y,p)=1 Di

Y

Suppose, as before, that Xi={a1, .., an}. The function f:W→Withat generates the

partition of Windexed by the worlds of Miis deﬁned as follows (so that Y∈Wiwill index

the cell f−1(Y) of the partition on W):

f(w) = {am|V(w, 32Am)=1}

It is immediate from the deﬁnition of Di

Ythat V(w, Di

f(w)) = 1 for every w∈W. Moreover,

because the DYare pairwise incompatible, each f(w) is the unique Yfor which V(w, Di

Y) =

1. It can be shown that the truth value of Di

pin Mat wis the same as the truth value of

pin Miat f(w), that is V(w, Di

p) = Vi(f(w), p). For V(w, Di

p) = 1 iﬀ some disjunct of Di

p

is true in Vat w, i.e. iﬀ V(w, Di

Y) = 1 for some Ysuch that Vi(Y, p) = 1. But f(w) is the

unique Ysuch that V(w, Di

Y) = 1, and so:

V(w, Di

p) = Vi(f(w), p)

To generalize this equivalence to an arbitrary sentence A(p1...pk):

V(w, A(Di

p1...Di

pk)) = Vi(f(w), A(p1...pk))

we show that fis a p-morphism:

Proposition 6. fis a p-morphism from the frame of Mto the frame of Mi

Proof. Suppose X={a1, ..., an}.

We ﬁrst we show that fpreserves the accessibility relation. Suppose wsees vin M.

Since the accessibility relation is transitive, anything possible at vis possible at w. So

f(v)⊆f(w) and hence f(w) sees f(v) in Mi.

To establish the reverse condition, suppose that f(w) sees a world Y∈P0(Xi), so that

Y⊆f(w). We must show that there exists a v⊆wsuch that f(v) = Y.

Suppose that f(w) = {a1, ..., am}. By deﬁnition, for each ak∈f(w), w32Ak. Thus

for each ak∈f(w) there is a terminal world {bk} ∈ Wsuch that {bk} ⊆ wand {bk}Ak.

Let v={bk|ak∈Y}. Clearly v⊆w. Moreover, we may show that f(v) = Y. For, given

any ak∈Y,{bk}2Akand {bk} ⊆ v, so v32Ak, and so ak∈f(v); and, conversely, if

v32Akthen bk∈vand so ak∈Y.

We can now prove our theorem. Here we will make use of the following fact, which is a

consequence of a more general result — proposition 38 — that will we prove later in section

5.2.11

Proposition 7. If A∈Med then V(X, A) = 1.

11In section 5.2 it is shown that Med is a coherent logic (see deﬁnition 10), and proposition 7 follows from

the observation that the function v, deﬁned by setting v(A) := V(X, A), is a meta-valuation (see deﬁnition

9).

12

Proposition 8. V(X, 2A) = 1 if and only V(X, iA)=1for every substitution i.

Proof. We will show that if V(X, 2A) = 0 then for some substitution j,V(X, jA) = 0

Suppose that V(X, 2A) = 0. So V(w, A) = 0 for some world w. If w=X, then

V(X, A) = 0, so jcan be the identity substitution. Otherwise V(w, A) = 0 for some world

w6=X. Since the subframe generated by wis a Medvedev frame, ¬Ais consistent in Med,

and so we may assume that wis the root of Wifor some i(that is, w=Xi). Moreover,

since f:W→Wiis a p-morphism and since Vi(f(w0), p) = V(w0, Di

p) for any letter pand

world w0, we know that V(f(w0), B) = V(w0, B[Di

p1/p1...Di

pn/pn]) for any sentence B. In

particular since f(X) = wand V(w, A) = 1, it follows that V(X, A[Di

p1/p1...Di

pn/pn]) = 1,

and so any substitution jmapping pkto Di

pkwill suﬃce.

Conversely, suppose that V(X, 2A) = 1. Since every consistent sentence of Med is true

at some world accessible to X, it follows that A∈Med. Since Med is closed under the rule

of substitution, iA ∈Med for any substitution i, and so V(X, iA) = 1 by proposition 7.

We may, ﬁnally, obtain a valuation vof the full substitution class by setting v(A) =

V(X, A).

Theorem 9. Every truth-value assignment v−may be extended to a valuation of the full

substitution class.

It should be observed that in the constructed valuation the truth of a possiblity sentence,

3A, may always be witnessed by a substitution of formulas of modal degree 2.

2.2 The uniqueness conjecture

Proposition 3 established that for any non-modal substitution class S, any truth assigment

to the sentence letters extends to a unique S-valuation. The analogue of proposition 3 for

the full substitution class is the principle we called conjecture 4:

Any truth-value assignment v−:L() → {0,1}extends to a unique S(¬ ∧2)-valuation.

Theorem 9 establishes the existence portion of the conjecture. The uniqueness conjecture is

important to the study of logical necessity because it is implies that there are interpretations

of 2in which 2means valid. Letting Sbe the full substitution class, conjecture 4 is

equivalent to:

For any S-valuation v,v(2A) = 1 iﬀ Ais valid with respect to S.

Supposing the uniqueness conjecture to be true, v(2A) = u(2A) for any pair of S-valuations,

and so v(2A) = 1 iﬀ u(2A) = 1 for all valuations u. Either side of the biconditional holds

iﬀ, for every valuation uand substitution i,u(iA) = 1. So under the supposition, 2Awill

coincide with validity with respect to the full substitution class. The converse holds too.

For suppose the previous claim holds. Then for any pair of S-valuations uand v,u(2A)=1

iﬀ Ais valid with respect to Siﬀ v(2A) = 1; and so if vand uadditionally agree on the

sentence letters, then u=v.

Notice that in the valuation constructed in theorem 9, v(2A) = 1 if and only if A∈Med.

Thus an extension of a truth-value assignment, v−, to a valuation for the full substitution

class may be deﬁned inductively as follows.

v(p) = v−(p)

13

v(A∧B) = min(A, B)

v(¬A)=1−v(A)

v(2A) = 1 iﬀ A∈Med

For an arbitrary valuation vof L, let ∆v={A|v(2A) = 1}. Then another way to state

conjecture 4 is:

For any valuation vof the full substitution class, ∆v=Med

While we are not able to settle the uniqueness conjecture, we will here establish the weaker

claim:

For any valuation vof the full substitution class, ∆v⊆Med

Moreover, in section 4.2 we’ll show that some principles distinctive to Med must also belong

to ∆v. Thus ∆vwould appear to be tightly sandwiched between Med and a strong subsystem

of Med, lending further credence to the uniqueness conjecture.

Let vbe a valuation for the full substitution class. For each n, pick a propositional

partition A1...An— i.e. npropositional formulas that are pairwise inconsistent and have a

tautologous disjunction. For any substitution, i, let f(i) = {m|v(32iAm)=1}.

Lemma 10. For any substitution iand Y⊆f(i), there exists a substitution jsuch that

f(j◦i) = Y.

Proof. In this proof we will frequently appeal to the fact that if Bis a formula, va valuation,

and ja substitution such that v(2jB) = 1, then there exists a >⊥ substitution ksuch that

v(2kB) = 1. For if v(2j A) = 1, for an arbitrary jthen the result of pre-composing jwith

any >⊥ substitution k0, yields >⊥ substitution, k0◦j, with the required property.

Suppose that Y⊆f(i). Let p1...prbe the letters appearing in A1...An, and henceforth

let kbe variable for a >⊥ substitution deﬁned only on p1...pr.

Since, by deﬁnition, v(32iAm) = 1 for each m∈f(i), there must exist for each m∈f(i)

a substitution ksuch that v(2kiAm) = 1. And without loss of generality we may assume that

kis a >⊥ substitution. For each m∈f(i) we shall pick a representative >⊥ substitution,

km, such that v(2kiAm) = 1.

Let Sbe the set of all >⊥ substitutions deﬁned on p1...pr, and let S0={km|m∈Y}.

Our goal will be to ﬁnd a substitution jthat ‘ﬁlters’ the substitutions in Sleaving all and

only substitutions in S0behind: i.e. for any k∈S,k◦jis equivalent to a substitution in

S0and every substitution in S0is equivalent to something of the form k◦jfor some k∈S.

Here two substitutions, kand k0are said to be equivalent when v(2(kp ↔k0p)) = 1 for any

sentence letter pfrom p1...pr.

Because A1...Anare pairwise inconsistent and have a tautologous disjunction, we also

know that for each >⊥ substitution kthere is a unique sentence Amsuch that v(2kiAm) = 1.

