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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 questions , 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 In particular, we present an hitherto unpublished solution to problems 41 and 42 from Friedman's 102 problems, which uses a different method of proof from the solution presented in the paper of Tadeusz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. 1 Given a language L containing 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.
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(AB) = 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 (AB)∈ 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 iSto 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 iSon Lcommutes with the logical
connectives, ,¬,2:
1. i(AB)=(iA iB)
2. i¬A=¬iA
3. i2A=2iA
Identity There is a substitution iSsuch that iA =Afor every sentence Aof L.
Composition For any two substitutions, i, j Sthere is as substitution kSsuch 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 iS. (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 mn.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(AB) = (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 kN.
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 ijiff 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(AB) = P(A)P(B), N(AB) = 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(AB) = min(v(A), v(B))
v(¬A)=1v(A)
v(2A) = 1 if and only if, for every iS,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(AB) = max(v(A), v(B))
6
v(>)=1
v()=0
v(3A) = 1 if and only if, for some iS,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 3is 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(3p3¬p2(p2p)) (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 iS(¬ ∧ 2). Similarly for any sentence letter q,v(¬2q) = 1: v(2q) = 0 since
v(r∧ ¬r) = 0 and for iS(¬ ∧ 2), iq =r∧ ¬r.
We shall show that upon substituting 2qfor pin (3p3¬p2(p2p)) we get
something true in v, and so 2qwitnesses the truth of 3(3p3¬p2(p2p)). (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(2q22q)) = 1. Take any iS(¬ ∧ 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 iS(¬ ∧ 2), and so v(2(2q22q) = 1.
Now let vbe any S(¬∧)-valuation, take any iS(¬∧), and let A:= ip. By definition A
is non-modal. Suppose for reductio that v(3A3¬A2(A2A)) = 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(A2A)) = 0.
This contradicts the assumption that v(3A3¬A2(A2A)) = 1.
Thus v(3ip 3¬ip 2(ip 2ip)) = 0 for every non-modal substitution i; and so
v(3(3p3¬p2(p2p))) = 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 jSsuch that
k=ji.
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 iSand 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 jS,V(ji, A) = 1. By the inductive hypothesis,
this holds iff v(jiA) = 1 for every jS, 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:WW0such that:
(i) For any x, y W, if Rxy then R0f(x)f(y).
(ii) For any xWand y0W0, if R0f(x)y0then there exists a yWsuch 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 xW— 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 vwhen
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 iSis non-
modal, iA will have modal degree n, and so v(2A) and v(3A) may and must be defined as
miniSv(iA) and maxiSv(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|AXn}. We define a valuation inductively as follows. Each
formula of X0is treated as true. For any positive substitution iand sentence AXn, 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 AXnand 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 AMed; 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 YXi={a1, ..., an}we may construct a formula Di
Ythat is
true at Yand only at Yin Mi:
Di
Y:= VamY32AmVam6∈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} | amY},
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 von all the sentence letters:
X:= SiωXi
W:= {wX|w=Xor wis finite and non-empty}
V(w, pj) = 1 iff wXifor 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 YP0(Xi) determine a partition
on the set of worlds Wthat is indexed by the worlds YP0(Xi) of the component model,
10A propositional partition of size nmay always be constructed. Pick sentence letters p1...pkwhere 2kn
and a surjection σ: 2kn.Ammay then be defined as Wσ(u)=m(Vu(r)=1 prVu(r)=0 ¬pr). A valuation
in which V({am}, Am) = 1 can be constructed by picking some representative element ufrom f1(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:WWithat generates the
partition of Windexed by the worlds of Miis defined as follows (so that YWiwill index
the cell f1(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 wW. 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 YP0(Xi), so that
Yf(w). We must show that there exists a vwsuch that f(v) = Y.
Suppose that f(w) = {a1, ..., am}. By definition, for each akf(w), w32Ak. Thus
for each akf(w) there is a terminal world {bk} ∈ Wsuch that {bk} ⊆ wand {bk}Ak.
