In all, but finitely many, possible worlds: model

In all, but finitely many, possible worlds: model-theoretic
investigations on ‘overwhelming majority’ default conditionals
Costas D. Koutras1
Christos Rantsoudis2
[email protected]
1
2
[email protected]
Department of Informatics and Telecommunications,
University of Peloponnese
end of Karaiskaki Street, 22100 Tripolis, Greece
Graduate Programme in Logic, Algorithms and Computation
M.P.L.A., Dept. of Mathematics, University of Athens
Panepistimioupolis, 157 84 Ilissia, Greece
November 20, 2015
Abstract
Defeasible conditionals of the form ‘if A then normally B’ are usually interpreted with
the aid of a ‘normality’ ordering between possible states of affairs: (A ⇒ B) is true if it
happens that in the most ‘normal ’ (least exceptional ) A-worlds, B is also true. Another
plausible interpretation of ‘normality’ introduced in nonmonotonic reasoning dictates that
(A ⇒ B) is true iff B is true in ‘most’ A-worlds. A formal account of ‘most’ in this
majority-based approach to default reasoning has been independently introduced by K.
Schlechta and V. Jauregui, through the notion of ‘weak filters’, a weakening of classical
filters used in Set Theory. In this paper, we investigate defeasible conditionals constructed
upon a notion of ‘overwhelming majority’, defined as ‘truth in a cofinite subset of ω’, the
first infinite ordinal. One approach employs the modal logic of the frame (ω, <), used
in the temporal logic of discrete linear time. We introduce and investigate conditionals,
defined modally over (ω, <); several modal definitions of the conditional connective are
examined, with an emphasis on the nonmonotonic ones. An alternative interpretation of
‘majority’ as sets cofinal (in ω) rather than cofinite (subsets of ω) is examined. For all
these modal approaches over (ω, <), a decision procedure readily emerges, as the modal
logic KD4LZ of this frame is well-known and a translation of the conditional sentences can
be mechanically checked for validity. A second approach employs the conditional version
of Scott-Montague semantics, in the form of ω-many possible worlds, endowed with neighborhoods populated by collections of cofinite subsets of ω. Again, different conditionals are
introduced and examined. Although it is difficult to obtain a complete axiomatization of
these, model-theoretically defined, logics (capturing ‘cofiniteness-in-ω’ syntactically seems
elusive) this research reveals the possible structure of ‘overwhelming majority’ conditionals,
whose relative strength is compared to (the conditional logic ‘equivalent’ of) KLM logics
and other conditional logics in the literature.
1
Introduction
Conditional Logic is primarily concerned with the logical and semantic analysis of the rich
class of conditional statements, identified with the sentences conforming with the ‘if X then
Y ’ structure. The topic has roots in antiquity and the medieval times but its contemporary
development seems to start with F. Ramsey in the ’30s and has blossomed after the late ’60s
[AC14]. There exist various conditionals of interest in Philosophy, Logic, Computer Science
and Artificial Intelligence, including counterfactual conditionals, causal conditionals, deontic
conditionals, normality conditionals (see [CdCH96] for a broad overview of applications); they
represent different linguistic constructions with a common structural form (‘if ... then’) and
the aim of the field is to provide a unifying formal logical account that accurately captures
their essential meaning.
Artificial Intelligence has been interested in conditional logics for default reasoning already from the ’80s (see the work of J. Delgrande [Del87, Del88]), in counterfactual conditionals (M. Ginsberg, [Gin86]) and in the ‘normality conditionals’ in nonmonotonic reasoning
[Bel90, Lam91, Bou94]. The reader is referred to the handbook article of J. Delgrande [Del98]
for a broad overview of conditional logics for defeasible reasoning. The investigations on the intimate relation of conditional logics to nonmonotonic reasoning have been further triggered by
the seminal work of S. Kraus, D. Lehmann and M. Magidor [KLM90, LM92], whose framework
(KLM) has become the ‘industry standard’ for nonmonotonic consequence relations. There
exist various possible-worlds semantics for conditional logics (see [Nut80, Del98]) and a connection to modal logic (known from D. Lewis’ work [Lew73]) which has been further explored
by the modal construction of ‘normality conditionals’ [Lam91, Bou92].
A logic of ‘normality conditionals’ for default reasoning, attempts to pin down the principles
governing the statements of the form ‘if A, then normally B is the case’. ‘Normally’ is
susceptible to a variety of interpretations. One is based on a ‘normality’ ordering between
possible worlds: (A ⇒ B) is true if it happens that in the most ‘normal ’ (least exceptional ) Aworlds, B is also true [Lam91, Bou94]. Another, more recent one [Jau08] interprets ‘normally’
as a ‘majority’ quantifier: (A ⇒ B) is true iff B is true in ‘most’ A-worlds. Questions of
‘size’ in preferential nonmonotonic reasoning have been firstly introduced by K. Schlechta
[Sch95, Sch97]; the notion of ‘weak filter ’1 that emerged (as a ‘core’ definition of a collection
of ‘large’ subsets) has been also employed in modal epistemic logics [KMZ14, AKZ12].
A majority-based account of default conditionals, depends heavily on what counts as a
‘majority’ of alternative situations, what is a ‘large’ set of possible worlds. It is difficult to
state a good definition that would work for both the finite and the infinite case; the notions
of weak filters and weak ultrafilters that have been used capture the minimum requirements
of such a notion [Sch97, Jau08]. In this paper, we experiment with a notion of ‘overwhelming
majority’, combined with the widely accepted intuition that (A ⇒ B) should mean that (A∧B)
is more plausible than (A ∧ ¬B). We define conditionals of this form to (essentially) mean that
(A ∧ B) is true in ‘almost all ’ (‘all, but finitely many’) points in the countable modal frame
(ω, <) (the first infinite ordinal, strictly ordered under <), whose modal axiomatization (the
normal modal logic KD4LZ) is known as the ‘future’ fragment of the temporal logic of discrete
1
This notion is not important for our work here. For the sake of completeness it suffices to say that it
is a weakening of the classical filters in Set Theory and Model Theory. Assuming a set W , a filter is a
collection of subsets of W (a subset of its powerset), which is upwards closed and closed under intersection. A
weak filter relaxes the second condition by just requiring that there do not exits pairwise disjoint sets in the
collection.
1
linear time [Gol92, Seg70]. This majority conditional is modally defined and this readily
provides a decision procedure, as a modal translation of conditional formulas can be checked
for validity in (ω, <) using any of the proof procedures known for KD4LZ. We examine the
properties of this conditional, in particular with respect to the (conditional incarnation of
the) ‘conservative core’ of defeasible reasoning set by the KLM framework. The paradigm of
‘overwhelming majority’ in our work is consistently represented with cofinite subsets of ω. En
route, we discuss variants: trying cofinal (rather than cofinite) subsets of ω, and/or varying
the modal definition of the conditional connective. Then, we discuss the possibility of defining
conditionals over cofinite subsets of ω in the neighborhood semantics for conditional logics;
we prove that the conditionals defined can be very weak, even compared to the conditionals
introduced in [Del06].
The paper is organized as follows: In Section 2 we provide the necessary background material, establishing notation and terminology. Section 3 contains the results of this paper: in
Section 3.1 we modally define several conditional logics over (ω, <) and examine their prop⇒
erties. In particular, we introduce the ‘majority-default’ conditional logics Ω (majority as
⇒
cofiniteness) and ω (majority as cofinality). Variants to the model and/or to the modal definition are also discussed, resulting unfortunately in logics satisfying monotonicity. Section 3.2
comprises of an investigation of some additional logics defined using the Scott-Montague semantics for Conditional Logic [Che75] and our results are summarized in a table at the end of
the Section. Finally, we conclude in Section 4 with some ideas for future research.
2
2.1
Background
Modal Logic
We assume a language L of classical propositional logic, built upon the known connectives
{¬, ∧, ∨, →, ≡} over a countable set Φ = {p1 , p2 , . . .} of propositional variables. The language
L2 of propositional modal logic extends L with a modal necessity operator 2A. We assume
that the reader is acquainted with the basics of propositional Modal Logic; for details consult
[BdRV01, Gol92]. A dual possibility operator is introduced by 3A ≡ ¬2¬A.
Modal Logics are sets of formulas from L2 , containing all propositional tautologies, and
closed under the rules of Modus Ponens (MP) A,A→B
and Uniform Substitution (US).
B
A normal modal logic is a modal logic containing the axiom
K. 2A ∧ 2(A → B) → 2B
and closed under the Rule of Necessitation
RN.
A
2A
We denote by KA1 A2 . . . An the normal modal logic axiomatized by the axioms A1 . . . An .
A formula B is a theorem of logic Λ (denoted by `Λ B) iff B ∈ Λ.
Normal Modal Logics are interpreted over relational possible-worlds models or Kripke
models. A Kripke frame F = (W, R) consists of a set of possible worlds W and a binary
relation R ⊆ W × W . We say that a world w ‘sees’ a world v iff wRv. A valuation V
determines which propositional variables are true inside each possible world. Given a valuation
2
V , a Kripke model M is the triple (W, R, V ). Within a world w, the propositional connectives
{¬, ∧, ∨, →, ≡} are interpreted classically, while 2A is true at w iff it is true in every world
‘seen’ by w, i.e. M, w |= 2A iff (∀v ∈ W )(wRv → M, v |= A). A formula A is said to be valid
in a model M (M |= A) iff it is true at every world w ∈ W of M. A formula A is valid in a
frame F (F |= A) iff it is valid in every model M of F. A formula A is valid in a class of frames
C iff it is valid in every frame F of C. The truth set of A, denoted by kAkM , is the set of worlds
w ∈ W in which A is true at model M, i.e. kAkM = {w ∈ W | M, w |= A}. If the model M is
obvious, the subscript is usually dropped and we denote it simply by kAk. General Frames
are possible-worlds frames F = (W, R, X), where X is a subset of 2W closed under union,
intersection, complement and the operator 2Y = {w ∈ W | (∀u ∈ W )(wRu → u ∈ Y )}. A
Kripke model M = (W, R, X, V ) (an admissible model) based on the general frame F requires
that V draws its valuations for propositional variables (and provably for every wff) from the
set X of admissible valuations (see [BdRV01]).
A normal modal logic Λ is sound with respect to a class C of frames iff `Λ A implies C |= A.
Λ is complete with respect to C iff C |= A implies `Λ A. A logic Λ is determined by a class
C of frames iff it is both sound and complete with respect to C.
We are going to make use of the following fact: it is well known [Gol92, Seg70] that the frame
(ω, <) of the natural numbers with their natural strict ordering is axiomatized by the logic Ω,
where Ω is an abbreviation for the normal modal logic K4DLZ comprising:
4. 2A → 22A
D. 2A → 3A
L. 2(A ∧ 2A → B) ∨ 2(B ∧ 2B → A)
Z. 2(2A → A) → (32A → 2A)
The logic Ω has been investigated in the context of axiomatizing the ‘future’ fragment of
discrete linear time. We will extensively exploit below that `Ω A iff (ω, <) |= A.
2.2
Conditional Logics
The language L⇒ of propositional conditional logic extends L with a binary conditional connective (A ⇒ B), interpreted for our purposes as ‘A normally implies B ’ (also interpreted in
the literature as a counterfactual conditional, causal conditional, indicative conditional, etc).
The reader is referred to [Nut80] for more details; see also [Poz10]. We wish to remind the
reader that throughout this paper we use the symbol → for the classical material implication
and reserve ⇒ for the default conditional.
The systems of conditional logic are defined as sets closed under certain rules and possessing
certain axioms. Important rules include:
3
RCEA.
A ≡ B
(A⇒C) ≡ (B⇒C)
RCK.
(A1 ∧ ... ∧ An ) → B
(C⇒A1 ∧ ... ∧ C⇒An ) → (C⇒B)
RCEC.
A ≡ B
(C⇒A) ≡ (C⇒B)
RCE.
A → B
A ⇒ B
Important axioms comprise:
ID.
A⇒A
CUT.