For a give kwe will call this sentence Ak. Now pick any surjection σ:S→S0and deﬁne a

substitution jas follows:

j(p) = _

σ(k)(p)=>

Ak

for each p∈ {p1...pr}

We now show that for any >⊥ substitution k,k◦jand σ(k) are equivalent. Firstly, we

show that for any >⊥ substitution kon p1...pn,σ(k) and k◦jare equivalent substitutions.

14

We will show for each letter p,v(2kjp) = 1, if and only if σ(k)(p) = >. Suppose that

σ(k)(p) = >. So Akis a disjunct of Wσ(k)(p)=>(Ak) = j(p), and so k(Ak) — which recall is

necessary in vby deﬁnition — is a disjunct kj(p). Conversely, suppose that v(2kjp) = 1.

Since the Akare pairwise incompatible, this can only happen if Akis one of the disjuncts

of j(p), and thus can only happen if σ(k)(p) = >.

This delivers us our desired result: for if m∈Ythen, because σis surjective, there

is some >⊥ substitution such that σ(k) = km, and kmand k◦jare equivalent. Since

v(2kmiAm) = 1, by deﬁnition, and kmis equivalent to k◦j,v(2kjiAm) = 1, and thus

v(32jiAm) = 1. So m∈f(j◦i). Conversely if m∈f(j◦i), this means v(32jiAm) = 1

and so there is substitution k(which we may assume without loss of generality to be >⊥)

such that v(2kjiAm). k◦jis equivalent to σ(k) which may be seen to equivalent to km.

Since v(2kmiAm) = 1, m∈Y.

The signiﬁcance of lemma 10 is this. Let (W, R, V ) be the Kripke model associated with

v(so Wis the class of substitutions, R={(i, j ◦i)|i, j ∈W}and V(i, p) = v(ip)). Then:

Corollary 11. f:W→P0(X)is a p-morphism of frames from (W, R)to (P0(X),⊇).

Proof. Since Rik iﬀ there exists a jsuch that k=j◦i, the back-and-forth condition amounts

to the claim that if f(i)⊇Ythen there exists a jsuch that f(j◦i) = Y. This is just lemma

10.

It remains to show that f(j◦i)⊆f(i) for every i, j ∈W. If m∈f(j◦i) then

v(32jiAm) = 1 and so there is a substitution ksuch that v(2kjiAm) = 1. In which case

the substitution k◦jwitnesses the truth of v(32iAm) = 1 so m∈f(i).

Let X={1, ..., n}. For a given Y⊆Xwe deﬁne, as before:

DY:= Vm∈Y32Am∧Vm6∈Y¬32Am

As before, DYand DZare inconsistent when Y6=Z. Observe, also, that for any substitution

i,v(iDf(i)) = 1 since, by deﬁnition, f(i) is just the set of msuch that v(32iAm) = 1.

Suppose that (P0(X),⊇, V ) is a Medvedev model, that X={1, ..., n}and that A1...An

are a partition of propositional formulas such that V({m}, Am) = 1. We may deﬁne a

substitution as follows:

i(p) = _

V(Y,p)=1

DY

Proposition 12. For any formula Aand substitution j,v(jiA) = V(f(j), A)

Proof. We may prove this by induction of formula complexity.

Base case: Suppose that V(f(j), p) = 1. So Df(j)is a disjunct of i(p), and since j

distributes over disjunctions, jDf(j)is a disjunct of jip. By our observation v(jDf(j)) = 1

so v(jip) = 1.

Conversely, suppose v(j ip) = 1. So v(jDY) = 1 for some Ysuch that V(Y, p) = 1.

By our observation we know that v(jDf(j)) = 1. Since the DZ’s are pairwise incompatible

(and, thus, so are their substitution instances) it follows that v(jDY) = v(jDf(j)) = 1 only

if f(j) = Y. So V(f(j), p) = V(Y , p) = 1

Inductive step: the negation and conjunction cases are straightforward, and the 2case

follows from the fact that fis a p-morphism. Explicitly: if v(2jiA) = 0 then v(kjiA)=0

for some substitution k. So V(f(kj ), A) = 0 by the inductive hypothesis.

15

Conversely, if V(f(j),2A) = 0 then V(Y , A) = 0 for some non-empty Y⊆f(j). By

lemma 10 there exists a ksuch that f(k◦j) = Y, and so by the inductive hypothesis,

v(kjiA) = V(f(k◦j), A) = V(Y, A) = 0.

Theorem 13. If vis a valuation of the full substitution class, ∆v⊆Med

Proof. It suﬃces to show that any sentence Cconsistent in Med is consistent in ∆v: there is

some substitution ksuch that v(kC) = 1. If Cis consistent in Med then there is a Medvedev

model (P0(X),⊇, V ) such that Cis true at some world Y⊆X. Construct a substitution i

as above, and pick some substitution jsuch that f(j) = Y(this is possible by lemma 10).

Then v(jiC ) = V(f(j), C ) = V(Y, C) = 1. j◦iis the required substitution.

3 The Logic of Logical Necessity

Which principles of modal logic are valid on the logical interpretation of 2? The existence

of an identity substitution immediately ensures the validity of the the Taxiom, 2A→A.

For given any substitution class S, and S-valuation, v, it follows that v(2A) = 1 only if

v(iA) = 1 for any substitution and, in particular, that v(ιA) = v(A) = 1 when ιis the

identity substitution. Similarly, the closure of the substitutions under composition ensures

the validity of the S4 axiom, since if v(2A) = 1 then, for any pair of substitution i, j ∈S,

their composition i◦jis also in S, and so v((i◦j)A) = 1. For ﬁxed j, if v(ijA) = 1 for

every i∈S, it follows that v(2jA) = 1; and since this holds for every j∈S,v(22A) = 1.

Indeed, McKinsey showed that for every substitution class S,L(S) contains the theorems

of S4.12

Theorem 14 (McKinsey).Let Sbe any substitution class.

1. The set of validities L(S)is closed under the rules of necessitation, modus ponens and

uniform substitution.

2. The formulas 2A→22A,2A→A,2(A→B)→(2A→2B)are all valid.

Thus every theorem of S4 is valid with respect to the class of S-valuations.

Proof. This result appears in McKinsey (1945). We may obtain it directly from proposition 2

by noting that the Kripke model associated with any S-valuation vis transitive and reﬂexive,

as ensured, respectively, by the Composition and Identity conditions on substitution classes.

It follows that for any Aand B,2A→22A,2A→Aand 2(A→B)→2A→2Bare

true in any valuation. Closure under the rules of necessitation, modus ponens and uniform

substitution follows straightforwardly from the deﬁnitions of validity and S-valuation.

3.1 The Brouwer and McKinsey axioms

Which further principles of modal logic might be valid under the logical interpretation of

necessity? Taking the logic of metaphysical necessity as our cue, one might wonder if the

Brouwerian axiom

12Drake (1962) improves on this result by showing that the intersection of the logics L(S) for every

substitution class Sis exactly S4.

16

BA→23A

is valid. The result of adding Bto S4 yields S5, a logic commonly supposed to be the

logic of metaphysical necessity. In correspondence, McKinsey oﬀers the following argument

against the truth of the Brouwerian axiom.13 While is certainly true that sugar is sweet,

and vinegar is not, the Brouwerian axiom, if it were true, would further imply that it’s

necessarily possible that sugar is sweet and vinegar is not sweet. But given the left-to-right

direction of the substitutional constraint, substituting sugar for vinegar, we may infer the

absurd conclusion that it’s possible that sugar is sweet and sugar is not sweet.

This conclusion is not peculiar to McKinsey’s deﬁnition of logical necessity either, since it

is also a consequence of our more general class of metalogical constraints on logical necessity.

For as we noted there, so long as our account of logical truth, ∆, is closed under the rule of

substitution we have the left-to-right direction of McKinsey’s substitutional constraint. (The

reader may have noticed that we actually do have a candidate notion of logical truth which

isn’t closed under the rule of substitution — pre-validity. We’ll investigate this interpretation

in section 3.2, where we will see that it is indeed consistent with the logic of S5.)

What is special about the substitution of ‘sugar’ for ‘vinegar’ in ‘sugar is sweet and

vinegar is not’ is that no further substitutions to the resulting sentence can change its truth

value. Let us sharpen this notion:

Deﬁnition 8. A>⊥-substitution is a substitution whose range on L() is {>,⊥}.