Let v={bk|akY}. Clearly vw. Moreover, we may show that f(v) = Y. For, given
any akY,{bk}2Akand {bk} ⊆ v, so v32Ak, and so akf(v); and, conversely, if
v32Akthen bkvand so akY.
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 AMed 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:WWiis 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 AMed. 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 vmay 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 AMed.
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(AB) = min(A, B)
v(¬A)=1v(A)
v(2A) = 1 iff AMed
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, ∆vMed
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 Yf(i), there exists a substitution jsuch that
f(ji) = 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, k0j, with the required property.
Suppose that Yf(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 mf(i), there must exist for each mf(i)
a substitution ksuch that v(2kiAm) = 1. And without loss of generality we may assume that
kis a >⊥ substitution. For each mf(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|mY}.
Our goal will be to find a substitution jthat ‘filters’ the substitutions in Sleaving all and
only substitutions in S0behind: i.e. for any kS,kjis equivalent to a substitution in
S0and every substitution in S0is equivalent to something of the form kjfor some kS.
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:SS0and define a
substitution jas follows:
j(p) = _
σ(k)(p)=>
Ak
for each p∈ {p1...pr}
We now show that for any >⊥ substitution k,kjand σ(k) are equivalent. Firstly, we
show that for any >⊥ substitution kon p1...pn,σ(k) and kjare 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 mYthen, because σis surjective, there
is some >⊥ substitution such that σ(k) = km, and kmand kjare equivalent. Since
v(2kmiAm) = 1, by definition, and kmis equivalent to kj,v(2kjiAm) = 1, and thus
v(32jiAm) = 1. So mf(ji). Conversely if mf(ji), 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). kjis equivalent to σ(k) which may be seen to equivalent to km.
Since v(2kmiAm) = 1, mY.
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:WP0(X)is a p-morphism of frames from (W, R)to (P0(X),).
Proof. Since Rik iff there exists a jsuch that k=ji, the back-and-forth condition amounts
to the claim that if f(i)Ythen there exists a jsuch that f(ji) = Y. This is just lemma
10.
It remains to show that f(ji)f(i) for every i, j W. If mf(ji) then
v(32jiAm) = 1 and so there is a substitution ksuch that v(2kjiAm) = 1. In which case
the substitution kjwitnesses the truth of v(32iAm) = 1 so mf(i).
Let X={1, ..., n}. For a given YXwe define, as before:
DY:= VmY32AmVm6∈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 Yf(j). By
lemma 10 there exists a ksuch that f(kj) = Y, and so by the inductive hypothesis,
v(kjiA) = V(f(kj), A) = V(Y, A) = 0.
Theorem 13. If vis a valuation of the full substitution class, vMed
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 YX. 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. jiis 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, 2AA.
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 ijis also in S, and so v((ij)A) = 1. For fixed j, if v(ijA) = 1 for
every iS, it follows that v(2jA) = 1; and since this holds for every jS,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 2A22A,2AA,2(AB)(2A2B)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,2A22A,2AAand 2(AB)2A2Bare
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
BA23A
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 iSis 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
2p022p0, 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:
M23A32A
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:
F23A23B3(AB)
which is only equivalent to Min the context of S4. The principle Mappears to originate from Cresswell and
Hughes (1996).
17
some terminal substitution iS,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 2AA
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 Afor 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. Ais 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 Ais a tautology.