(A∧B ⇒ C) ∧ (A⇒B) → (A⇒C)
AC.
(A⇒B) ∧ (A⇒C) → (A∧B ⇒ C)
CC.
(A⇒B) ∧ (A⇒C) → (A ⇒ B∧C)
Loop.
(A0 ⇒A1 ∧ . . . ∧ Ak ⇒A0 ) → (A0 ⇒Ak )
OR.
(A⇒C) ∧ (B⇒C) → (A∨B ⇒ C)
CV.
(A⇒B) ∧ ¬(A⇒¬C) → (A∧C ⇒ B)
CSO.
(A⇒B) ∧ (B⇒A) → ((A⇒C) ≡ (B⇒C))
CM.
(A ⇒ B∧C) → (A⇒B) ∧ (A⇒C)
MP.
(A⇒B) → (A→B)
MOD.
(¬A⇒A) → (B⇒A)
CA.
(A⇒B) ∧ (C⇒B) → (A∧C ⇒ B)
CS.
(A ∧ B) → (A⇒B)
CEM.
(A⇒B) ∨ (A⇒¬B)
SDA.
(A∨B ⇒ C) → (A⇒C) ∧ (B⇒C)
Principles of interest to the study of ‘normality’ conditionals:
4
Transitivity
(A⇒B) ∧ (B⇒C) → (A⇒C)
Weak Transitivity
(A⇒B) ∧ (B⇒C) ⇒ (A⇒C)
Monotonicity
(A⇒B) → (A∧C ⇒ B)
Weak Monotonicity
(A⇒B) ⇒ (A∧C ⇒ B)
Modus Ponens
A ∧ (A⇒B) → B
Weak Modus Ponens
A ∧ (A⇒B) ⇒ B
Here are some important logics from the literature (see [Nut80]):
• CU
=
RCEC + RCEA + RCK + ID + AC + CUT
• CE
=
RCEC + RCEA + RCK + ID + AC + CUT + CA
•
=
RCEC + RCEA + RCK + ID + AC + CUT + CA + CV
2.3
V
KLM Logics
The KLM approach [KLM90, LM92] to nonmonotonic reasoning (NMR) emerged in the
early ’90s from a study of nonmonotonic consequence relations and their connection
to the preferential semantics for NMR. We only provide the basic facts in the interests of
demonstrating their connection to the conditional logics defined; see the original articles for
details.
In this section, we work with the language L of propositional logic and the new formation
rule (in the metalanguage): A ∼ B denotes a (KLM) default conditional or a nonmonotonic
consequence relation (A, B ∈ L). The following rules are sometimes called Gabbay - Makinson
conditions:
A ∼A
REF.
RW.
CM.
À→B
C
C
∼B
∼B
A
A
A∧B
Loop.
RM.
A0
A
∼A
∼C
... Ak
A0
∼Ak
A
À≡B
A
B
∼C
CUT.
A∧B
∼A0
AND.
A
∼
C
OR.
A
∼C
∼A1
∼C
LLE.
A∧B
∼¬B
∼C A ∼B
A ∼C
∼B A ∼C
A ∼B∧C
∼C
A∨B
5
∼C
B
∼C
∼C
The KLM logics comprise the following systems, from the strongest to the weakest: R (Rational ), P (Preferential ), CL (Loop-Cumulative) and C (Cumulative).
• R consists of the propositional calculus plus REF, LLE, RW, CM, AND, OR, RM.
• P results from R by dropping RM.
• CL results from P by dropping OR and adding Loop and CUT.
• C, the weakest, results from CL by dropping Loop.
The following correspondences are known; [CL92, Poz10], see also Table 1, page 53:
• R corresponds to the ‘flat’ fragment of conditional logic V (CE + CV).
• P corresponds to the ‘flat’ fragment of conditional logic named CE.
• C corresponds to the ‘flat’ fragment of CU.
A ‘translation’ of the KLM conditions to conditional logic principles is known from [CL92]. It
is readily checked in the correspondences shown in Table 1, at page 53.
2.4
Modal approaches to Default Conditionals
A very influential modal approach to normality conditionals has been suggested in the early
’90s, independently by Craig Boutilier and Philippe Lamarre [Bou92, Lam91]. Starting from
similar considerations on the ‘normality’ ordering between possible worlds, they arrived to
conditional logics ‘equivalent’ to the well-known modal logic S4 which is axiomatized by
K. 2A ∧ 2(A → B) → 2B, T. 2A → A, 4. 2A → 22A and is characterized by the
possible-words models with a preorder (reflexive and transitive) relation. The relation is seen
as a normality ordering, in the sense that ‘world u sees the world v ’ is interpreted as ‘v is more
normal (or less exceptional) than w ’.
The normality conditional (A ⇒ B) is defined as follows:
(A ⇒ B)
≡def
2(A → 3(A ∧ 2(A → B)))
Through this definition, the desired interpretation of ‘B is true in the most normal worlds where
A is also true’ becomes clear and the conditional logic of normality which emerges is shown
to be equivalent to S4. The translation from conditional to modal sentences is achieved by a
technique dating from D. Lewis’ work: defining the necessity operator 2A to be a shorthand
for (¬A ⇒ A) and, accordingly, the possibility operator 3A to be a shorthand for ¬(A ⇒ ¬A).
The conditional logics CT4 [Bou92] and C4 [Lam91] defined this way are very interesting.
They contain many essential axioms and invalidate many undesirable ones. A quick glimpse
on their properties can be made through Table 1, page 53.
6
3
‘Overwhelming Majority ’ Conditionals
We wish to define default conditionals of the form ‘A normally implies B ’. The fundamental
question is to provide a concrete interpretation of the statement ‘normally’. Earlier approaches
resort to ‘normality’ orderings ([Del88, Lam91, Bou94]: (A ⇒ B) is true iff B is true in the
most normal A-worlds), and considerations of ‘size’ ([Jau08]: (A ⇒ B) is true iff B is true in
‘many’ (‘most’) A-worlds). Another fundamental intuition dictates that (A ∧ B) is ‘preferred ’
over (A ∧ ¬B) for (A ⇒ B) to be true; incidentally, another similar view, is that (A → B) is
more ‘normal ’ than (A → ¬B) [Boc01].
In this paper, we design ‘majority default’ conditionals based on the following intuition - note
that we consistently work with the infinite set ω of possible worlds:
• (A ⇒ B) is an ‘overwhelming majority’ default conditional, in the sense that B is true
in a very large set of A-worlds. We consider as ‘large’ the cofinite subsets of ω (and
‘small ’ the finite ones). Obviously, this is an (extreme, but) intuitively acceptable form
of ‘overwhelming majority’. We remind the reader that a set is cofinite iff its complement
is finite (with respect to ω in this context).
• (A ⇒ B) is true, either vacuously (if there are no ‘many’ A-worlds) or essentially: iff
kA ∧ Bk is much bigger (it is a cofinite subset of ω) than kA ∧ ¬Bk.
Throughout this section, we will be working with the set ω of countably many possible worlds,
with the aim of providing different accounts of ‘A normally implies B ’ (A ⇒ B) as ‘B is true
in all, but finitely many, A-worlds’. To keep the paper at a reasonable length, we do not
provide the full proof in many of the results below; all the details can be found at the lengthy
full report [KR15a].
3.1
Conditionals modally defined over (ω, <)
Our first approach is to define a ‘majority’ conditional over the frame (ω, <) of natural numbers, strictly ordered under <. Conforming to the intuition(s) expressed above, we will define
(A ⇒ B) as shorthand for:
(A ⇒ B)
≡def
32¬A ∨ 32(A ∧ B)
The import of such a modal definition over (ω, <) is that either there do not exist ‘many’
A-worlds (A settles down to be false, at some ‘point’ onwards) or there exist ‘overwhelmingly
many’ ‘points’ in which A∧B is true (A∧B is true in a cofinite subset of ω). To state properly
the conditional logic induced by this definition , we proceed to define the following translation
of conditionals to the (mono)modal language L2 :
Definition 3.1 We recursively define the following translation ()∗ : L⇒ → L2
(i) (p)∗ = p , if p ∈ Φ (p is a propositional variable)
(ii) (A ◦ B)∗ = (A)∗ ◦ (B)∗ for ◦ ∈ {∧, ∨, →, ≡}
7
(iii) (¬A)∗ = ¬(A)∗
(iv) (A ⇒ B)∗ = 32¬(A)∗ ∨ 32(A∗ ∧ B ∗ )
⇒
We proceed to define the logic Ω of ‘majority default conditionals’ over (ω, <):
⇒
⇒
Definition 3.2 [Conditional Logic Ω]. The logic Ω consists of all formulae A ∈ L⇒ , such
that:
⇒
A ∈ Ω iff (ω, <) |= A∗ iff `Ω A∗
The second equivalence follows from the completeness of Ω = K4DLZ with respect to the
frame (ω, <).
Fact 3.3 Let M be a model of F = (ω, <) and n ∈ ω an arbitrary world. We will say that
M, n |= (A ⇒ B) iff one of the following holds:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= A ∧ B
⇒
Some comments on the definition of Ω are in order. This model-theoretic modal definition of
the conditional has the advantage that it is a clear ‘majority’ definition, easy to understand,
with an intuitively acceptable ‘largeness’ condition. It captures ‘cofinite’ subsets of ω in an
easy manner, in contrast to the difficulty of capturing this axiomatically. Further on, and
perhaps more important, a decision procedure readily emerges from the definition: to check
⇒
whether a conditional (A ⇒ B) is in Ω, simply check whether (A ⇒ B)∗ has a tableaux proof
in K4DLZ; such a proof procedure exists, see [Gor99].
On the other hand, the ordering in (ω, <) has not any clear ‘preference’ meaning here. It
implicitly provides a ‘temporal ’ change flavour in the conditional defined, see below on the
invalidity of ID.
⇒
Let us proceed to check the properties of Ω. Throughout the proofs, F refers to (ω, <) and all
models M are based on that frame.
⇒
Theorem 3.4 The logic Ω:
1. is closed under the rules RCEA, RCK and RCEC
8
2. contains the axioms CUT, AC, CC, Loop, OR, CSO, CM, CA, Transitivity,
Weak Transitivity and Weak Modus Ponens
Proof. RCEA: Let M |= (A ≡ B). We have to show that
M |= (A ⇒ C) ≡ (B ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ C). Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A, or
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ∧ C)
Case (i): By M |= (A ≡ B) we also have that (∀n2 > n1 ) M, n2 |= (¬A ≡ ¬B). This gives
that (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬B, and M, n |= (B ⇒ C) follows.
Case (ii): Similarly, by M |= (A ≡ B) we obtain
(∀n4 > n3 ) M, n4 |= (A ∧ C ≡ B ∧ C)
It follows that (∃n3 > n)(∀n4 > n3 ) M, n4 |= (B ∧ C), and thus M, n |= (B ⇒ C).
So, if M, n |= (A ⇒ C), then M, n |= (B ⇒ C), which gives that
M, n |= (A ⇒ C) → (B ⇒ C)
Similarly, M, n |= (B ⇒ C) → (A ⇒ C). Since the world n was arbitrarily chosen, the proof
is complete.
RCK: Let M |= (A1 ∧ . . . ∧ An ) → B. We have to show that
M |= (C ⇒ A1 ∧ . . . ∧ C ⇒ An ) → (C ⇒ B)
Assume an arbitrary state n ∈ ω, such that M, n |= (C ⇒ A1 ∧ . . . ∧ C ⇒ An ). Obviously,
M, n |= (C ⇒ A1 ) and . . . and M, n |= (C ⇒ An ). Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬C, or
(ii) (∃n3 > n)(∀n4 > n3 )(M, n4 |= (C ∧ A1 ) and . . . and M, n4 |= (C ∧ An )), which means
that (∃n3 > n)(∀n4 > n3 ) M, n4 |= (C ∧ A1 ∧ . . . ∧ An )
Case (i): By definition, M, n |= (C ⇒ B) follows.