A substitution i∈Sis terminal with respect to Sif and only if v(2iA)=1or v(2¬iA) =

1for every sentence Aof Land S-valuation v. A terminal substitution, i,matches to a

>⊥-substitution jif and only if v(2ip) = 1 when jp =>and v(2¬ip)=1 when jp =⊥.

Finally, say that a substitution class contains all terminal substitutions when it has a

terminal substitution matching any >⊥ substitution.

Clearly any substitution class containing a >⊥ substitution, such as S(>⊥...) or H(the

class of substitutions that leave letters alone or replace them with >or ⊥) has a terminal

substitution. However, many other formulas can play the role of >, such as (p0∨ ¬p0) or

2p0→22p0, and similarly many formulas can play the role of ⊥. So one need not have >

and ⊥in the range of substitutions in order to have terminal substitutions.

The presence of terminal substitutions imposes its own distinctive modal logic. The

McKinsey axiom says:

M23A→32A

It characterises the Kripke frames in which each world sees a terminal world: a world that

sees only itself.14

Proposition 15 (McKinsey).Let Sbe a substitution class containing a terminal substitu-

tion. Then the theorems of S4M are valid over the class of S-valuations.

Proof. Since, by theorem 14, the validities of any substititution class are closed under modus

ponens and necessitation and contain the axioms of S4, it remains only to show that the

McKinsey axiom Mis valid. If 23Ais true in some arbitrary S-valuation v, then for

13See Anderson and Belnap (1975) p122-123.

14McKinsey actually considers the formula:

F23A∧23B→3(A∧B)

which is only equivalent to Min the context of S4. The principle Mappears to originate from Cresswell and

Hughes (1996).

17

some terminal substitution i∈S,v(3iA) = 1. Since iis terminal, v(2iA) = 1, and so

v(32A) = 1.

The McKinsey axiom is compatible with Brouwer’s axiom. However, the only consistent

modal logic containing S4, McKinsey’s axiom, and Brouwer’s axiom is the trivial modal logic

Triv in which 2just means true.15 The modal logic Triv has as its characteristic axiom:

Triv 2A↔A

Triv is the logic of the trivial substitution class that has the identity substitution ιas its

sole element.

Indeed, according to the Brouwerian axiom, truths of the form 23Aare quite common

place: for any sentence A, either 23Aor 23¬Ais true, depending on whether Aor ¬A

is true. By contrast, in a substitution class containing all terminal substitutions, truths of

the form 23Aare quite special: it may be shown that the sentence 23Ais true only if

Ais a theorem of Triv, or equivalently, if the result of deleting all modal operators from A

is a tautology. Let us write A−for the propositional formula that results from deleting all

modal operators from A. Then:

Proposition 16. Suppose Sis a substitution class containing all terminal substitutions.

Then the following are equivalent.

1. 23Ais true in some S-valuation.

2. A−is a tautology.

3. Ais a theorem of Triv

Proof. The equivalence of 2 and 3 is straightforward. We show that 1 and 2 are equivalent.

We may show by a straightforward induction that if iis a terminal substitution, then

v(iB) = v(i(B−)) for any B. The only non-trivial case is to show that v(i2C) = v(i(2C)−).

v(i2C) = v(2iC) = 1 iﬀ v(iC ) = 1 since iis terminal. Moreover, by the inductive hypoth-

esis, v(iC) = 1 iﬀ v(i(C−)) = 1. But v(i(2C)−) = v(iC−). So v(i2C) = v(i((2C)−)).

If v(23A) = 1 then for every terminal substitution i,v(3iA) = v(iA) = 1. So v(iA−) =

1 for every terminal substitution i, which clearly implies that A−is a tautology.

Conversely, suppose A−is a tautology and jan arbitrary Ssubstitution. Let ibe any

terminal substitution. v(ijA) = v(i((jA)−)) from before. But if A−is a tautology, then

(jA)−is a substitution instance of A−of the letters by sentences of propositional logic

(speciﬁcally the substitution j−where j−(pk) = (jpk)−). So (jA)−is a tautology too, and

hence v(ijA) = v(i((jA)−) = 1. Thus iwitnesses the truth of v(3jA) = 1; and since jwas

an arbitrary Ssubstitution, v(23A)=1

15

1. A→23A(an instance of B)

2. 23A→32A(an instance of M)

3. 32A→2A(an equivalent of S5)

4. A→2Afrom 1-3

5. 2A→Aby T.

18

3.2 Carnap’s Theory of Logical Necessity

Earlier we reported a general argument, due to McKinsey, that the logic of logical necessity

does not contain the Brouwerian axiom. It rested on the proposed constraint connecting

logical necessity and logical truth, along with the assumption that logical truth is closed

under the rule of substitution. Might we resist this argument by adopting a conception of

logic that is not closed under the rule of substitution? Interestingly, an early example of

a theory of logical necessity in Carnap (1946) and Carnap (1947) (pp173-177) results in a

notion of logical truth that fails to be closed under the rule of substitution (see Cresswell

(2013)), whilst simultaneously validating all the theorems of S5, including all instances of

Brouwer’s axiom.

Indeed, we have encountered our own candidate notion of logical truth not closed under

substitution: pre-validity. This raises the question of whether there might be a substitutional

interpretation of 2as pre-validity:

Is there a substitution class Ssuch that, for any S-valuation v,v(2A) = 1 if and only

if Ais pre-valid with respect to S?

Indeed, we will show that there is and that the pre-validities of this class coincide with

Carnap’s notion of logical truth for propositional modal logic.16

Recall from example 5 in section 1.1 that Kis the class of substitutions isuch that i(pk)

is either pkitself, or the result of preﬁxing a ﬁnite string of negations in front of pk. Note

that there is in eﬀect no diﬀerence between substitutions with the same parity: if for each

k,ipkand jpkagree on whether they contain an even or odd number of negations, then for

the purposes of valuation we might as well treat these substitutions as the same. However

this redundancy is necessary to meet the formal requirement that a substitution class be

closed under substitution.

Proposition 17. For any K-valuation v,v(2A)=1iﬀ Ais pre-valid with respect to K

Proof. Given any K-valuation vand substitution i∈K, let (iv) be the unique K-valuation

deﬁned by setting (iv)(pk) = v(ipk) (so (iv)(pk) = 1 iﬀ ipkhas an even number of negations

and v(pk) = 1, or ipkhas an odd number of negations and v(pk) = 0).

We show by induction on the modal degree of Athat for any vand i∈K:

v(iA)=(iv)(A)

The identity holds of the sentence letters by construction and is clearly inherited over truth-

functional compounds. So suppose that for sentences of modal degree n,v(iA) = (iv)(A)

for any K-valuation vand substitution i∈S. Recall that v(2iA) = 1 iﬀ v(jiA) = 1 for

every substitution j∈K. Now for any j, we may apply the inductive hypothesis to vand

the substitution j◦ito get v(jiA) = (jiv)(A); and by similarly applying the inductive

hypothesis to iv and j, we get that (jiv)(A)=(iv)(j A). So v(jiA) = 1 for every j∈Siﬀ

(iv)(jA) = 1 for every j∈S, which holds iﬀ (iv)(2A) = 1, as required.

We may now prove the proposition. Note that for a ﬁxed v, every K-valuation uis of the

form iv for some substitution i∈K. (Set i(pk) = pkif u(pk) = v(pk), and let i(pk) = ¬pkif

u(pk)6=v(pk).) So v(2A) = 1 iﬀ v(iA) = 1 for every i∈K, iﬀ (iv)(A) = 1 for every i∈K,

iﬀ u(A) = 1 for every K-valuation, iﬀ Ais pre-valid with respect to K.

16See Carnap (1946). The propositional fragment of Carnap’s logic has been treated in Cresswell (2013),

Thomason (1973), Hendry and Pokriefka (1985). A related project is carried out in Cocchiarella (1974); see

also Carroll (1978).

19

It is also easy to verify that all the theorems of S5 are valid with respect the class of K

valuations.

We now compare the pre-logic of our substitution class with the propositional modal

logic of Carnap’s theory of logical necessity.17 Formulas of propositional modal logic are

evaluated with respect to truth-value assignments to sentence letters, v−:L() → {0,1}.

We deﬁne what it means for a sentence Aof propositional modal logic, L(∧¬2), to be true

at a truth-value assignment v−, written v−|=A, as follows:

v−|=pkiﬀ v−(pk)=1

v−|=A∧Biﬀ v−|=Aand v−|=B

v−|=¬Aiﬀ v−6|=A

v−|=2Aiﬀ u−|=Afor every truth-value assignment u−.