Conversely, suppose Ais a tautology and jan arbitrary Ssubstitution. Let ibe any
terminal substitution. v(ijA) = v(i((jA))) from before. But if Ais a tautology, then
(jA)is a substitution instance of Aof the letters by sentences of propositional logic
(specifically the substitution jwhere 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. A23A(an instance of B)
2. 23A32A(an instance of M)
3. 32A2A(an equivalent of S5)
4. A2Afrom 1-3
5. 2AAby 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 iK, 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 iK:
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 iS. Recall that v(2iA) = 1 iff v(jiA) = 1 for
every substitution jK. Now for any j, we may apply the inductive hypothesis to vand
the substitution jito 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 jSiff
(iv)(jA) = 1 for every jS, 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 iK. (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 iK, iff (iv)(A) = 1 for every iK,
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|=ABiff v|=Aand v|=B
v|=¬Aiff v6|=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 vfor corresponding the truth-value assignment:
vL(). By proposition 3 every truth-value assignment is identical to vfor 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 iK,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 iK. Thus v(2A) = 1. Conversely, since
every K-valuation is of the form iv for some iK, 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(A2A)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(A2A)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: ijwhen there exists a ksuch that j=ki(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) ji=i, so
v(jiA) = 0 or else (b) ji>i, in which case v(2jiA) = 1, for v(kjiA) = 1 for any kH,
since iwas the maximal substitution making Afalse and kjiji > 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=p2p.
Proof. Let vbe a S(>⊥)-valuation with a true and a false letter, pand q. We will show that
v(2(2(A2A)A)) = 1 and v(A) = 0 when A=p2p. Let ibe any substitution in
S(>⊥), and suppose v(i(p2p)) = 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(p2p)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(q2q) = 1 since qis false, and v(ji2A) = v(2(q2q)) = 0 since
p2pis a substitution instance of q2qand is false. Thus v(jiA 2jiA) = 0, and
so v(2(iA 2iA)) = 0. So we have shown that, for an arbitrary substitution iS(>⊥),
when v(iA) = 0 then v(2(iA 2iA)) = 0 and hence that v(2(2(A2A)A)) = 1.
Finally, v(p2p) = 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 iS(¬∧)such that iA is equivalent to B.
When B∈ L(>⊥¬), there will exist a substitution iS(>⊥¬)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 iS(>⊥¬) 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(p2p)) = 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(ji) for any iand jin H. If f(i)(pk) is
defined and = >then ipk=>and so jipk=>, and thus f(ji)(pk) = >. Similarly if
f(i)(pk) is defined and = then f(ji)(pk) = , so f(i)f(ji).
The second condition amounts to showing that, for any iHand w∈ {p1...pn}*
{>,⊥}, if f(i)wthen there exists a substitution jHsuch that f(ji) = 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 kHsuch 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 jH,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)
(3pm3¬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 jH,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 3jpm3¬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 PQ.
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 a2Z, let Babe the formula Va(pm)=1 pm
Va(pm)=0 ¬pm.
24
Define a function f:C→ {(P, a)|P2X, a P}as follows:
f(i)=({a2X|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(ji) for any iand jin C. Suppose
abelongs to the first component of f(ji), so that v(3jiBa) = 1. So for some kC,
v(kjiBa) = 1. kjC, 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(ji).
Now suppose that f(i)(Q, e), where f(i) = (P, d). We wish to find a substitution j
such that f(ji)=(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 a2{q1...qr}and a propositional
formula Ain the letters q1...qrwe write a(A) for As truth-value under the assignment
a. Let ai:{p1...pn} → {0,1}be the function pm7→ a(ipm), and let Ybe the set
{a2{q1...qr}|aiQ}. (Note that every element of Pis of the form aifor some
aX.) For any a2{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
a2{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 cQthen c=bifor some bYand
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 cQ,v(3jiB c) = 1, and thus Q⊆ {c2X|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 aX. So 1 = v(kajiBc) = σ(a)(iBc)=(σ(a)i)(Bc), and since
σ(a)Y,σ(a)iQ. But (σ(a)i)(Bc) = 1 iff σ(a)i=c,cQ.
We would further like v(jiBe) = 1, so that f(ji)=(Q, e) as required of a p-morphism.
Since eQ, 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 kjis 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) = vi(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
kCsuch 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 jH,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 := ^
aP
3Ba^
a2X\P
¬3BaBd
Where, as before, Ba=Va(pm)=1 pmVa(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 jC,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 Bvjand vj(Bvj) = 1 (since a(Ba) = 1 for any
a2X).