Case (ii): By M |= (A1 ∧ . . . ∧ An ) → B we obtain
(∀n4 > n3 ) M, n4 |= (C ∧ A1 ∧ . . . ∧ An ) → C ∧ B
It follows that (∃n3 > n)(∀n4 > n3 ) M, n4 |= (C ∧ B), and thus M, n |= (C ⇒ B).
So, if M, n |= (C ⇒ A1 ∧ . . . ∧ C ⇒ An ), then M, n |= (C ⇒ B), which gives that
M, n |= (C ⇒ A1 ∧ . . . ∧ C ⇒ An ) → (C ⇒ B)
9
Since the world n was arbitrarily chosen, the proof is complete.
RCEC: Let M |= (A ≡ B). We have to show that
M |= (C ⇒ A) ≡ (C ⇒ B)
Assume an arbitrary state n ∈ ω, such that M, n |= (C ⇒ A). Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬C, or
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= (C ∧ A)
Case (i): By definition, M, n |= (C ⇒ B) follows.
Case (ii): By M |= (A ≡ B) we obtain
(∀n4 > n3 ) M, n4 |= (C ∧ A ≡ C ∧ B)
It follows that (∃n3 > n)(∀n4 > n3 ) M, n4 |= (C ∧ B), and thus M, n |= (C ⇒ B).
So, if M, n |= (C ⇒ A), then M, n |= (C ⇒ B), which gives that
M, n |= (C ⇒ A) → (C ⇒ B)
Similarly, M, n |= (C ⇒ B) → (C ⇒ A). Since the world n was arbitrarily chosen, the proof
is complete.
CUT: We have to show that
F |= (A ∧ B ⇒ C) ∧ (A ⇒ B) → (A ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ∧ B ⇒ C) ∧ (A ⇒ B), where M is a
model of F. Obviously, M, n |= (A ∧ B ⇒ C) and M, n |= (A ⇒ B). Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A, or
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ∧ B ∧ C)
Case (i): By definition, M, n |= (A ⇒ C) follows.
Case (ii): We also have that (∀n4 > n3 ) M, n4 |= (A ∧ C), and thus M, n |= (A ⇒ C).
So, if M, n |= (A ∧ B ⇒ C) ∧ (A ⇒ B), then M, n |= (A ⇒ C), which gives that
M, n |= (A ∧ B ⇒ C) ∧ (A ⇒ B) → (A ⇒ C)
Since the world n and model M were arbitrarily chosen, the proof is complete.
AC: We have to show that
F |= (A ⇒ B) ∧ (A ⇒ C) → (A ∧ B ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ B) ∧ (A ⇒ C), where M is a model
of F. Obviously, M, n |= (A ⇒ B) and M, n |= (A ⇒ C). Then, either:
10
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A, or
(ii) (∃n3 > n)(∀n4 > n3 )(M, n4 |= (A ∧ B) and M, n4 |= (A ∧ C)), which means that
(∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ∧ B ∧ C)
Case (i): We also have that (∀n2 > n1 ) M, n2 |= ¬(A ∧ B), and thus M, n |= (A ∧ B ⇒ C).
Case (ii): By definition, M, n |= (A ∧ B ⇒ C) follows.
So, if M, n |= (A ⇒ B) ∧ (A ⇒ C), then M, n |= (A ∧ B ⇒ C), which gives that
M, n |= (A ⇒ B) ∧ (A ⇒ C) → (A ∧ B ⇒ C)
Since the world n and model M were arbitrarily chosen, the proof is complete.
CC: We have to show that
F |= (A ⇒ B) ∧ (A ⇒ C) → (A ⇒ B ∧ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ B) ∧ (A ⇒ C), where M is a model
of F. Obviously, M, n |= (A ⇒ B) and M, n |= (A ⇒ C). Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A, or
(ii) (∃n3 > n)(∀n4 > n3 )(M, n4 |= (A ∧ B) and M, n4 |= (A ∧ C)), which means that
(∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ∧ B ∧ C)
Case (i): By definition, M, n |= (A ⇒ B ∧ C) follows.
Case (ii): Similarly, by definition, M, n |= (A ⇒ B ∧ C) follows.
So, if M, n |= (A ⇒ B) ∧ (A ⇒ C), then M, n |= (A ⇒ B ∧ C), which gives that
M, n |= (A ⇒ B) ∧ (A ⇒ C) → (A ⇒ B ∧ C)
Since the world n and model M were arbitrarily chosen, the proof is complete.
Loop: We have to show that
F |= (A0 ⇒ A1 ∧ . . . ∧ Ak ⇒ A0 ) → (A0 ⇒ Ak )
Assume an arbitrary state n ∈ ω, such that M, n |= (A0 ⇒ A1 ∧ . . . ∧ Ak ⇒ A0 ), where M is
a model of F. Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A0 , or
(ii) (∃n3 > n)(∀n4 > n3 )(M, n4 |= (A0 ∧ A1 ) and . . . and M, n4 |= (Ak ∧ A0 )), which means
that (∃n3 > n)(∀n4 > n3 )M, n4 |= (A0 ∧ A1 ∧ . . . ∧ Ak )
Case (i): By definition, M, n |= (A0 ⇒ Ak ) follows.
Case (ii): We also have that (∀n4 > n3 ) M, n4 |= (A0 ∧ Ak ), and thus M, n |= (A0 ⇒ Ak ).
11
So, if M, n |= (A0 ⇒ A1 ∧ . . . ∧ Ak ⇒ A0 ), then M, n |= (A0 ⇒ Ak ), which gives that
M, n |= (A0 ⇒ A1 ∧ . . . ∧ Ak ⇒ A0 ) → (A0 ⇒ Ak )
Since the world n and model M were arbitrarily chosen, the proof is complete.
OR: We have to show that
F |= (A ⇒ C) ∧ (B ⇒ C) → (A ∨ B ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ C) ∧ (B ⇒ C), where M is a model
of F. Obviously, M, n |= (A ⇒ C) and M, n |= (B ⇒ C). Then one of the following must
hold:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= (¬A ∧ ¬B)
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= (¬A ∧ B ∧ C)
(iii) (∃n5 > n)(∀n6 > n5 ) M, n6 |= (¬B ∧ A ∧ C)
(iv) (∃n7 > n)(∀n8 > n7 ) M, n8 |= (A ∧ B ∧ C)
Case (i): Equivalently, we have that (∀n2 > n1 ) M, n2 |= ¬(A ∨ B), and M, n |= (A ∨ B ⇒ C)
follows.
Cases (ii) - (iv): We also have that (∀n4,6,8 > n3,5,7 ) M, n4,6,8 |= (A ∨ B) ∧ C, and thus
M, n |= (A ∨ B ⇒ C).
So, if M, n |= (A ⇒ C) ∧ (B ⇒ C), then M, n |= (A ∨ B ⇒ C), which gives that
M, n |= (A ⇒ C) ∧ (B ⇒ C) → (A ∨ B ⇒ C)
Since the world n and model M were arbitrarily chosen, the proof is complete.
CSO: We have to show that
F |= (A ⇒ B) ∧ (B ⇒ A) → ((A ⇒ C) ≡ (B ⇒ C))
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ B) ∧ (B ⇒ A), where M is a model
of F. Obviously, M, n |= (A ⇒ B) and M, n |= (B ⇒ A). Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= (¬A ∧ ¬B), or
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ∧ B)
Case (i): By definition, M, n |= (A ⇒ C) and M, n |= (B ⇒ C), so M, n |= (A ⇒ C) ≡ (B ⇒
C) follows.
Case (ii): Let M, n |= (A ⇒ C). Then it follows that (∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ∧ C),
because it cannot be the case that (∃n3 > n)(∀n4 > n3 ) M, n4 |= ¬A. By (∃n3 > n)(∀n4 >
n3 ) M, n4 |= (A ∧ B) we obtain
(∃n3 > n)(∀n4 > n3 ) M, n4 |= (B ∧ C)
12
It follows that M, n |= (B ⇒ C), and thus M, n |= (A ⇒ C) → (B ⇒ C). Similarly,
M, n |= (B ⇒ C) → (A ⇒ C), and thus M, n |= (A ⇒ C) ≡ (B ⇒ C).
So, if M, n |= (A ⇒ B) ∧ (B ⇒ A), then M, n |= (A ⇒ C) ≡ (B ⇒ C), which gives that
M, n |= (A ⇒ B) ∧ (B ⇒ A) → ((A ⇒ C) ≡ (B ⇒ C))
Since the world n and model M were arbitrarily chosen, the proof is complete.
CM: We have to show that
F |= (A ⇒ B ∧ C) → (A ⇒ B) ∧ (A ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ B ∧ C), where M is a model of F.
Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A, or
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ∧ B ∧ C)
Case (i): By definition, M, n |= (A ⇒ B) and M, n |= (A ⇒ C), so M, n |= (A ⇒ B)∧(A ⇒ C)
follows.
Case (ii): We also have that (∀n4 > n3 ) M, n4 |= (A ∧ B) and (∀n4 > n3 ) M, n4 |= (A ∧ C),
so M, n |= (A ⇒ B) and M, n |= (A ⇒ C), and thus M, n |= (A ⇒ B) ∧ (A ⇒ C).
So, if M, n |= (A ⇒ B ∧ C), then M, n |= (A ⇒ B) ∧ (A ⇒ C), which gives that
M, n |= (A ⇒ B ∧ C) → (A ⇒ B) ∧ (A ⇒ C)
Since the world n and model M were arbitrarily chosen, the proof is complete.
CA: We have to show that
F |= (A ⇒ B) ∧ (C ⇒ B) → (A ∧ C ⇒ B)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ B) ∧ (C ⇒ B), where M is a model
of F. Obviously, M, n |= (A ⇒ B) and M, n |= (C ⇒ B). Then one of the following must
hold:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= (¬A ∧ ¬C)
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= (¬A ∧ C ∧ B)
(iii) (∃n5 > n)(∀n6 > n5 ) M, n6 |= (¬C ∧ A ∧ B)
(iv) (∃n7 > n)(∀n8 > n7 ) M, n8 |= (A ∧ C ∧ B)
Cases (i) - (iii): We also have that (∀n2,4,6 > n1,3,5 ) M, n2,4,6 |= ¬(A ∧ C), and thus M, n |=
(A ∧ C ⇒ B).
Case (iv): By definition, M, n |= (A ∧ C ⇒ B) follows.
13
So, if M, n |= (A ⇒ B) ∧ (C ⇒ B), then M, n |= (A ∧ C ⇒ B), which gives that
M, n |= (A ⇒ B) ∧ (C ⇒ B) → (A ∧ C ⇒ B)
Since the world n and model M were arbitrarily chosen, the proof is complete.
Transitivity: We have to show that
F |= (A ⇒ B) ∧ (B ⇒ C) → (A ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ B) ∧ (B ⇒ C), where M is a model
of F. Obviously, M, n |= (A ⇒ B) and M, n |= (B ⇒ C). Then one of the following must
hold:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= (¬A ∧ ¬B)
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= (¬A ∧ B ∧ C)
(iii) (∃n5 > n)(∀n6 > n5 ) M, n6 |= (A ∧ B ∧ C)
Cases (i) - (ii): We also have that (∀n2,4 > n1,3 ) M, n2,4 |= ¬A, and M, n |= (A ⇒ C) follows.
Case (iii): Similarly, we also have that (∀n6 > n5 ) M, n6 |= (A∧C), and thus M, n |= (A ⇒ C).
So, if M, n |= (A ⇒ B) ∧ (B ⇒ C), then M, n |= (A ⇒ C), which gives that
M, n |= (A ⇒ B) ∧ (B ⇒ C) → (A ⇒ C)
Since the world n and model M were arbitrarily chosen, the proof is complete.