Thus 2Acan be taken to mean in an extended sense that Ais a tautology. A sentence Ais

C-valid iﬀ v−|=Afor every truth-value assignment v−.C-validity is not closed under the

rule of substitution: for instance, ¬2p0is C-valid, while ¬2¬(p0∧ ¬p0) is not. Much like

our proposition 17, a sentence of the form 2Ais true at a propositional valuation iﬀ Ais

C-valid. Indeed, we can show:

Proposition 18. Ais C-valid iﬀ Ais pre-valid with respect to K.

Proof. Given a K-valuation v, we write v−for corresponding the truth-value assignment:

vL(). By proposition 3 every truth-value assignment is identical to v−for some unique K-

valuation v, and so we lose no generality by restricting attention to truth-value assignments

of the form v−. We shall show by induction that for every K-valuation v,v(A) = 1 iﬀ

v−|=A.

The identity clearly holds for sentence letters. The clauses for the truth functional

connectives are straightforward. v−|=2Aiﬀ u−|=Afor every truth-value assignment

u−. So by the inductive hypothesis, u(A) = 1 for every K-valuation u. Now for every

substitution i∈K,iv (as deﬁned in proposition 17) is a K-valuation, so (iv)(A) = 1 and

since (iv)(A) = v(iA), v(iA) = 1 for every i∈K. Thus v(2A) = 1. Conversely, since

every K-valuation is of the form iv for some i∈K, if v(2A) = 1, then u(A) = 1 for every

valuation, so by the inductive hypothesis u−|=Afor every uand so v−|=2A.

As we saw, Ais pre-valid iﬀ v(2A) = 1. By the above, v(2A) = 1 iﬀ v−|=2Aiﬀ Ais

C-valid. This completes the proof.

4 The Logic of Speciﬁc Substitution Classes

The foregoing remarks give us some idea of what sort of logical principles must be a part of

the logic of logical necessity. The situation becomes more intricate when we look at speciﬁc

substitution classes. We shall see that the logics of some, though not all, substitution classes

includes the Grzegorczyk axiom Grz

Grz 2(2(A→2A)→A)→A

a principle that is not already a theorem of S4M.

We shall also see that the logics of some, though not all, substitution classes include an

axiom we call “the subset principle” that is distinctive of Med.

17A presentation of the propositional fragment of Carnap’s theory may be found in Cresswell (2013).

20

4.1 The logic of non-modal substitution classes

In this section we investigate various classes of of non-modal substitutions — substitutions

that map letters to non-modal propositional formulas. In his book Philosophical Applications

of Modal Logic, Lloyd Humberstone provides an interpretation of McKinsey’s theory of

modality in terms of (what we have called) the class of Humberstone substitutions, H.

These are substitutions that either map a letter to itself, >or ⊥. After noting that every

theorem of S4M is valid with respect to this substitution class (Proposition 14), he poses

the following questions (p.168):

1. Are the theorems of S4M exactly the validities with respect to the class H?

2. How sensitive is the logic to the exact substitution class used? Is the logic of Hthe

same as the logic of S(>⊥) or the logic of the full substitution class?

The ﬁrst question may be answered negatively. In particular, the Grzegorczyk axiom Grz

above is valid with respect to the class of H-valuations but is not a theorem of S4M.18

Proposition 19. Grz is valid with respect to H-valuations.

Proof. Let vbe an arbitrary H-valuation and Aan arbitrary formula. We suppose that

v(A) = 0 and then show that v(2(2(A→2A)→A)) = 0. Our goal is thus to ﬁnd a

substitution isuch that v(iA) = 0 and v(2(iA →2iA)) = 1.

Without loss of generality, we may restrict our attention to substitutions that map

sentence letters not appearing in Ato >. Partially order these remaining substitutions as

follows: i≤jwhen there exists a ksuch that j=k◦i(or, equivalently, for all n,j(pn) = >

whenever i(pn) = >and j(pn) = ⊥whenever i(pn) = ⊥).

Because there are only ﬁnitely many letters in A, there are only ﬁnitely many substi-

tutions in this ordering. Since v(A) = 0, there must be a maximal substitution, i, with

respect to this ordering such that v(iA) = 0. This means that v(iA) = 0 and that for any

j > i v(jA) = 1. We now show that v(2(iA →2iA) = 1. For any j, either (a) j◦i=i, so

v(jiA) = 0 or else (b) j◦i>i, in which case v(2jiA) = 1, for v(kjiA) = 1 for any k∈H,

since iwas the maximal substitution making Afalse and k◦j◦i≥j◦i > i. Either way,

v(jiA →2jiA) = 1 and since jwas arbitrary, v(2(iA →2iA) = 1. This completes the

argument.

We can also use Grz to make progress with Humberstone’s second question regarding

S(>⊥). Recall that the diﬀerence between S(>⊥) and His that while Hsubstitutions

must map pto itself or to >or ⊥, an S(>⊥) substitution can map pto any other letter, in

addition to >and ⊥. We can now show:

Proposition 20. Grz 6∈ L(S(>⊥)) and, speciﬁcally, Grz may be invalidated through the

instance A=p→2p.

Proof. Let vbe a S(>⊥)-valuation with a true and a false letter, pand q. We will show that

v(2(2(A→2A)→A)) = 1 and v(A) = 0 when A=p→2p. Let ibe any substitution in

S(>⊥), and suppose v(i(p→2p)) = 0. It straightforardly follows that ip cannot be >or

18To see that Grz is not a theorem of S4M, consider the frame ({0,1,2},{(0,1),(1,0),(0,2),(1,2)} ∪

{(w, w)|w= 0,1,2}) and the model over it in which pis true at 0 and 2, but not at 1. Since it is

transitive, reﬂexive and every world sees a terminal world, the theorems of S4M hold in this model, but

2(2(p→2p)→p) holds at 1 while pdoes not.

21

⊥, and so it must be some sentence letter r. Let jbe a substitution such that jr =q, the

false sentence letter.

Then v(jiA) = v(q→2q) = 1 since qis false, and v(ji2A) = v(2(q→2q)) = 0 since

p→2pis a substitution instance of q→2qand is false. Thus v(jiA →2jiA) = 0, and

so v(2(iA →2iA)) = 0. So we have shown that, for an arbitrary substitution i∈S(>⊥),

when v(iA) = 0 then v(2(iA →2iA)) = 0 and hence that v(2(2(A→2A)→A)) = 1.

Finally, v(p→2p) = 0 since pis true and 2pfalse in v.

We can extend the above line of argument to show that Grz is invalid with respect to

the class of all non-modal substitutions S(¬∧). In order to do this, we need to appeal to

the following fact about propositional logic:

Lemma 21. Suppose that Ais a sentence of the propositional calculus that is neither

tautologous nor contradictory, and that Bis any other sentence (possibly involving modal

operators). Then there exists a substitution i∈S(¬∧)such that iA is equivalent to B.

When B∈ L(>⊥¬), there will exist a substitution i∈S(>⊥¬)such that iA is equivalent

to B.

Proof. Let p1...pnbe the sentence letters in Aand consider the truth-value assignments to

p1...pn. Since Ais neither tautologous nor contradictory, there are assignments vand v0

making Atrue and false respectively. Since v0may be obtained from vby picking a letter pk

from p1...pnand ﬂipping its truth value, and repeating this as many times as necessary, there

must exist some assignment uand letter pksuch that umakes Atrue, but the assignment

that results from ﬂipping the truth value of pkin umakes Afalse. We can then deﬁne the

desired substitution as follows:

ipm=

Bif m=kand u(pk)=1

¬Bif m=kand u(pk)=0

¬(p0∧ ¬p0) if m6=kand u(pk)=1

p0∧ ¬p0if m6=kand u(pk)=0

For B∈ L(>⊥¬) we may analogously obtain a substitution i∈S(>⊥¬) such that iA is

equivalent to Bby replacing ¬(p0∧ ¬p0) with >and p0∧¬p0with ⊥in the above deﬁnition.

Proposition 22. Grz 6∈ L(S(¬∧)) and Grz 6∈ L(S(>⊥¬)).

Proof. The proof is completely parallel to the proof of proposition 17, except that when

we assume that iis a S(¬∧) or S(>⊥¬) substitution and that v(i(p→2p)) = 0 we may

infer that iis neither tautologous nor contradictory and can apply lemma 21 to obtain the

relevant substitution j.

We end our discussion of these non-modal substitution classes by relating them to Kripke

semantics. For the substitution class Hwe will work with the class of partial function frames

(W, ⊆) where:

W:= X * {>,⊥} (the set of partial functions from a ﬁnite set Xto a two valued

set).