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(AB)2A2B
T2AA
5¬2A2¬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
AMed. 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 AMed. So for any S(¬ ∧ 2)-valuation v,v(2A) = 1 by the conjecture
and theorem 9. So v(iA) = 1 for every substitution iS(¬ ∧ 2). Since vwas arbitrary,
Ais valid. Conversely, suppose that Ais valid, so that v(iA) = 1 for every substitution
iS(¬ ∧ 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 AMed.
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 YP0(X)
that characterize the worlds of certain sorts of Medvedev models of the form (P0(X),, V ).
Write Z0Y(Z0Y) 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 YX, we will define a formula DX
Y(or simply DYwhen Xis clear from context).
When |Y|= 1:
D{m}=32AmVk6=m¬32Ak
When DZis defined for |Z| ≤ kand |Y|=k+ 1:
DY=VZ0Y3DZ2(VZ0Y3DZWZ0YDZ).
The subset principle is then:20
Sub WYP0(X)DY
Observe that Sub implies the formula 2(VZ0Y3DZWZ0YDZ) for each non-empty
YX, 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 wP0(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 mX, so w32Amand w¬32Akfor m6=k. That is, wD{m}.
Suppose the claim is true when |f(w)| ≤ k. To show:
1. w3DZfor each Zf(w)
2. w2(VZ0f(w)3DZWZ0f(w)DZ)
For 1, note that for any Z0f(w), wsees at least a world vsuch that f(v) = Z, namely
{aw| ∃kZ, {a}Ak}. By the inductive hypothesis vDf(v)where f(v) = Z, and
so w3DZ.
For 2, suppose that wv. If f(v) = f(w) then wVZ0f(w)3DZby the previous
argument. Otherwise f(v)f(w), and by the inductive hypothesis vDf(v), and so
vWZ0f(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:
(Vim32Ai∧ ¬3Wi6=jAiAj)3(Vik132AiVkjm¬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(iWYXDY) = 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 Zf(i) and
(ii) v(2(VZf(i)3iDZWZf(i)iDZ)) = 1.
For (i), we may appeal to lemma 10 to obtain a jsuch that f(ji) = Z. Since |f(ji)| ≤
nwe may apply the inductive hypothesis and conclude v(j iDf(ji)) = 1, and thus that
v(3iDf(ji)) = 1 = v(3iDZ).
Now we show (ii). For any j,f(ji)f(i). If f(ji) = f(i) then v(3jiDZ) = 1 for
each Zf(i) = f(ji) by repeating the reasoning for part (i) (using jiinstead of i). If
f(ji)f(i) then v(jiDf(ji)) = 1 by the inductive hypothesis. So for any substitution j,
v(j(VZf(i)3iDZWZf(i)iDZ)) = 1 and so v(2(VZf(i)3iDZWZf(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) DY3(DYVZ0YWk2(DZBk))
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 ZX={1, ..., n}. The formulas A1...Anpartition
the terminal worlds {{m} | mw}according to which Akthey make true. For each formula
Akof A1...Ansuch that w32Ak, pick a single representative mkw, 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|kZ}. Because each DZis true in exactly one
world seen by {m1, ..., mn}, it follows that {m1, ..., mn}DY2VZ0YWk2(DZBk).
Another is a rule under which Med is closed:
(R1) If `(DX
XVZ0XWk2(DX
ZSk)) 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: p1p2,p1∧ ¬p2,
¬p1p2, and ¬p1∧ ¬p2.
In any H-valuation, it is easy to see that p1p2,p1∧ ¬p2,¬p1p2, 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(p1p2) and 32(¬p1∧ ¬p2), and also ¬32(p1∧ ¬p2) and
¬32(¬p1p2). Yet any Hsubstitution according to which p1p2is 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, ¬p1p2and 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 AMed, 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 2pp, with a single sentence of propositionally quan-
tified modal logic, in this case X(2XX). 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: 2X(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 2X(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 wg(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(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 k1
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 wW.
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 wW, 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 nN, 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 iS(¬ ∧ 2) and, since v
is a S(¬ ∧ 2)-valuation, it follows that v(2A) = 1, and so Av. 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 iS(¬ ∧ 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