Weak Transitivity: We have to show that
F |= (A ⇒ B) ∧ (B ⇒ C) ⇒ (A ⇒ C)
Assume an arbitrary state n ∈ ω and M a model of F. We have that M, n |= (A ⇒ B) ∧ (B ⇒
C) ⇒ (A ⇒ C) iff one of the following holds:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬(A ⇒ B) ∨ ¬(B ⇒ C), or
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ⇒ B) ∧ (B ⇒ C) ∧ (A ⇒ C)
Let (i) be false, that is let (∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ⇒ B) ∧ (B ⇒ C) (∗). We will
show that (ii) has to be true. By (∗) we have that (∀n1 > n)(∃n2 > n1 ) such that both the
following disjunctions hold:
• (∃n5 > n2 )(∀n6 > n5 ) M, n6 |= ¬A or (∃n5 > n2 )(∀n6 > n5 ) M, n6 |= (A ∧ B)
• (∃n5 > n2 )(∀n6 > n5 ) M, n6 |= ¬B or (∃n5 > n2 )(∀n6 > n5 ) M, n6 |= (B ∧ C)
This means that one of the following must hold:
(a) (∀n1 > n)(∃n5 > n1 )(∀n6 > n5 ) M, n6 |= (¬A ∧ ¬B)
14
(b) (∀n1 > n)(∃n5 > n1 )(∀n6 > n5 ) M, n6 |= (¬A ∧ B ∧ C)
(c) (∀n1 > n)(∃n5 > n1 )(∀n6 > n5 ) M, n6 |= (A ∧ B ∧ C)
All of these cases give us that
(∀n1 > n) M, n1 |= (A ⇒ B) ∧ (B ⇒ C) ∧ (A ⇒ C)
Consequently, we also have that
(∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ⇒ B) ∧ (B ⇒ C) ∧ (A ⇒ C)
which is exactly (ii). So one of (i) or (ii) must hold, which means that
M, n |= (A ⇒ B) ∧ (B ⇒ C) ⇒ (A ⇒ C)
Since the world n and model M were arbitrarily chosen, the proof is complete.
Weak Modus Ponens: We have to show that
F |= A ∧ (A ⇒ B) ⇒ B
Assume an arbitrary state n ∈ ω and M a model of F. We have that M, n |= A∧(A ⇒ B) ⇒ B
iff one of the following holds:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A ∨ ¬(A ⇒ B), or
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= A ∧ (A ⇒ B) ∧ B
Let (i) be false, that is let (∀n1 > n)(∃n2 > n1 ) M, n2 |= A ∧ (A ⇒ B) (∗). We will show that
(ii) has to be true. By (∗) we have that (∀n1 > n)(∃n2 > n1 ) such that M, n2 |= A and
(∃n5 > n2 )(∀n6 > n5 ) M, n6 |= ¬A or (∃n5 > n2 )(∀n6 > n5 ) M, n6 |= (A ∧ B)
This means that
(∀n1 > n)(∃n5 > n1 )(∀n6 > n5 ) M, n6 |= (A ∧ B)
This gives us that
(∀n1 > n) M, n1 |= (A ⇒ B)
From the last two, we also obtain
(∃n3 > n)(∀n4 > n3 ) M, n4 |= A ∧ (A ⇒ B) ∧ B
which is exactly (ii). So one of (i) or (ii) must hold, which means that
M, n |= A ∧ (A ⇒ B) ⇒ B
Since the world n and model M were arbitrarily chosen, the proof is complete.
⇒
Having the intention to describe Ω in full detail, we proceed now to identify the rules and
axioms not present in this majority-default conditional logic.
15
⇒
Theorem 3.5 The logic Ω:
1. is not closed under the rule RCE
2. does not contain the axioms ID, CV, MP, MOD, CS, CEM, SDA, Monotonicity,
Weak Monotonicity and Modus Ponens
Proof. RCE: We have to show that
M |= (A → B) and M 6|= (A ⇒ B)
for some model M of F.
Let M = (ω, <, V ) be a model of F such that V (A) = {n | n is even} and V (B) = ω. Then
(∀n ∈ ω) M, n |= (A → B), because (∀n ∈ ω) M, n |= B. Thus M |= (A → B).
But for an arbitrary state n ∈ ω we have that
(∀n1 > n)(∃n2 > n1 ) M, n2 |= A and (∀n3 > n)(∃n4 > n3 ) M, n4 |= ¬A
It follows then that M, n 6|= (A ⇒ B) and consequently M 6|= (A ⇒ B).
ID: We have to show that
F 6|= (A ⇒ A)
Let M = (ω, <, V ) be a model of F such that V (A) = {n | n is even}. Then for an arbitrary
state n ∈ ω we have that
(∀n1 > n)(∃n2 > n1 ) M, n2 |= A and (∀n3 > n)(∃n4 > n3 ) M, n4 |= ¬A
It follows then that M, n 6|= (A ⇒ A) and the proof is complete.
CV: We have to show that
F 6|= (A ⇒ B) ∧ ¬(A ⇒ ¬C) → (A ∧ C ⇒ B)
Let M = (ω, <, V ) be a model of F such that V (A) = V (B) = ω and V (C) = {n | n is even}.
Then (∀n ∈ ω) M, n |= (A ∧ B), so for an arbitrary state n ∈ ω we have that
(∃n1 > n)(∀n2 > n1 ) M, n2 |= (A ∧ B)
and thus M, n |= (A ⇒ B).
Additionally, for an arbitrary world n ∈ ω we have that
(∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ C)
which gives M, n 6|= (A ⇒ ¬C), or equivalently M, n |= ¬(A ⇒ ¬C).
16
Consequently we have
M, n |= (A ⇒ B) ∧ ¬(A ⇒ ¬C)
But for n ∈ ω, M, n 6|= (A ∧ C ⇒ B), because we have that
(∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ C) and (∀n5 > n)(∃n6 > n5 ) M, n6 |= ¬C
It follows then that M, n 6|= (A ⇒ B) ∧ ¬(A ⇒ ¬C) → (A ∧ C ⇒ B) and the proof is complete.
MP: We have to show that
F 6|= (A ⇒ B) → (A → B)
Let M = (ω, <, V ) be a model of F such that V (A) = ω and V (B) = ω − {n}. Then we have
that
(∃n1 > n)(∀n2 > n1 ) M, n2 |= (A ∧ B)
so by definition M, n |= (A ⇒ B).
But M, n 6|= (A → B), because M, n |= (A ∧ ¬B).
It follows then that M, n 6|= (A ⇒ B) → (A → B) and the proof is complete.
MOD: We have to show that
F 6|= (¬A ⇒ A) → (B ⇒ A)
Let M = (ω, <, V ) be a model of F such that V (A) = {n ∈ ω | n > n1 } and V (B) = {n | n is
even}. Then we have that
(∃n2 > n1 )(∀n3 > n2 ) M, n3 |= A
so by definition M, n1 |= (¬A ⇒ A).
But M, n1 6|= (B ⇒ A), because
(∀n4 > n1 )(∃n5 > n4 )M, n5 |= B and (∀n6 > n1 )(∃n7 > n6 )M, n7 |= ¬B
It follows then that M, n1 6|= (¬A ⇒ A) → (B ⇒ A) and the proof is complete.
CS: We have to show that
F 6|= (A ∧ B) → (A ⇒ B)
Let M = (ω, <, V ) be a model of F such that V (A) = ω and V (B) = {n | n is even}. Then for
an arbitrary even world n we have that M, n |= (A ∧ B). But M, n 6|= (A ⇒ B), because
(∀n1 > n)(∃n2 > n1 ) M, n2 |= A and (∀n3 > n)(∃n4 > n3 ) M, n4 |= ¬B
It follows then that M, n 6|= (A ∧ B) → (A ⇒ B) and the proof is complete.
CEM: We have to show that
F 6|= (A ⇒ B) ∨ (A ⇒ ¬B)
Let M = (ω, <, V ) be a model of F such that V (A) = ω and V (B) = {n | n is even}. Then for
an arbitrary state n ∈ ω we have that M, n 6|= (A ⇒ B) and M, n 6|= (A ⇒ ¬B), because
17
(∀n1 > n)(∃n2 > n1 ) M, n2 |= A
(∀n3 > n)(∃n4 > n3 ) M, n4 |= B
(∀n5 > n)(∃n6 > n5 ) M, n6 |= ¬B
It follows then that M, n 6|= (A ⇒ B) ∨ (A ⇒ ¬B) and the proof is complete.
SDA: We have to show that
F 6|= (A ∨ B ⇒ C) → (A ⇒ C) ∧ (B ⇒ C)
Let M = (ω, <, V ) be a model of F such that V (A) = {n | n is even} and V (B) = V (C) = ω.
Then for an arbitrary state n ∈ ω we have that M, n |= (A ∨ B ⇒ C), because
(∃n1 > n)(∀n2 > n1 ) M, n2 |= (B ∧ C)
and consequently
(∃n1 > n)(∀n2 > n1 ) M, n2 |= (A ∨ B) ∧ C
But M, n 6|= (A ⇒ C), because
(∀n3 > n)(∃n4 > n3 ) M, n4 |= A and (∀n5 > n)(∃n6 > n5 ) M, n6 |= ¬A
and thus M, n 6|= (A ⇒ C) ∧ (B ⇒ C).
It follows then that M, n 6|= (A ∨ B ⇒ C) → (A ⇒ C) ∧ (B ⇒ C) and the proof is complete.
Monotonicity: We have to show that
F 6|= (A ⇒ B) → (A ∧ C ⇒ B)
Let M = (ω, <, V ) be a model of F such that V (A) = V (B) = ω and V (C) = {n | n is even}.
Then (∀n ∈ ω) M, n |= (A ∧ B), so for an arbitrary state n ∈ ω we have that
(∃n1 > n)(∀n2 > n1 ) M, n2 |= (A ∧ B)
and thus M, n |= (A ⇒ B).
But M, n 6|= (A ∧ C ⇒ B), because
(∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ C) and (∀n5 > n)(∃n6 > n5 ) M, n6 |= ¬C
It follows then that M, n 6|= (A ⇒ B) → (A ∧ C ⇒ B) and the proof is complete.
Weak Monotonicity: We have to show that
F 6|= (A ⇒ B) ⇒ (A ∧ C ⇒ B)
Let M = (ω, <, V ) be a model of F such that V (A) = V (B) = ω and V (C) = {n | n is even}.
For an arbitrary state n ∈ ω we have that M, n |= (A ⇒ B) ⇒ (A ∧ C ⇒ B) iff one of the
following holds:
18
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬(A ⇒ B)
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ⇒ B) ∧ (A ∧ C ⇒ B)
Case (i): By construction we have that (∀n ∈ ω) M, n |= (A ∧ B), so by definition
(∀n ∈ ω) M, n |= (A ⇒ B)
This gives us that (∀n ∈ ω) M, n 6|= ¬(A ⇒ B) and the case (i) cannot hold.
Case (ii): By construction we have that
(∀n5 > n)(∃n6 > n5 ) M, n6 |= (A ∧ C) and (∀n7 > n)(∃n8 > n7 ) M, n8 |= ¬C
so by definition (∀n ∈ ω) M, n 6|= (A ∧ C ⇒ B) and the case (ii) cannot hold.
It follows then that M, n 6|= (A ⇒ B) ⇒ (A ∧ C ⇒ B) and the proof is complete.
Modus Ponens: We have to show that
F 6|= A ∧ (A ⇒ B) → B
Let M = (ω, <, V ) be a model of F such that V (A) = {n} and V (B) = ∅. Then M, n |= A
and by definition M, n |= (A ⇒ B), which gives M, n |= A ∧ (A ⇒ B).
But M, n 6|= B by construction.
It follows then that M, n 6|= A ∧ (A ⇒ B) → B and the proof is complete.
⇒
Observe that Ω does not contain the ID axiom. This might appear strange; after all ‘reflexivity seems to be satisfied universally by any kind of reasoning based on some notion of
consequence’ [KLM90, p. 177] and defeasible conditionals are designed to incarnate some form
of defeasible consequence. Moreover, as observed also in [KLM90], conditionals that do not
satisfy it ‘probably express some notion of theory change’. It seems that failure of ID is due
to the unavoidable ‘temporal ’ flavour of (ω, <), whose ordering directly reminds the setting
of discrete linear time. This might seem appropriate for conditionals incorporating a notion
of ‘temporal’ causation, in the form ‘if X, then normally it should be the case that Y holds
in the future’ - “normally, a strong earthquake implies a permanent change in future building
codes”. Nevertheless, failure of ID is not a happy incident and it seems natural to consider
alternative modal definitions of the conditional connective that would enforce the validity of
ID. Below, we demonstrate that some plausible attempts to validate ID, unfortunately result
into a monotonic conditional logic.