22

We call the logic of these frames the logic of ﬁnite partial functions.

Recall that the Kripke frame (WH, RH) associated with His deﬁned by letting WH=H

and RH={(i, j ◦i)|i, j ∈H}, and the valuation Von (WH, RH) associated with a H-

valuation vis deﬁned by V(i, pk) = v(ipk)..

Proposition 23. For any ﬁnite set Xthere is a surjective p-morphism, f, from the Kripke

frame associated with H,(WH, RH), to the ﬁnite partial function frame (X * {>,⊥},⊆).

Moreover, for any H-valuation vand letters p1...pnwhere n=|X|, there exists a valu-

ation Ufor which the truth-values of p1...pnare preserved by f:

V(i, pk) = U(f(i), pk)for k= 1...n

here Vis the valuation associated with v, deﬁned by V(i, pk) = v(ipk).

Proof. For convenience we will let X={p1...pn}. Deﬁne a function f:H→({p1...pn}*

{>,⊥}) as follows:

f(i)(pk) = (undeﬁned if ipk=pk

ipkotherwise

It remains to show that fsatisﬁes the two conditions for being a p-morphism. To establish

the ﬁrst condition we must show that f(i)⊆f(j◦i) for any iand jin H. If f(i)(pk) is

deﬁned and = >then ipk=>and so jipk=>, and thus f(j◦i)(pk) = >. Similarly if

f(i)(pk) is deﬁned and = ⊥then f(j◦i)(pk) = ⊥, so f(i)⊆f(j◦i).

The second condition amounts to showing that, for any i∈Hand w∈ {p1...pn}*

{>,⊥}, if f(i)⊆wthen there exists a substitution j∈Hsuch that f(j◦i) = w.j(pk) may

be deﬁned as w(pk) if w(pk) is deﬁned, and as pkotherwise.

Given a H-valuation v, we can now deﬁne a valuation Uon ({p1...pk}*{>,⊥}):

U(w, pk) =

v(pk) if w(pk) is undeﬁned

1 if w(pk) = >

0 if w(pk) = ⊥

To establish that it respects the letters p1...pnwe must show that v(ipk) = V(f(i), pk) for

each k= 1, ..., n. If ipk=pk, then f(i)(pk) is undeﬁned and so by the deﬁnition of V,

V(f(i), pk) = v(pk). But since ipk=pk,V(f(i), pk) = v(ipk) as required. Otherwise ipk=

>or ipk=⊥. In the former case V(f(i), pk) = 1 by deﬁnition of V, and v(ipk) = v(>) = 1,

so they are identical as required. The latter case is proved in the same manner.

Theorem 24. L(H)is the logic of ﬁnite partial functions.

Proof. We shall show that a sentence is consistent with L(H) iﬀ it is satisﬁable in some

ﬁnite partial function frame.

If Ais consistent with L(H) then for some (arbitrary) substitution instance A0and some

H-valuation, v,v(A0) = 1. Let p1...pnbe the letters in A0. It is immediate from proposition

23 that there is a valuation Uover the partial function frame ({p1...pn}*{>,⊥},⊆) such

that U(f(ι), A0) = 1 and so A0is satisﬁable in a ﬁnite partial function frame. Athus must

be also consistent in the logical of partial function frames, for otherwise its substitution

instances would be inconsistent too.

23

Now suppose that Ais true at some world win a partial function model (X * {>,⊥},⊆

, U ): U(w, A) = 1. By 23 there exists a surjective p-morphism from (WH, RH) to (X *

{>,⊥},⊆). Choose some substitution k∈Hsuch that f(k) = w.

Here we follow the proof of proposition 12. Let vbe any H-valuation and Vthe associated

valuation on (WH, RH). We can construct a substitution, i(which need not necessarily

belong to H), such that for any j∈H,V(j, ipm) = U(f(j), pm). Then, because fis a p-

morphism, V(j, iB) = U(f(j), B ) for any sentence B(see proposition 12). So in particular,

V(k, iA) = U(f(k), A) = U(w, A) = 1. Thus some substitution instance of A, namely kiA,

is true in a H-valuation, namely v, and thus Ais consistent in L(H).

For each w:X * {>,⊥} deﬁne a sentence

Cw:= ^

w(pm)=>

2pm∧^

w(pm)=⊥

2¬pm∧^

pm6∈dom(w)

(3pm∧3¬pm)

where pmranges over the letters p1...pn. Now consider the substitution:

ipm:= _

U(w,pm)=1

Cw

Observe ﬁrstly that Cwand C0

ware inconsistent in the propositional calculus when w6=w0.

First we establish that for any j∈H,V(j, Cf(j)) = v(jCf(j)) = 1. There is a conjunct

of jCf(j)for each pmwhere m∈ {1, ..., n}: we will show that each such conjunct is true.

If f(j)(pm) is undeﬁned, then jpm=pmand the relevant conjunct 3jpm∧3¬jpmis true

according to v. If f(j)(pm) = >or ⊥then jpm=>or ⊥, respectively, and the relevant

conjunct, 2jpmor 2¬jpmrespectively, is true according to v.

Now V(j, ipm) = 1 iﬀ v(jipm) = v(jWU(w,pm)=1 Cw) = 1 iﬀ for some wsuch that

U(w, pm) = 1, v(j Cw) = 1. v(jCf(j)= 1 and the Cs are pairwise incompatible, w=f(j).

So the last statement holds iﬀ U(f(j, pm) = 1 as required.

Next we turn to the logic of the non-modal substitution class. Let Cdenote the set of

substitutions of letters for arbitrary non-modal sentences (i.e. S(¬∧)). Consider the class

of frames FX= (W, ≤), where Xis a ﬁnite set and:

W:= {(P, a)|P⊆ {0,1}X, a ∈P}

(P, a)≤(Q, b) iﬀ P⊇Q.

The logic of the substitution class Cmay similarly be related to the logic of this class of

frames.

Proposition 25. For any ﬁnite set Xthere is a surjective p-morphism, f, from the Kripke

frame associated with C,(WC, RC), to the frame FX.

Moreover, for any C-valuation vand letters p1...pnwhere n=|X|, there exists a valua-

tion Ufor which the truth-values of p1...pnare preserved by f:

V(i, pk) = U(f(i), pk)for k= 1...n

Proof. Let vbe any C-valuation. Without loss of generality we will suppose that X=

{p1...pn}. Given a ﬁnite set of letters, Z, and a∈2Z, let Babe the formula Va(pm)=1 pm∧

Va(pm)=0 ¬pm.

24

Deﬁne a function f:C→ {(P, a)|P⊆2X, a ∈P}as follows:

f(i)=({a∈2X|v(3iBa)=1}, v ◦iX)

It remains to show that fsatisﬁes the two conditions for being a p-morphism. To

establish the ﬁrst condition we must show that f(i)≤f(j◦i) for any iand jin C. Suppose

abelongs to the ﬁrst component of f(j◦i), so that v(3jiBa) = 1. So for some k∈C,

v(kjiBa) = 1. k◦j∈C, so v(3iBa) = 1, which means that abelongs to the ﬁrst component

of f(i). Since awas arbitrary, we have shown that f(i)≤f(j◦i).

Now suppose that f(i)≤(Q, e), where f(i) = (P, d). We wish to ﬁnd a substitution j

such that f(j◦i)=(Q, b). Let q1...qrbe the letters appearing in ip1...ipn, and call members

of 2{q1...qr}truth-value assignments to q1...qr. Given a∈2{q1...qr}and a propositional

formula Ain the letters q1...qrwe write a(A) for As truth-value under the assignment

a. Let a◦i:{p1...pn} → {0,1}be the function pm7→ a(ipm), and let Ybe the set

{a∈2{q1...qr}|a◦i∈Q}. (Note that every element of Pis of the form a◦ifor some

a∈X.) For any a∈2{q1...qr}there is a corresponding >⊥ substitution, ka, deﬁned

on {q1...qr}deﬁned by ka(qm) = >if a(qm) = 1 and ka(qm) = ⊥otherwise. For any

propositional formula Ain the letters q1...qrit is clear that v(kaA) = a(A).

Given a surjection σ: 2{q1...qr}→Y, we can deﬁne a substitution on q1...qr(as in lemma

10):

j(qm) = _

σ(a)(qm)=1

Ba

where aranges over members of 2{q1...qr}. Notice that σ(a)(pm) = 1 iﬀ Bais a disjunct of

j(qm), iﬀ a(jqm) = 1, iﬀ v(kajqm) = 1. So we have σ(a)(qm) = a(jqm) = v(kajqm), for any

a∈2{q1...qr}and m∈ {1, ..., r}, and so for any propositional formula Ain letters q1...qr:

σ(a)(A) = a(jA) = v(kajA)

Since iBcis a propositional formula, we get thatσ(a)(iBc) = a(jiBc) = v(kajiBc).

We ﬁrstly show that {a|v(3jiBa) = 1}=Q. If c∈Qthen c=b◦ifor some b∈Yand

so there exists an asuch that σ(a) = bsince σis surjective. So v(kajiBc) = σ(a)(iBc) =

b(iBc) = c(Bc) = 1. So for any c∈Q,v(3jiB c) = 1, and thus Q⊆ {c∈2X|v(3jiBc) =

1}. Conversely, if v(3jiBc) = 1 then for some substitution k,v(kjiBc) = 1. Indeed, since

jiBcis a propositional formula, we may assume without loss of generality that kis a >⊥-

substitution, kafor some a∈X. So 1 = v(kajiBc) = σ(a)(iBc)=(σ(a)◦i)(Bc), and since

σ(a)∈Y,σ(a)◦i∈Q. But (σ(a)◦i)(Bc) = 1 iﬀ σ(a)◦i=c,c∈Q.

We would further like v(jiBe) = 1, so that f(j◦i)=(Q, e) as required of a p-morphism.

Since e∈Q, we know there exists a substitution ksuch that v(kjiBe) = 1. Moreover,

since jiBeis a propositional formula, any substitution k0such that v(kqm) = v(k0qm) will

also be such that v(k0jiBe) = 1. So we may assume without loss of generality that kis a

substitution that maps qmto itself or ¬qmfor m= 1...r. We claim that k◦jis the required

substitution. By construction, v(kjiBe) = 1, but also, since we have replaced literals for

literals, {a|v(3kjiBa)=1}={a|v(3jiBa)=1}=Q.

Now deﬁne a valuation Uon FX. For m= 1...n:

U((P, d), pm) = d(pm)

Umay be set arbitrarily on the remaining letters. It is immediate fpreserves the truth-

values of p1...pn:f(i)=(P, v ◦i) for some P, so V(i, pm) = v(ipm) = v◦i(pm) = U((P , v ◦

i), pm).

25

Theorem 26. L(C)is the logic of the frames FXfor ﬁnite sets X.

Proof. We shall show that a sentence is consistent with L(C) iﬀ it is satisﬁable in some

frame FX.

If Ais consistent with L(C) then for some (arbitrary) substitution instance A0and some

C-valuation, v,v(A0) = 1. Let X={p1...pn}be the set of letters in A0. It is immediate

from proposition 25 that there is a valuation Uover the frame FXsuch that U(f(ι), A0)=1

and so A0is satisﬁable in FX.Amust be also consistent in the logical of partial function

frames, for otherwise its substitution instances would be inconsistent too.

Now suppose that Ais true at some world (P, d) in a model (FX, U ): U((P, d), A) = 1.

By 25 there exists a surjective p-morphism from (WC, RC) to FX. Choose some substitution

k∈Csuch that f(k)=(P, d).

Here we follow the proof of proposition 12. Let vbe any C-valuation and Vthe associated

valuation on (WC, RC). We can construct a substitution, i(which need not necessarily

belong to C), such that for any j∈H,V(j, ipm) = U(f(j), pm). Then, because fis a

p-morphism, V(j, iB) = U(f(j), B ) for any sentence Bin letters p1...pn. So in particular,

V(k, iA) = U(f(k), A) = U((P, d), A) = 1. Thus some substitution instance of A, namely

kiA, is true in a C-valuation, namely v, and thus Ais consistent in L(C).

For each world (P, d) of FXdeﬁne a sentence:

CP,d := ^

a∈P

3Ba∧^

a∈2X\P

¬3Ba∧Bd

Where, as before, Ba=Va(pm)=1 pm∧Va(pm)=0 ¬pmand mranges from 1 to n. Now

consider the substitution:

ipm:= _

U((P,d),pm)=1

CP,d

Observe ﬁrstly that CP,d and CP0,d0are inconsistent in the propositional calculus when

(P, d)6= (P0, d0).

First it can be shown that for any j∈C,V(j, Cf(j)) = v(jCf(j)) = 1. In fact, this is

trivial from the deﬁnition of f: the conjuncts of the form 3jBain Cf(j)are deﬁned as those

where v(3jBa) = 1, the conjuncts of the form ¬3jBain Cf(j)are deﬁned as those where

v(3jBa)6= 1, and the ﬁnal conjunct is Bv◦jand v◦j(Bv◦j) = 1 (since a(Ba) = 1 for any

a∈2X).

Now V(j, ipm) = 1 iﬀ v(jipm) = v(jWU((P ,d),pm)=1 CP,d) = 1 iﬀ for some (P, d) such that

U((P, d), pm) = 1, v(jCP,d ) = 1. v(jCf(j)= 1 and since the Cs are pairwise incompatible,

(P, d) = f(j). So the last statement holds iﬀ U(f(j), pm) = 1 as required.

The ﬁnal non-modal substitution class we will consider is K. In section 3.2 we showed

that the pre-validities of Kwere identical to the C-valid sentences — the sentences valid

according to Carnap’s interpretation of propositional modal logic. In Thomason (1973)

an axiom system extending S5 is presented, and it is shown to be complete with respect

to the C-valid sentences. It consists of the result of closing all instances of the following

axioms under modus ponens and the rule of necessitation (but not the rule of uniform

substitution):19

19This system is closely related to S13 from Cocchiarella (1974), the main diﬀerence being that Coccia-

rella’s system doesn’t have a primitive notion of negation.

26

PC Any instance of a propositional tautology.

K2(A→B)→2A→2B

T2A→A

5¬2A→2¬2A

Log ¬2Awhen Ais a propositional formula that is not tautologous.

It follows from proposition 17 and Thomason’s completeness result that this system com-

pletely axiomatizes the pre-validities of K.

What of the validities of the substitution class K,L(K)? In light of the fact that C-

validity is not closed under the rule of substitution, Cresswell (2013) has proposed that we

treat a sentence as valid according to the Carnapian interpretation of 2only if all of its

substitution instances are C-valid. Thus Cresswell’s proposed notion of validity stands to

Carnap’s as the present notion of validity with respect to a substitution class stands to pre-

validity (as in deﬁnitions 2 and 3). Cresswell proves that the validities of Carnap’s theory

are exactly the theorems of S5. Thus we have

Corollary 27. L(K) = S5

Proof. From proposition 17 and theorem 3.3 of Cresswell (2013).

4.2 The logic of the full substitution class

Section 2.1 established the existence of a valuation for the full substitution class S(¬ ∧ 2),

and so we know that the logic of this class is non-trivial. While we do not know what the

logic of the full substitution class is, we will show in this section that, given a variant of

Friedman’s conjecture (conjecture 4 above), it is exactly Med, and that even if this conjecture

is false, the logic of the full substitution class contains Sub, a distinctive principle belonging

to Med.

The uniqueness conjecture tells us that for any S(¬ ∧ 2)-valuation v,v(2A) = 1 iﬀ

A∈Med. We may now see how this conjecture settles the logic of the full substitution

class:

Proposition 28. Given the uniqueness conjecture, L(S(¬ ∧ 2)) = Med.

Proof. Suppose A∈Med. So for any S(¬ ∧ 2)-valuation v,v(2A) = 1 by the conjecture

and theorem 9. So v(iA) = 1 for every substitution i∈S(¬ ∧ 2). Since vwas arbitrary,

Ais valid. Conversely, suppose that Ais valid, so that v(iA) = 1 for every substitution

i∈S(¬ ∧ 2) and S(¬ ∧ 2)-valuation v. By theorem 9 we know there is at least one such

v, and by the condition for necessity, we know v(2A) = 1. Moreover, according to the

valuation constructed in theorem 9 v(2A) = 1 only if A∈Med.

Note that we only appealed to the conjecture in proving one of the two directions,

allowing us to establish the following without assuming the conjecture to be true.

Proposition 29. S4M ⊆L(S(¬∨∧)) ⊆Med

27

This rules out some extensions of S4M, but obviously leaves open any modal logic be-

tween S4M and Med.