As a first attempt, it seems natural to constrain the set of valuations, using the known
modal recipe of general frames. For the rest of this subsection, to facilitate the exposition,
we will use the symbol (A V B) for the monotonic conditional(s) we define below.
Definition 3.6 Let F = (ω, <, X) be the general frame obtained by setting the set X of
admissible valuations to be the set of finite and cofinite subsets of ω. We assume again that
A V B as previously to be a shorthand for 32¬A ∨ 32(A ∧ B). That is
AVB
iff
32¬A ∨ 32(A ∧ B)
19
Note that X is well-defined, as it is closed under the boolean algebra operators and the modal
operator (see [BdRV01, p. 30]).
Proposition 3.7 The conditional defined in Definition 3.6 satisfies Monotonicity, i.e.
F |= (A V B) → (A ∧ C V B)
Proof. Assume an arbitrary state n ∈ ω, such that M, n |= (A V B), where M is a model
of F. Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A, or
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ∧ B)
Case (i): We also have that (∀n2 > n1 ) M, n2 |= ¬(A ∧ C), and thus M, n |= (A ∧ C V B).
Case (ii): For a proposition C, since kCk ∈ X, one of the following must hold:
(a) (∃n5 > n)(∀n6 > n5 ) M, n6 |= ¬C
(b) (∃n7 > n)(∀n8 > n7 ) M, n8 |= C
In (a), we also have that (∀n6 > n5 ) M, n6 |= ¬(A ∧ C), and thus M, n |= (A ∧ C V B).
In (b), by (ii) we obtain (∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ∧ B ∧ C), and M, n |= (A ∧ C V B)
follows.
So, if M, n |= (A V B), then M, n |= (A ∧ C V B), which gives that
M, n |= (A V B) → (A ∧ C V B)
Since the world n and model M were arbitrarily chosen, the proof is complete.
This phenomenon persists, even if we go back to (ω, <) and attempt to provide alternative
modal definitions of the conditional. Not all of them conform to the ‘true in all, but finitely
many, worlds’ intuition, but they demonstrate the possibility of different variants.
Theorem 3.8 Assume the following definitions:
(i) A V B ≡def 23¬A ∨ 23(A ∧ B)
(ii) A V B ≡def 23¬A ∨ 32(A ∧ B)
(iii) A V B ≡def 32(A → B)
(iv) A V B ≡def 23(A → B)
All V conditionals defined above satisfy Monotonicity.
20
Proof. In each case, we have to show that
F |= (A V B) → (A ∧ C V B)
(i): Assume an arbitrary state n ∈ ω, such that M, n |= (A V B), where M is a model of F.
Then, either:
(i) (∀n1 > n)(∃n2 > n1 ) M, n2 |= ¬A, or
(ii) (∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ B)
Case (i): We also have that (∃n2 > n1 ) M, n2 |= ¬(A ∧ C), and thus M, n |= (A ∧ C V B).
Case (ii): For a proposition C one of the following must hold:
(a) (∀n5 > n)(∃n6 > n5 ) M, n6 |= ¬C
(b) (∃n7 > n)(∀n8 > n7 ) M, n8 |= C
In (a), we also have that (∃n6 > n5 ) M, n6 |= ¬(A ∧ C), and thus M, n |= (A ∧ C V B).
In (b), by (ii) we obtain (∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ B ∧ C), and M, n |= (A ∧ C V B)
follows.
So, if M, n |= (A V B), then M, n |= (A ∧ C V B), which gives that
M, n |= (A V B) → (A ∧ C V B)
Since the world n and model M were arbitrarily chosen, the proof is complete.
(ii): Assume an arbitrary state n ∈ ω, such that M, n |= (A V B), where M is a model of F.
Then, either:
(i) (∀n1 > n)(∃n2 > n1 ) M, n2 |= ¬A, or
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ∧ B)
Case (i): We also have that (∃n2 > n1 ) M, n2 |= ¬(A ∧ C), and thus M, n |= (A ∧ C V B).
Case (ii): For a proposition C one of the following must hold:
(a) (∀n5 > n)(∃n6 > n5 ) M, n6 |= ¬C
(b) (∃n7 > n)(∀n8 > n7 ) M, n8 |= C
In (a), we also have that (∃n6 > n5 ) M, n6 |= ¬(A ∧ C), and thus M, n |= (A ∧ C V B).
In (b), by (ii) we obtain (∃n3 > n)(∀n4 > n3 ) M, n4 |= (A ∧ B ∧ C), and M, n |= (A ∧ C V B)
follows.
So, if M, n |= (A V B), then M, n |= (A ∧ C V B), which gives that
M, n |= (A V B) → (A ∧ C V B)
Since the world n and model M were arbitrarily chosen, the proof is complete.
21
(iii): Assume an arbitrary state n ∈ ω, such that M, n |= (A V B), where M is a model of F.
By definition, this means that
(∃n1 > n)(∀n2 > n1 ) M, n2 |= (A → B)
By propositional logic, we have that (∀n ∈ ω) M, n |= (A → B) → (A ∧ C → B).
From these, we obtain
(∃n1 > n)(∀n2 > n1 ) M, n2 |= (A ∧ C → B)
and M, n |= (A ∧ C V B) follows.
So, if M, n |= (A V B), then M, n |= (A ∧ C V B), which gives that
M, n |= (A V B) → (A ∧ C V B)
Since the world n and model M were arbitrarily chosen, the proof is complete.
(iv): Assume an arbitrary state n ∈ ω, such that M, n |= (A V B), where M is a model of F.
By definition, this means that
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A → B)
By propositional logic, we have that (∀n ∈ ω) M, n |= (A → B) → (A ∧ C → B).
From these, we obtain
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ C → B)
and M, n |= (A ∧ C V B) follows.
So, if M, n |= (A V B), then M, n |= (A ∧ C V B), which gives that
M, n |= (A V B) → (A ∧ C V B)
Since the world n and model M were arbitrarily chosen, the proof is complete.
3.1.1
An alternative: cofinal vs cofinite in (ω, <)
In this subsection, we discuss a possible alternative. Instead of working with the (obviously
large) cofinite subsets of ω, we will attempt to work with cofinal subsets: S ⊆ ω is cofinal in
ω iff for every n ∈ ω there exists an s ∈ S, such that n < s.
We proceed to define the conditional (A ⇒ B) as follows:
(A ⇒ B)
≡def
32¬A ∨ 23(A ∧ B)
⇒
⇒
Definition 3.9 [Conditional Logic ω]. The logic ω consists of all formulae A ∈ L⇒ , such
that:
⇒
A ∈ ω iff (ω, <) |= A∗ iff `K4DLZ A∗
where A∗ is the obvious translation defined similarly to Definition 3.1.
22
Fact 3.10 Let M be a model of F = (ω, <) and n ∈ ω an arbitrary world. We will say that
M, n |= (A ⇒ B) iff one of the following holds:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A
(ii) (∀n3 > n)(∃n4 > n3 ) M, n4 |= A ∧ B
⇒
The logic ω turns out to be quite interesting.
⇒
Theorem 3.11 The logic ω:
1. is closed under the rules RCEA, RCEC and RCE
2. contains the axioms ID, CUT, Loop, OR, CV, CM, MOD, CEM and
Weak Modus Ponens
Proof. RCEA: Let M |= (A ≡ B). We have to show that
M |= (A ⇒ C) ≡ (B ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ C). Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A, or
(ii) (∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ C)
Case (i): By M |= (A ≡ B) we also have that (∀n2 > n1 ) M, n2 |= (¬A ≡ ¬B). This gives
that (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬B, and M, n |= (B ⇒ C) follows.
Case (ii): Similarly, by M |= (A ≡ B) we obtain
(∃n4 > n3 ) M, n4 |= (A ∧ C ≡ B ∧ C)
It follows that (∀n3 > n)(∃n4 > n3 ) M, n4 |= (B ∧ C), and thus M, n |= (B ⇒ C).
So, if M, n |= (A ⇒ C), then M, n |= (B ⇒ C), which gives that
M, n |= (A ⇒ C) → (B ⇒ C)
Similarly, M, n |= (B ⇒ C) → (A ⇒ C). Since the world n was arbitrarily chosen, the proof
is complete.
RCEC: Let M |= (A ≡ B). We have to show that
M |= (C ⇒ A) ≡ (C ⇒ B)
Assume an arbitrary state n ∈ ω, such that M, n |= (C ⇒ A). Then, either:
23
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬C, or
(ii) (∀n3 > n)(∃n4 > n3 ) M, n4 |= (C ∧ A)
Case (i): By definition, M, n |= (C ⇒ B) follows.
Case (ii): By M |= (A ≡ B) we obtain
(∃n4 > n3 ) M, n4 |= (C ∧ A ≡ C ∧ B)
It follows that (∀n3 > n)(∃n4 > n3 ) M, n4 |= (C ∧ B), and thus M, n |= (C ⇒ B).
So, if M, n |= (C ⇒ A), then M, n |= (C ⇒ B), which gives that
M, n |= (C ⇒ A) → (C ⇒ B)
Similarly, M, n |= (C ⇒ B) → (C ⇒ A). Since the world n was arbitrarily chosen, the proof
is complete.
RCE: Let M |= (A → B). We have to show that
M |= (A ⇒ B)
Assume an arbitrary state n ∈ ω. Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A, or
(ii) (∀n3 > n)(∃n4 > n3 ) M, n4 |= A
Case (i): By definition, M, n |= (A ⇒ B) follows.
Case (ii): By M |= (A → B) we obtain
(∀n3 > n)(∃n4 > n3 )(M, n4 |= A and M, n4 |= A → B)
It follows that (∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ B), and thus M, n |= (A ⇒ B).
So in either case M, n |= (A ⇒ B). Since the world n was arbitrarily chosen, the proof is
complete.
ID: We have to show that
F |= (A ⇒ A)
Assume an arbitrary state n ∈ ω and M a model of F. Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A, or
(ii) (∀n3 > n)(∃n4 > n3 ) M, n4 |= A
In both cases, by definition M, n |= (A ⇒ A). Since the world n and model M were arbitrarily
chosen, the proof is complete.
24
CUT: We have to show that
F |= (A ∧ B ⇒ C) ∧ (A ⇒ B) → (A ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ∧ B ⇒ C) ∧ (A ⇒ B), where M is a
model of F. Obviously, M, n |= (A ∧ B ⇒ C) and M, n |= (A ⇒ B). Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A, or
(ii) (∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ B ∧ C)
Case (i): By definition, M, n |= (A ⇒ C) follows.
Case (ii): We also have that (∃n4 > n3 ) M, n4 |= (A ∧ C), and thus M, n |= (A ⇒ C).
So, if M, n |= (A ∧ B ⇒ C) ∧ (A ⇒ B), then M, n |= (A ⇒ C), which gives that
M, n |= (A ∧ B ⇒ C) ∧ (A ⇒ B) → (A ⇒ C)
Since the world n and model M were arbitrarily chosen, the proof is complete.
Loop: We have to show that
F |= (A0 ⇒ A1 ∧ . . . ∧ Ak ⇒ A0 ) → (A0 ⇒ Ak )
Assume an arbitrary state n ∈ ω, such that M, n |= (A0 ⇒ A1 ∧ . . . ∧ Ak ⇒ A0 ), where M is
a model of F. Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A0 , or
(ii) (∀n3 > n)(∃n4 > n3 ) M, n4 |= (A0 ∧ A1 ) and . . . and (∀n3 > n)(∃n4 > n3 ) M, n4 |=
(Ak ∧ A0 )
Case (i): By definition, M, n |= (A0 ⇒ Ak ) follows.
Case (ii): We need only that (∀n3 > n)(∃n4 > n3 ) M, n4 |= (A0 ∧ Ak ), and thus M, n |=
(A0 ⇒ Ak ).