One distinctive feature of Med is that it contains something we will call the “subset prin-

ciple”. To state this principle we will begin by describing some formulas DYfor Y∈P0(X)

that characterize the worlds of certain sorts of Medvedev models of the form (P0(X),⊇, V ).

Write Z⊆0Y(Z⊂0Y) to mean that Zis a non-empty (proper) subset of Y. To each set,

X={1, ..., n}, associate in some canonical way a propositional partition A1...An. For non-

empty Y⊆X, we will deﬁne a formula DX

Y(or simply DYwhen Xis clear from context).

When |Y|= 1:

D{m}=32Am∧Vk6=m¬32Ak

When DZis deﬁned for |Z| ≤ kand |Y|=k+ 1:

DY=VZ⊂0Y3DZ∧2(VZ⊂0Y3DZ∨WZ⊂0YDZ).

The subset principle is then:20

Sub WY∈P0(X)DY

Observe that Sub implies the formula 2(VZ⊂0Y3DZ∨WZ⊂0YDZ) for each non-empty

Y⊆X, and indeed, these formulas provide an equivalent formulation of Sub.

Proposition 30. Every instance of Sub is in Med

Proof. Consider a Medvedev model (P0(Y),⊇, V ), and for each w∈P0(Y) let f(w) = {k∈

X|w32Ak}for X={1,2, ..., n}.

Claim: wDf(w)

The proof is by induction on the cardinality of f(w). If |f(w)|= 1 then f(w) = {m}for

some m∈X, so w32Amand w¬32Akfor m6=k. That is, wD{m}.

Suppose the claim is true when |f(w)| ≤ k. To show:

1. w3DZfor each Z⊂f(w)

2. w2(VZ⊂0f(w)3DZ∨WZ⊂0f(w)DZ)

For 1, note that for any Z⊂0f(w), wsees at least a world vsuch that f(v) = Z, namely

{a∈w| ∃k∈Z, {a}Ak}. By the inductive hypothesis vDf(v)where f(v) = Z, and

so w3DZ.

For 2, suppose that w⊇v. If f(v) = f(w) then wVZ⊂0f(w)3DZby the previous

argument. Otherwise f(v)⊂f(w), and by the inductive hypothesis vDf(v), and so

vWZ⊂0f(w)DZ.

Note that fis a p-morphism from the present Medvedev frame on P0(Y) to the frame

on P0(X).

It may also be shown that all instances of Sub are valid for the full substitution class:

Proposition 31. Sub ∈L(S(¬ ∧ 2))

20It can be see to be a generalization of an axiom schema from Holliday (2017) and Hamkins et al. (2015),

where 1 < k ≤m:

(Vi≤m32Ai∧ ¬3Wi6=jAi∧Aj)→3(Vi≤k−132Ai∧Vk≤j≤m¬32Aj)

Our axiom is strict stronger than this axiom in the presence of S4.

28

Proof. Let X={1, ..., n}, and let vbe an arbitrary valuation of the full substitution class.

We will show that v(iDf(i)) = 1. It follows that v(iWY⊆XDY) = 1 for every substitution

i, securing the validity of (A1).

Base: suppose |f(i)|= 1. So f(i) = {m}, which means that v(32iAm) = 1 and

v(32iAr) = 0 for r6=m.

Inductive Step: suppose the inductive hypothesis holds when |f(i)| ≤ m, and suppose

that |f(i)|=m+ 1.

We must show

(i) v(3iDZ) = 1 for each Z⊂f(i) and

(ii) v(2(VZ⊂f(i)3iDZ∨WZ⊂f(i)iDZ)) = 1.

For (i), we may appeal to lemma 10 to obtain a jsuch that f(j◦i) = Z. Since |f(j◦i)| ≤

nwe may apply the inductive hypothesis and conclude v(j iDf(j◦i)) = 1, and thus that

v(3iDf(j◦i)) = 1 = v(3iDZ).

Now we show (ii). For any j,f(j◦i)⊆f(i). If f(j◦i) = f(i) then v(3jiDZ) = 1 for

each Z⊂f(i) = f(j◦i) by repeating the reasoning for part (i) (using j◦iinstead of i). If

f(j◦i)⊂f(i) then v(jiDf(j◦i)) = 1 by the inductive hypothesis. So for any substitution j,

v(j(VZ⊂f(i)3iDZ∨WZ⊂f(i)iDZ)) = 1 and so v(2(VZ⊂f(i)3iDZ∨WZ⊂f(i)iDZ)) = 1.

Thus we have a tight bound on the logic of the full substitution class:

S4MSub ⊆L(S(¬ ∧ 2)) ⊆Med.

There are, however, further valid principles of Med that we have not been able to verify

to be valid for the full substitution class. One is:

(A1) DY→3(DY∧VZ⊆0YWk2(DZ→Bk))

where B1...Bris an arbitrary partition of propositional formulas.

(A1) may be seen to be valid as follows. Let (P0(Y),⊇, V ) be a model over a Medvedev

frame. Suppose DZis true at w, where Z⊆X={1, ..., n}. The formulas A1...Anpartition

the terminal worlds {{m} | m∈w}according to which Akthey make true. For each formula

Akof A1...Ansuch that w32Ak, pick a single representative mk∈w, such that {mk}

Ak. It is clear that wsees {m1...mn}, and that for any Z⊆ {1, ..., n},DZis true only at the

corresponding subset of {m1, ..., mn}:{mk|k∈Z}. Because each DZis true in exactly one

world seen by {m1, ..., mn}, it follows that {m1, ..., mn}DY∧2VZ⊆0YWk2(DZ→Bk).

Another is a rule under which Med is closed:

(R1) If `(DX

X∧VZ⊆0XWk2(DX

Z→Sk)) →Cfor for every ﬁnite set X, then `C

Where Skrange over all possible consistent conjunctions of literals in the letters appearing

in Cand DX. Indeed, it may be shown, without too much diﬃculty, that Med is the smallest

modal logic containing S4,Sub and closed under (R1). Thus if we could show that ∆vis

closed under (R1) whenever vis a valuation of the full substitution class, the uniqueness

conjecture would be solved.

We may also use the fact that Sub belongs to the logic of the full substitution class

to fully resolve Humberstone’s second question: the logic of the full substitution class is

diﬀerent from the logic of H.

Proposition 32. L(H)does not contain all instances of Sub.

29

Proof. Let A1...A4be the four way partition of propositional formulas: p1∧p2,p1∧ ¬p2,

¬p1∧p2, and ¬p1∧ ¬p2.

In any H-valuation, it is easy to see that p1∧p2,p1∧ ¬p2,¬p1∧p2, and ¬p1∧ ¬p2are

all possibly necessary by considering the four possible ways of substituting >and ⊥for p1

and p2. Thus v(DY) = 0 for any proper subset Yof {1,2,3,4}and valuation v. So if Sub

were valid, then v(D{1,2,3,4}) = 1; and so, in particular, v(3D{1,4}) = 1 by the deﬁnition of

D{1,2,3,4}.

But 3D{1,4}implies 32(p1∧p2) and 32(¬p1∧ ¬p2), and also ¬32(p1∧ ¬p2) and

¬32(¬p1∧p2). Yet any Hsubstitution according to which p1∧p2is possibly necessary and

¬p1∧ ¬p2is possibly necessary, p1and p2must be mapped to themselves, and not to >or

⊥, since it must remain possible to change the truth values of both p1and p2. But relative

to any substitution in which p1and p2are mapped to themselves, ¬p1∧p2and p1∧ ¬p2

must also be possibly necessary, since p1and p2may be replaced with >or ⊥at will.

We end this section with a weaker conjecture. It’s interesting to note that in the S(¬ ∧

2)-valuations that we constructed by setting v(2A) = 1 iﬀ A∈Med, every possibility

is witnessed by a substitution of sentence letters with sentences of modal degree 2 (the

sentences Di

p) — for short, a substitution of modal degree 2. So if vis such a valuation,

and if v(3A) = 1 then there is some substitution iof modal degree 2 such that v(iA) = 1.

(Note that the substitutions of modal degree 2 do not themselves form a substitution class,

since they are not closed under composition, despite the fact that they are well-behaved in

this context.) This suggests the following more general claim:

Conjecture 33. If vis a S(¬ ∧ 2)-valuation and v(3A) = 1, then v(iA) = 1 for some

substitution of modal degree 2.

Given the above observation, this claim clearly follows from conjecture 4, but is a question

that might be more easily tackled directly.