So, if M, n |= (A0 ⇒ A1 ∧ . . . ∧ Ak ⇒ A0 ), then M, n |= (A0 ⇒ Ak ), which gives that
M, n |= (A0 ⇒ A1 ∧ . . . ∧ Ak ⇒ A0 ) → (A0 ⇒ Ak )
Since the world n and model M were arbitrarily chosen, the proof is complete.
OR: We have to show that
F |= (A ⇒ C) ∧ (B ⇒ C) → (A ∨ B ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ C) ∧ (B ⇒ C), where M is a model
of F. Obviously, M, n |= (A ⇒ C) and M, n |= (B ⇒ C). Then one of the following must
hold:
25
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= (¬A ∧ ¬B)
(ii) (∃n3 > n)(∀n4 > n3 ) M, n4 |= ¬A and (∀n3 > n)(∃n4 > n3 ) M, n4 |= (B ∧ C)
(iii) (∃n5 > n)(∀n6 > n5 ) M, n6 |= ¬B and (∀n5 > n)(∃n6 > n5 ) M, n6 |= (A ∧ C)
(iv) (∀n7 > n)(∃n8 > n7 ) M, n8 |= (A ∧ C) and (∀n7 > n)(∃n8 > n7 ) M, n8 |= (B ∧ C)
Case (i): Equivalently, we have that (∀n2 > n1 ) M, n2 |= ¬(A ∨ B), and M, n |= (A ∨ B ⇒ C)
follows.
Cases (ii) - (iv): We also have that (∃n4,6,8 > n3,5,7 ) M, n4,6,8 |= (A ∨ B) ∧ C, and thus
M, n |= (A ∨ B ⇒ C).
So, if M, n |= (A ⇒ C) ∧ (B ⇒ C), then M, n |= (A ∨ B ⇒ C), which gives that
M, n |= (A ⇒ C) ∧ (B ⇒ C) → (A ∨ B ⇒ C)
Since the world n and model M were arbitrarily chosen, the proof is complete.
CV: We have to show that
F |= (A ⇒ B) ∧ ¬(A ⇒ ¬C) → (A ∧ C ⇒ B)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ B) ∧ ¬(A ⇒ ¬C), where M is a
model of F. Obviously, M, n |= (A ⇒ B) and M, n 6|= (A ⇒ ¬C). The only way this can be
achieved is if
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ B) and (∃n3 > n)(∀n4 > n3 ) M, n4 |= (¬A ∨ C)
But then we also have
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ B ∧ C)
and, by definition, M, n |= (A ∧ C ⇒ B) follows.
So, if M, n |= (A ⇒ B) ∧ ¬(A ⇒ ¬C), then M, n |= (A ∧ C ⇒ B), which gives that
M, n |= (A ⇒ B) ∧ ¬(A ⇒ ¬C) → (A ∧ C ⇒ B)
Since the world n and model M were arbitrarily chosen, the proof is complete.
CM: We have to show that
F |= (A ⇒ B ∧ C) → (A ⇒ B) ∧ (A ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ B ∧ C), where M is a model of F.
Then, either:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A, or
(ii) (∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ B ∧ C)
26
Case (i): By definition, M, n |= (A ⇒ B) and M, n |= (A ⇒ C), so M, n |= (A ⇒ B)∧(A ⇒ C)
follows.
Case (ii): We also have that (∃n4 > n3 ) M, n4 |= (A ∧ B) and (∃n4 > n3 ) M, n4 |= (A ∧ C),
so M, n |= (A ⇒ B) and M, n |= (A ⇒ C), and thus M, n |= (A ⇒ B) ∧ (A ⇒ C).
So, if M, n |= (A ⇒ B ∧ C), then M, n |= (A ⇒ B) ∧ (A ⇒ C), which gives that
M, n |= (A ⇒ B ∧ C) → (A ⇒ B) ∧ (A ⇒ C)
Since the world n and model M were arbitrarily chosen, the proof is complete.
MOD: We have to show that
F |= (¬A ⇒ A) → (B ⇒ A)
Assume an arbitrary state n ∈ ω, such that M, n |= (¬A ⇒ A), where M is a model of F. The
only way this can be achieved is if
(∃n1 > n)(∀n2 > n1 ) M, n2 |= A
For the proposition B one of the following must hold:
(i) (∃n3 > n)(∀n4 > n3 ) M, n4 |= ¬B, or
(ii) (∀n5 > n)(∃n6 > n5 ) M, n6 |= B
Case (i): By definition, M, n |= (B ⇒ A) follows.
Case (ii): By (∃n1 > n)(∀n2 > n1 ) M, n2 |= A we also have that (∀n5 > n)(∃n6 > n5 ) M, n6 |=
(B ∧ A), and thus M, n |= (B ⇒ A).
So, if M, n |= (¬A ⇒ A), then M, n |= (B ⇒ A), which gives that
M, n |= (¬A ⇒ A) → (B ⇒ A)
Since the world n and model M were arbitrarily chosen, the proof is complete.
CEM: We have to show that
F |= (A ⇒ B) ∨ (A ⇒ ¬B)
Assume an arbitrary state n ∈ ω, such that M, n 6|= (A ⇒ B), where M is a model of F. The
only way this can be achieved is if
(∀n1 > n)(∃n2 > n1 ) M, n2 |= A and (∃n3 > n)(∀n4 > n3 ) M, n4 |= (¬A ∨ ¬B)
But then we also have
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ ¬B)
and, by definition, M, n |= (A ⇒ ¬B) follows.
27
So, if M, n |= ¬(A ⇒ B), then M, n |= (A ⇒ ¬B), which gives that
M, n |= ¬(A ⇒ B) → (A ⇒ ¬B)
Since the world n and model M were arbitrarily chosen, the proof is complete.
Weak Modus Ponens: We have to show that
F |= A ∧ (A ⇒ B) ⇒ B
Assume an arbitrary state n ∈ ω and M a model of F. We have that M, n |= A∧(A ⇒ B) ⇒ B
iff one of the following holds:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬A ∨ ¬(A ⇒ B), or
(ii) (∀n3 > n)(∃n4 > n3 ) M, n4 |= A ∧ (A ⇒ B) ∧ B
Let (i) be false, that is let (∀n1 > n)(∃n2 > n1 ) M, n2 |= A ∧ (A ⇒ B) (∗). We will show that
(ii) has to be true. By (∗) we have that (∀n1 > n)(∃n2 > n1 ) such that M, n2 |= A and
(∃n5 > n2 )(∀n6 > n5 ) M, n6 |= ¬A or (∀n5 > n2 )(∃n6 > n5 ) M, n6 |= (A ∧ B)
This means that
(∀n5 > n)(∃n6 > n5 ) M, n6 |= (A ∧ B)
This gives us that
(∀n1 > n) M, n1 |= (A ⇒ B)
From the last two, we also obtain
(∀n3 > n)(∃n4 > n3 ) M, n4 |= A ∧ (A ⇒ B) ∧ B
which is exactly (ii). So one of (i) or (ii) must hold, which means that
M, n |= A ∧ (A ⇒ B) ⇒ B
Since the world n and model M were arbitrarily chosen, the proof is complete.
⇒
Theorem 3.12 The logic ω:
1. is not closed under the rule RCK
2. does not contain the axioms AC, CC, CSO, MP, CA, CS, SDA, Transitivity,
Weak Transitivity, Monotonicity, Weak Monotonicity and Modus Ponens
28
Proof. RCK: For n = 2: We have to show that
M |= (A1 ∧ A2 → B) and M 6|= (C ⇒ A1 ) ∧ (C ⇒ A2 ) → (C ⇒ B)
for some model M of F.
Let M = (ω, <, V ) be a model of F such that V (A1 ) = {n | n is even}, V (A2 ) = {n | n
is odd}, V (B) = ∅ and V (C) = ω. Then (∀n ∈ ω) M, n |= (A1 ∧ A2 → B), because
(∀n ∈ ω) M, n |= (¬A1 ∨ ¬A2 ). Thus M |= (A1 ∧ A2 → B).
For an arbitrary state n ∈ ω we have that
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (C ∧ A1 ) and (∀n3 > n)(∃n4 > n3 ) M, n4 |= (C ∧ A2 )
This gives M, n |= (C ⇒ A1 ) and M, n |= (C ⇒ A2 ), which means that
M, n |= (C ⇒ A1 ) ∧ (C ⇒ A2 )
But M, n 6|= (C ⇒ B), because we have that
(∀n5 > n)(∃n6 > n5 ) M, n6 |= C and (∃n7 > n)(∀n8 > n7 ) M, n8 |= (¬C ∨ ¬B)
It follows then that M, n 6|= (C ⇒ A1 ) ∧ (C ⇒ A2 ) → (C ⇒ B) and consequently M 6|= (C ⇒
A1 ) ∧ (C ⇒ A2 ) → (C ⇒ B).
AC: We have to show that
F 6|= (A ⇒ B) ∧ (A ⇒ C) → (A ∧ B ⇒ C)
Let M = (ω, <, V ) be a model of F such that V (A) = ω, V (B) = {n | n is even} and V (C) =
{n | n is odd}. Then for an arbitrary state n ∈ ω we have that
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ B) and (∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ C)
This gives
M, n |= (A ⇒ B) and M, n |= (A ⇒ C)
and consequently
M, n |= (A ⇒ B) ∧ (A ⇒ C)
But M, n 6|= (A ∧ B ⇒ C), because
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ B) and (∃n5 > n)(∀n6 > n5 ) M, n6 |= ¬(A ∧ B ∧ C)
It follows then that M, n 6|= (A ⇒ B) ∧ (A ⇒ C) → (A ∧ B ⇒ C) and the proof is complete.
CC: We have to show that
F 6|= (A ⇒ B) ∧ (A ⇒ C) → (A ⇒ B ∧ C)
Let M = (ω, <, V ) be a model of F such that V (A) = ω, V (B) = {n | n is even} and V (C) =
{n | n is odd}. Then for an arbitrary state n ∈ ω we have that
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ B) and (∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ C)
29
This gives
M, n |= (A ⇒ B) and M, n |= (A ⇒ C)
and consequently
M, n |= (A ⇒ B) ∧ (A ⇒ C)
But M, n 6|= (A ⇒ B ∧ C), because
(∀n5 > n)(∃n6 > n5 ) M, n6 |= A and (∃n7 > n)(∀n8 > n7 ) M, n8 |= ¬(A ∧ B ∧ C)
It follows then that M, n 6|= (A ⇒ B) ∧ (A ⇒ C) → (A ⇒ B ∧ C) and the proof is complete.
CSO: We have to show that
F 6|= (A ⇒ B) ∧ (B ⇒ A) → ((A ⇒ C) ≡ (B ⇒ C))
Let M = (ω, <, V ) be a model of F such that V (A) = ω, V (B) = {n | n is even} and V (C) =
{n | n is odd}. Then for an arbitrary state n ∈ ω we have that
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ B) and (∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ C)
This gives
M, n |= (A ⇒ B), M, n |= (B ⇒ A) and M, n |= (A ⇒ C)
and consequently
M, n |= (A ⇒ B) ∧ (B ⇒ A) and M, n |= A ⇒ C
But M, n 6|= (B ⇒ C), because
(∀n5 > n)(∃n6 > n5 ) M, n6 |= B and (∃n7 > n)(∀n8 > n7 )M, n8 |= (¬B ∨ ¬C)
It follows then that M, n 6|= (A ⇒ B) ∧ (B ⇒ A) → ((A ⇒ C) → (B ⇒ C)) and the proof is
complete.
MP: We have to show that
F 6|= (A ⇒ B) → (A → B)
Let M = (ω, <, V ) be a model of F such that V (A) = ω and V (B) = ω − {n}. Then we have
that
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ B)
so by definition M, n |= (A ⇒ B).
But M, n 6|= (A → B), because M, n |= (A ∧ ¬B).
It follows then that M, n 6|= (A ⇒ B) → (A → B) and the proof is complete.
CA: We have to show that
F 6|= (A ⇒ B) ∧ (C ⇒ B) → (A ∧ C ⇒ B)
30
Let M = (ω, <, V ) be a model of F such that V (A) = ω − {n = 3k | k ∈ ω}, V (B) = ω − {n =
3k − 1 | k ∈ ω} and V (C) = ω − {n = 3k − 2 | k ∈ ω}. Then for an arbitrary state n ∈ ω we
have that
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ B) and (∀n3 > n)(∃n4 > n3 ) M, n4 |= (C ∧ B)
This gives
M, n |= (A ⇒ B) and M, n |= (C ⇒ B)
and consequently
M, n |= (A ⇒ B) ∧ (C ⇒ B)
But M, n 6|= (A ∧ C ⇒ B), because
(∀n5 > n)(∃n6 > n5 ) M, n6 |= (A ∧ C) and (∃n7 > n)(∀n8 > n7 ) M, n8 |= ¬(A ∧ B ∧ C)
It follows then that M, n 6|= (A ⇒ B) ∧ (C ⇒ B) → (A ∧ C ⇒ B) and the proof is complete.
CS: We have to show that
F 6|= (A ∧ B) → (A ⇒ B)
Let M = (ω, <, V ) be a model of F such that V (A) = {n | n is even} and V (B) = {n}, where
n is an arbitrary even world. Then we have that M, n |= (A ∧ B). But M, n 6|= (A ⇒ B),
because
(∀n1 > n)(∃n2 > n1 ) M, n2 |= A and (∀n3 > n)(∃n4 > n3 ) M, n4 |= (¬A ∨ ¬B)
It follows then that M, n 6|= (A ∧ B) → (A ⇒ B) and the proof is complete.
SDA: We have to show that
F 6|= (A ∨ B ⇒ C) → (A ⇒ C) ∧ (B ⇒ C)
Let M = (ω, <, V ) be a model of F such that V (A) = {n | n is even}, V (B) = ω and V (C) =
{n | n is odd}. Then for an arbitrary state n ∈ ω we have that M, n |= (A ∨ B ⇒ C), because
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (B ∧ C)
and consequently
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∨ B) ∧ C
But M, n 6|= (A ⇒ C), because
(∀n3 > n)(∃n4 > n3 ) M, n4 |= A and (∃n5 > n)(∀n6 > n5 ) M, n6 |= ¬A ∨ ¬C
and thus M, n 6|= (A ⇒ C) ∧ (B ⇒ C).
It follows then that M, n 6|= (A ∨ B ⇒ C) → (A ⇒ C) ∧ (B ⇒ C) and the proof is complete.
Transitivity: We have to show that
F 6|= (A ⇒ B) ∧ (B ⇒ C) → (A ⇒ C)
31
Let M = (ω, <, V ) be a model of F such that V (A) = {n | n is even}, V (B) = ω and V (C) =
{n | n is odd}. Then for an arbitrary state n ∈ ω we have that
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ B) and (∀n3 > n)(∃n4 > n3 ) M, n4 |= (B ∧ C)
This gives
M, n |= (A ⇒ B) and M, n |= (B ⇒ C)
and consequently
M, n |= (A ⇒ B) ∧ (B ⇒ C)
But M, n 6|= (A ⇒ C), because
(∀n5 > n)(∃n6 > n5 ) M, n6 |= A and (∃n7 > n)(∀n8 > n7 ) M, n8 |= ¬(A ∧ C)
It follows then that M, n 6|= (A ⇒ B) ∧ (B ⇒ C) → (A ⇒ C) and the proof is complete.
Weak Transitivity: We have to show that
F 6|= (A ⇒ B) ∧ (B ⇒ C) ⇒ (A ⇒ C)
Let M = (ω, <, V ) be a model of F such that V (A) = {n | n is even}, V (B) = ω and V (C) =
{n | n is odd}. For an arbitrary state n ∈ ω we have that M, n |= (A ⇒ B) ∧ (B ⇒ C) ⇒
(A ⇒ C) iff one of the following holds:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬(A ⇒ B) ∨ ¬(B ⇒ C)
(ii) (∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ⇒ B) ∧ (B ⇒ C) ∧ (A ⇒ C)
Case (i): By construction we have that
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ B) and (∀n3 > n)(∃n4 > n3 ) M, n4 |= (B ∧ C)
so by definition
(∀n ∈ ω) M, n |= (A ⇒ B) ∧ (B ⇒ C)
This gives us that (∀n ∈ ω) M, n 6|= ¬(A ⇒ B) ∨ ¬(B ⇒ C) and the case (i) cannot hold.
Case (ii): By construction we have that
(∀n5 > n)(∃n6 > n5 ) M, n6 |= A and (∃n7 > n)(∀n8 > n7 ) M, n8 |= ¬(A ∧ C)
so by definition (∀n ∈ ω) M, n 6|= (A ⇒ C) and the case (ii) cannot hold.
It follows then that M, n 6|= (A ⇒ B) ∧ (B ⇒ C) ⇒ (A ⇒ C) and the proof is complete.
Monotonicity: We have to show that
F 6|= (A ⇒ B) → (A ∧ C ⇒ B)
Let M = (ω, <, V ) be a model of F such that V (A) = ω, V (B) = {n | n is even} and V (C) =
{n | n is odd}. Then for an arbitrary state n ∈ ω we have that
(∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ B)
32
and thus M, n |= (A ⇒ B).
But M, n 6|= (A ∧ C ⇒ B), because
(∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ C) and (∃n5 > n)(∀n6 > n5 ) M, n6 |= ¬(A ∧ B ∧ C)
It follows then that M, n 6|= (A ⇒ B) → (A ∧ C ⇒ B) and the proof is complete.
Weak Monotonicity: We have to show that
F 6|= (A ⇒ B) ⇒ (A ∧ C ⇒ B)
Let M = (ω, <, V ) be a model of F such that V (A) = ω, V (B) = {n | n is even} and V (C) =
{n | n is odd}. For an arbitrary state n ∈ ω we have that M, n |= (A ⇒ B) ⇒ (A ∧ C ⇒ B)
iff one of the following holds:
(i) (∃n1 > n)(∀n2 > n1 ) M, n2 |= ¬(A ⇒ B)
(ii) (∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ⇒ B) ∧ (A ∧ C ⇒ B)
Case (i): By construction we have that (∀n1 > n)(∃n2 > n1 ) M, n2 |= (A ∧ B), so by definition
(∀n ∈ ω) M, n |= (A ⇒ B)
This gives us that (∀n ∈ ω) M, n 6|= ¬(A ⇒ B) and the case (i) cannot hold.
Case (ii): By construction we have that
(∀n3 > n)(∃n4 > n3 ) M, n4 |= (A ∧ C) and (∃n5 > n)(∀n6 > n5 ) M, n6 |= ¬(A ∧ B ∧ C)
so by definition (∀n ∈ ω) M, n 6|= (A ∧ C ⇒ B) and the case (ii) cannot hold.
It follows then that M, n 6|= (A ⇒ B) ⇒ (A ∧ C ⇒ B) and the proof is complete.
Modus Ponens: We have to show that
F 6|= A ∧ (A ⇒ B) → B
Let M = (ω, <, V ) be a model of F such that V (A) = {n} and V (B) = ∅. Then M, n |= A
and by definition M, n |= (A ⇒ B), which gives M, n |= A ∧ (A ⇒ B).
But M, n 6|= B by construction.
It follows then that M, n 6|= A ∧ (A ⇒ B) → B and the proof is complete.
3.2
Majority Conditionals over ω equipped with neighborhoods of cofinite
subsets
In this section, we take a different route. We proceed to define more majority-default conditionals, but not through modal logic. The approach is still model-theoretic, we resume our
original ‘cofinite-as-large’ intuition and work in the language of conditional logic. We resort to
the Scott-Montague semantics for conditionals introduced by B. Chellas [Che75] (called
‘minimal models’ by the author). The set of possible words is still ω and each world (natural
33
number or finite ordinal, if you prefer) is associated to a neigborhood consisting of cofinite
subsets of ω. We denote by F = (ω, f ) the frame we use, where
f : ω × 2ω → 22
ω
maps worlds (n ∈ ω) and propositions (sets of possible worlds, also subsets of ω), to neighborhoods of cofinite subsets of ω. It is interesting that the conditional logics defined this way are
rather weak compared to the previous ones but also to other known logics in the literature;
see Table 1, page 53 for a quick glimpse.
Our first approach is designed upon the following truth assignment: the conditional (A ⇒ B)
is true at a world n, iff either the proposition expressed by ¬A in M (k¬Ak) or the proposition
expressed by (A ∧ B) in M (kA ∧ Bk) are entailed by a proposition ‘necessary’ at the specific
neighborhood of n which corresponds to the proposition expressed by A in M (kAk). For those
acquainted with non-normal modal logics, this is a truth assignment which reminds a way to
define truth in a Scott-Montague model for the regular modal logics [Che80].
⇒
⇒
Definition 3.13 [Conditional Logic m1 ]. Let m1 be the logic consisting of all A ∈ L⇒
valid in F = (ω, f ), where a conditional is evaluated as follows:
For a model M over F, M, n |= (A ⇒ B) iff either
(i) there exists S ⊆ k¬Ak such that S ∈ f (n, kAk), or
(ii) there exists T ⊆ kA ∧ Bk such that T ∈ f (n, kAk)
⇒
Theorem 3.14 The logic m1 :
1. is closed under the rules RCEA and RCEC
2. contains the axiom CM
Proof. RCEA: Let M |= (A ≡ B). This means that kAk = kBk, k¬Ak = k¬Bk and
kA ∧ Ck = kB ∧ Ck. We have to show that
M |= (A ⇒ C) ≡ (B ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ C). Then, either:
(i) there exists S ⊆ k¬Ak such that S ∈ f (n, kAk), or
(ii) there exists T ⊆ kA ∧ Ck such that T ∈ f (n, kAk)
Case (i): By kAk = kBk and k¬Ak = k¬Bk we also have that there exists S ⊆ k¬Bk such
that S ∈ f (n, kBk) and by definition, M, n |= (B ⇒ C) follows.
34
Case (ii): Similarly, by kAk = kBk and kA ∧ Ck = kB ∧ Ck we obtain that there exists
T ⊆ kB ∧ Ck such that T ∈ f (n, kBk) and thus M, n |= (B ⇒ C).
So, if M, n |= (A ⇒ C), then M, n |= (B ⇒ C), which gives that
M, n |= (A ⇒ C) → (B ⇒ C)
Similarly, M, n |= (B ⇒ C) → (A ⇒ C). Since the world n was arbitrarily chosen, the proof
is complete.
RCEC: Let M |= (A ≡ B). This means that kC ∧ Ak = kC ∧ Bk. We have to show that
M |= (C ⇒ A) ≡ (C ⇒ B)
Assume an arbitrary state n ∈ ω, such that M, n |= (C ⇒ A). Then, either:
(i) there exists S ⊆ k¬Ck such that S ∈ f (n, kCk), or
(ii) there exists T ⊆ kC ∧ Ak such that T ∈ f (n, kCk)
Case (i): By definition, we also have that M, n |= (C ⇒ B).
Case (ii): By kC ∧ Ak = kC ∧ Bk we obtain that there exists T ⊆ kC ∧ Bk such that T ∈
f (n, kCk) and thus M, n |= (C ⇒ B).
So, if M, n |= (C ⇒ A), then M, n |= (C ⇒ B), which gives that
M, n |= (C ⇒ A) → (C ⇒ B)
Similarly, M, n |= (C ⇒ B) → (C ⇒ A). Since the world n was arbitrarily chosen, the proof
is complete.
CM: We have to show that
F |= (A ⇒ B ∧ C) → (A ⇒ B) ∧ (A ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ B ∧ C), where M is a model of F.
Then, either:
(i) there exists S ⊆ k¬Ak such that S ∈ f (n, kAk), or
(ii) there exists T ⊆ kA ∧ B ∧ Ck such that T ∈ f (n, kAk)
Case (i): By definition, M, n |= (A ⇒ B) and M, n |= (A ⇒ C), so M, n |= (A ⇒ B)∧(A ⇒ C)
follows.