5 Other Theories of Logical Necessity

In the introduction we considered a couple of alternatives to McKinsey’s constraint. The

ﬁrst replaces the substitutional analysis of logical truth with an arbitrary logic ∆. The

propositional analogue of logical truth, so conceived, is an interpretation of the modal

operator 2under which a sentence 2Ais true just in case Ais a member of ∆. In the second,

we replaced Bolzano’s substitutional analysis with a Tarski-style analysis in a higher-order

logic, in which a sentence A(c1...cn) is logically true just in case the ∀x1...xn.A(x1...xn) is

true where c1...cnenumerate the non-logical constants appearing in A.21 The corresponding

constraint on logical necessity can then by stated by a single biconditional:22

2A(c1...cn)↔ ∀x1...xn.A(x1...xn)

In this section we will investigate both of these options.

21Tarski’s theory of logical truth was originally formulated in a higher-order logic, as opposed to the set-

theoretic model theory it has come to be associated with. See also the notion of ‘Metaphysical Universality’

from Williamson (2013).

22See the principle Logical Necessity from Bacon (2020) for applications of this theory of logical necessity

in the context of theorizing about fundamentality.

30

5.1 The Tarskian Constraint

According to the Tarskian analysis, one can express the logical truth of a sentence of a

propositional modal logic, such as 2p→p, with a single sentence of propositionally quan-

tiﬁed modal logic, in this case ∀X(2X→X). More generally, the logical truth of a closed

sentence A(p1...pn) in sentential constants p1...pnshould be equivalent to the truth of the

quantiﬁed sentence ∀X1...Xn.A(X1...Xn); and so one might articulate the idea that a propo-

sitional operator expresses the worldly analogue of logical truth under this conception by

means of the following schema:

2A(p1...pn)↔ ∀X1...Xn.A(X1...Xn)

This is clearly a restriction of what we earlier called the Tarskian Constraint to a sublanguage

of higher-order logic.

We will ﬁnd it helpful to consider the dualized form of this schema:

3A(p1...pn)↔ ∃X1...Xn.A(X1...Xn)

Employing standard deﬁnitions for ∃and 3.

This schema may be regarded as expressing a principle of recombinatorialism with re-

spect to the atomic propositions, a position once popular among the early logical atomists.

For let us suppose that the sentential constants p1, ..., pnof the language express atomic

propositions. Since A(X1...Xn) contains propositional variables, not sentence letters, it

expresses a relation among propositions deﬁned in purely logical terms. So the right-to-

left direction of the equivalence says that the particular atomic propositions p1, ..., pncan

instantiate any logical pattern that is in fact instantiated by some propositions, and the

left-to-right direction tells us that these exhaust the logical patterns that the atomic propo-

sitions can instantiate. For instance, since there are truths and falsehoods, two instances of

the schema (3p0↔ ∃X.X and 3¬p0↔ ∃X.¬X) tell us that every atomic proposition must

be contingent. Similarly, since there are logically necessary propositions and logically false

propositions, we may infer that every atomic proposition is possibly necessary and possibly

necessarily false: 32p0and 32¬p0.23 It is also worth noting that for this very reason we

should not expect the schema itself to be necessary: we have already established that it is

possible that p0is necessary, and also that certain instances of the schema imply that p0is

contingent and so if all instances of the schema were necessary it would be possible for p0

to be both necessary and contingent.

Here we show that there are indeed interpretations of propositionally quantiﬁed modal

logic under which the schema is true. The language of propositionally quantiﬁed modal

logic augments the language of propositional modal logic with an inﬁnite set of variables

that may occupy sentence position, X1, X2, ..., and a quantiﬁer ∀that can bind sentence

variables. In addition to the usual syntactic clauses for propositional modal logic we also

stipulate that sentence variables are formulas, and that ∀Xk.A is a formula whenever Ais.

We will consider the “full” sorts of Kripke models for this language in which the propo-

sitional quantiﬁers range over all sets of worlds (see Fine (1970)). Given an ordinary Kripke

23The existence of truths and falsehoods may be derived given the usual axioms for the propositional

quantiﬁers. E.g. existential generalization lets one move from >to ∃X.X and ¬⊥ to ∃X.¬X. The existence

of logically necessary propositions and logically false propositions can also be derived directly: 2∀X(X∨

¬X)↔ ∀X(X∨ ¬X) is an instance of the schema (where n= 0), and since the usual quantiﬁcational

axioms secure ∀X(X∨ ¬X), we may infer 2∀X(X∨¬X) from which we may conclude ∃X2Xby existential

generalization. The existence of logically false propositions follows by a similar argument.

31

model for propositional modal logic, (W, R, V ), we interpret an arbitrary sentence of propo-

sition modal as follows. Let gand g0range over variable assignments mapping each propo-

sitional variable Xkto a subset of W:

Vg(w, pk) = V(w, pk)

Vg(w, Xk) = 1 iﬀ w∈g(Xk)

Vg(w, A ∧B) = min(Vg(w, A), V g(w , B))

Vg(w, ¬A) = 1 −Vg(w , A)

Vg(w, 2A) = 1 iﬀ Vg(w0, A) = 1 for every w0such that Rww0.

Vg(w, ∀XkA) = 1 iﬀ Vg0(w, A) = 1 for every g0for which gand g0agree on every

variable except, possibly, on Xk

We may now construct a model of the Tarskian constraint.24

Theorem 34. There is a model of the Tarskian constraint.

Proof. Consider the Kripke frame:

W:= N<ω (ﬁnite sequences of naturals)

Rww0iﬀ wis a (proper or improper) initial segment of w0.

The frame can be visualised as an inﬁnitary tree, that has the empty sequence as its root,

and that branches inﬁnitely many times at each node. Enumerate the sentences satisﬁable

at the root of this frame A1, A2, A3.... For each n, let (W, R, Vn) be a Kripke model over

this frame in which Anis true at the root. We construct a global model over (W, R) as

follows:

V((), pi) may be set arbitrarily

V((n1, ..., nk), pi) = Vn1((n2, ..., nk), pi) when k≥1

It is clear by construction that Anis true at the world (n), and thus 3Anis true at the root

world () for every n. Let F ↑ wbe the subframe of Fgenerated by w. Then it is also clear

that (W, R)↑wis isomorphic to (W, R) for any world w∈W.

Now if ∀X1...Xn.A(X1...Xn) is true at the root world () then Ais valid over the frame

(W, R). But since for any world w∈W, the generated frame (W, R)↑wis isomorphic to

(W, R) it follows that Ais valid over the frame (W, R)↑w, and so true at the submodel

of (W, R, V ) generated by w. This means that Ais true at win (W, R, V ), and since

wwas arbitrary, 2Ais true at the base world (). Conversely, if 2Ais true at (), then

Amust be true at (n) for each n∈N, and thus ¬Ais not satisﬁable in (W, R). So

∀X1...Xn.A[X1/p1...Xn/pn]) is true at ().

24For a version of this sort of construction for a fully ﬂedged higher-order logic, see Bacon (2020). There

it is also connected to the substitutional analysis (Bacon (2019)).

32

5.2 The Metalogical Constraint

We observed previously that in the valuation constructed in theorem 9, v(2A) = 1 if and

only if Awas a member of a certain modal logic, Med. Let ∆v={A|v(2A) = 1}. Then

we have the following general result:

Proposition 35. If vis a S(¬ ∧ 2)-valuation then ∆vis a modal logic extending S4M.

Proof. Propositions 14 and 15 already establish that whenever Ais an instance of K,T,4

or Mthen Ais valid in v, i.e. v(iA) = 1 for every substitution i∈S(¬ ∧ 2) and, since v

is a S(¬ ∧ 2)-valuation, it follows that v(2A) = 1, and so A∈∆v. Since every instance

of 4is true in v, ∆vis closed under the rule of necessitation. Finally, ∆vis closed under

the rule of substitution since if v(2A) = 1, then v(22A) = 1, and so v(2iA) = 1 for any

substitution i∈S(¬ ∧ 2).

Note that for an arbitrary substitution class, S, and S-valuation v, ∆vwill contain all

instances of K,T,4and will be closed under the rule of necessitation. However, it may not

be a modal logic because it may not be closed under the rule of substitution, but will only

be guaranteed to be closed under substitutions in S.

It follows that the interpretation of 2in a S(¬ ∧ 2)-valuation will satisfy an instance of

the Meta-Logical Constraint considered in the introduction, according to which the inter-

pretation of logical necessity is identiﬁed with a ﬁxed modal logic ∆. We make this precise

as follows:

Deﬁnition 9 (Meta-valuations).Let ∆be a normal modal logic. A function v:L