Case (ii): We also have that there exists T ⊆ kA ∧ B ∧ Ck ⊆ kA ∧ Bk such that T ∈ f (n, kAk)
and there exists T ⊆ kA ∧ B ∧ Ck ⊆ kA ∧ Ck such that T ∈ f (n, kAk). By definition then,
M, n |= (A ⇒ B) and M, n |= (A ⇒ C), and thus M, n |= (A ⇒ B) ∧ (A ⇒ C).
So, if M, n |= (A ⇒ B ∧ C), then M, n |= (A ⇒ B) ∧ (A ⇒ C), which gives that
M, n |= (A ⇒ B ∧ C) → (A ⇒ B) ∧ (A ⇒ C)
Since the world n and model M were arbitrarily chosen, the proof is complete.
35
⇒
Theorem 3.15 The logic m1 :
1. is not closed under the rules RCK and RCE
2. does not contain the axioms ID, CUT, AC, CC, Loop, OR, CV, CSO, MP,
MOD, CA, CS, CEM, SDA, Transitivity, Weak Transitivity, Monotonicity,
Weak Monotonicity, Modus Ponens and Weak Modus Ponens
Proof. RCK: For n = 2: We have to show that
M |= (A1 ∧ A2 → B) and M 6|= (C ⇒ A1 ) ∧ (C ⇒ A2 ) → (C ⇒ B)
for some model M of F.
Let M = (ω, f, V ) be a model of F such that V (A1 ) = ω − {n1 }, V (A2 ) = ω − {n2 }, V (B) =
ω − {n1 , n2 }, V (C) = ω and for an arbitrary world n ∈ ω let f (n, kCk) = {ω − {n1 }, ω − {n2 }}.
Then (∀n ∈ ω) M, n |= (A1 ∧ A2 → B), because
M, n1 |= ¬A1 , M, n2 |= ¬A2 and (∀n 6= n1 , n2 ) M, n |= B
and thus M |= (A1 ∧ A2 → B).
By definition, we also have that M, n |= (C ⇒ A1 ) and M, n |= (C ⇒ A2 ), which means that
M, n |= (C ⇒ A1 ) ∧ (C ⇒ A2 )
But M, n 6|= (C ⇒ B), because we have that
there is no S ⊆ k¬Ck such that S ∈ f (n, kCk) and
there is no T ⊆ kC ∧ Bk such that T ∈ f (n, kCk)
It follows then that M, n 6|= (C ⇒ A1 ) ∧ (C ⇒ A2 ) → (C ⇒ B) and consequently M 6|= (C ⇒
A1 ) ∧ (C ⇒ A2 ) → (C ⇒ B).
RCE: We have to show that
M |= (A → B) and M 6|= (A ⇒ B)
for some model M of F.
Let M = (ω, f, V ) be a model of F such that V (A) = {n | n is even} and V (B) = ω. Then
(∀n ∈ ω) M, n |= (A → B), because (∀n ∈ ω) M, n |= B. Thus M |= (A → B).
But we have that both k¬Ak and kA ∧ Bk are not co-finite, so for an arbitrary world n ∈ ω
there is no S ⊆ k¬Ak such that S ∈ f (n, kAk) and
there is no T ⊆ kA ∧ Bk such that T ∈ f (n, kAk)
36
It follows then that M, n 6|= (A ⇒ B) and consequently M 6|= (A ⇒ B).
ID: We have to show that
F 6|= (A ⇒ A)
Let M = (ω, f, V ) be a model of F such that V (A) = {n | n is even}. Then we have that both
kAk and k¬Ak are not co-finite, so for an arbitrary world n ∈ ω
there is no S ⊆ k¬Ak such that S ∈ f (n, kAk) and
there is no T ⊆ kAk such that T ∈ f (n, kAk)
It follows then that M, n 6|= (A ⇒ A) and the proof is complete.
CUT: We have to show that
F 6|= (A ∧ B ⇒ C) ∧ (A ⇒ B) → (A ⇒ C)
Let M = (ω, f, V ) be a model of F such that V (A) = ω, V (B) = ω−{n1 }, V (C) = ω−{n2 } and
for an arbitrary world n ∈ ω let f (n, kA ∧ Bk) = {ω − {n1 , n2 }} and f (n, kAk) = {ω − {n1 }}.
By definition then, we have that
M, n |= (A ∧ B ⇒ C) and M, n |= (A ⇒ B)
and thus M, n |= (A ∧ B ⇒ C) ∧ (A ⇒ B).
But M, n 6|= (A ⇒ C), because
there is no S ⊆ k¬Ak such that S ∈ f (n, kAk) and
there is no T ⊆ kA ∧ Ck such that T ∈ f (n, kAk)
It follows then that M, n 6|= (A ∧ B ⇒ C) ∧ (A ⇒ B) → (A ⇒ C) and the proof is complete.
CC: We have to show that
F 6|= (A ⇒ B) ∧ (A ⇒ C) → (A ⇒ B ∧ C)
Let M = (ω, f, V ) be a model of F such that V (A) = ω, V (B) = ω − {n1 }, V (C) = ω − {n2 }
and for an arbitrary world n ∈ ω let f (n, kAk) = {ω − {n1 }, ω − {n2 }}. By definition then, we
have that
M, n |= (A ⇒ B) and M, n |= (A ⇒ C)
and thus M, n |= (A ⇒ B) ∧ (A ⇒ C).
But M, n 6|= (A ⇒ B ∧ C), because
there is no S ⊆ k¬Ak such that S ∈ f (n, kAk) and
there is no T ⊆ kA ∧ B ∧ Ck such that T ∈ f (n, kAk)
37
It follows then that M, n 6|= (A ⇒ B) ∧ (A ⇒ C) → (A ⇒ B ∧ C) and the proof is complete.
Loop: For k = 2: We have to show that
F 6|= (A0 ⇒ A1 ) ∧ (A1 ⇒ A2 ) ∧ (A2 ⇒ A0 ) → (A0 ⇒ A2 )
Let M = (ω, f, V ) be a model of F such that V (A0 ) = ω, V (A1 ) = ω − {n1 }, V (A2 ) = ω − {n2 }
and for an arbitrary world n ∈ ω let f (n, kA0 k) = {ω − {n1 }}, f (n, kA1 k) = {ω − {n1 , n2 }}
and f (n, kA2 k) = {ω − {n2 }}. By definition then, we have that
M, n |= (A0 ⇒ A1 ), M, n |= (A1 ⇒ A2 ) and M, n |= (A2 ⇒ A0 )
and thus M, n |= (A0 ⇒ A1 ) ∧ (A1 ⇒ A2 ) ∧ (A2 ⇒ A0 ).
But M, n 6|= (A0 ⇒ A2 ), because
there is no S ⊆ k¬A0 k such that S ∈ f (n, kA0 k) and
there is no T ⊆ kA0 ∧ A2 k such that T ∈ f (n, kA0 k)
It follows then that M, n 6|= (A0 ⇒ A1 ) ∧ (A1 ⇒ A2 ) ∧ (A2 ⇒ A0 ) → (A0 ⇒ A2 ) and the proof
is complete.
CSO: We have to show that
F 6|= (A ⇒ B) ∧ (B ⇒ A) → ((A ⇒ C) ≡ (B ⇒ C))
Let M = (ω, f, V ) be a model of F such that V (A) = ω, V (B) = ω−{n1 }, V (C) = ω−{n2 } and
for an arbitrary world n ∈ ω let f (n, kAk) = {ω − {n1 }, ω − {n2 }} and f (n, kBk) = {ω − {n1 }}.
By definition then, we have that
M, n |= (A ⇒ B), M, n |= (B ⇒ A) and M, n |= (A ⇒ C)
and thus M, n |= (A ⇒ B) ∧ (B ⇒ A) and M, n |= (A ⇒ C).
But M, n 6|= (B ⇒ C), because
there is no S ⊆ k¬Bk such that S ∈ f (n, kBk) and
there is no T ⊆ kB ∧ Ck such that T ∈ f (n, kBk)
It follows then that M, n 6|= (A ⇒ B) ∧ (B ⇒ A) → ((A ⇒ C) → (B ⇒ C)) and the proof is
complete.
MP: We have to show that
F 6|= (A ⇒ B) → (A → B)
Let M = (ω, f, V ) be a model of F such that V (A) = ω, V (B) = ω − {n} and for the world n
let f (n, kAk) = {ω − {n}}. By definition then, we have that M, n |= (A ⇒ B).
38
But M, n 6|= (A → B), because M, n |= A ∧ ¬B.
It follows then that M, n 6|= (A ⇒ B) → (A → B) and the proof is complete.
CS: We have to show that
F 6|= (A ∧ B) → (A ⇒ B)
Let M = (ω, f, V ) be a model of F such that V (A) = V (B) = {n} and for the world n let
f (n, kAk) = {ω}. By definition then, we have that M, n |= (A ∧ B).
But M, n 6|= (A ⇒ B), because
there is no S ⊆ k¬Ak such that S ∈ f (n, kAk) and
there is no T ⊆ kA ∧ Bk such that T ∈ f (n, kAk)
It follows then that M, n 6|= (A ∧ B) → (A ⇒ B) and the proof is complete.
CEM: We have to show that
F 6|= (A ⇒ B) ∨ (A ⇒ ¬B)
Let M = (ω, f, V ) be a model of F such that V (A) = V (B) = {n | n is even} and for an
arbitrary state n ∈ ω let f (n, kAk) = {ω}. Then we have that M, n 6|= (A ⇒ B) and
M, n 6|= (A ⇒ ¬B), because
there is no S ⊆ k¬Ak such that S ∈ f (n, kAk)
there is no T ⊆ kA ∧ Bk such that T ∈ f (n, kAk)
there is no R ⊆ kA ∧ ¬Bk such that R ∈ f (n, kAk)
It follows then that M, n 6|= (A ⇒ B) ∨ (A ⇒ ¬B) and the proof is complete.
Transitivity: We have to show that
F 6|= (A ⇒ B) ∧ (B ⇒ C) → (A ⇒ C)
Let M = (ω, f, V ) be a model of F such that V (A) = ω, V (B) = ω − {n1 }, V (C) = ω − {n2 }
and for an arbitrary world n ∈ ω let f (n, kAk) = {ω − {n1 }} and f (n, kBk) = {ω − {n1 , n2 }}.
By definition then, we have that
M, n |= (A ⇒ B) and M, n |= (B ⇒ C)
and thus M, n |= (A ⇒ B) ∧ (B ⇒ C).
But M, n 6|= (A ⇒ C), because
there is no S ⊆ k¬Ak such that S ∈ f (n, kAk) and
39
there is no T ⊆ kA ∧ Ck such that T ∈ f (n, kAk)
It follows then that M, n 6|= (A ⇒ B) ∧ (B ⇒ C) → (A ⇒ C) and the proof is complete.
Weak Transitivity: We have to show that
F 6|= (A ⇒ B) ∧ (B ⇒ C) ⇒ (A ⇒ C)
Let M = (ω, f, V ) be a model of F such that V (A) = ω − {n1 }, V (B) = ω − {n2 }, V (C) = ω
and for n∗ ∈ ω let the following hold:
(i) (∀n ∈ ω − {n∗ }) f (n, kAk) = {ω − {n1 }, ω − {n1 , n2 }}, while f (n∗ , kAk) = {ω}
(ii) (∀n ∈ ω − {n∗ }) f (n, kBk) = {ω − {n2 }}, while f (n∗ , kBk) = {ω}
(iii) (∀n ∈ ω) f (n, k(A ⇒ B) ∧ (B ⇒ C)k) = {ω}
By definition then, we have that
kA ⇒ Bk = kB ⇒ Ck = kA ⇒ Ck = ω − {n∗ }
This means that
k(A ⇒ B) ∧ (B ⇒ C) ∧ (A ⇒ C)k = ω − {n∗ } (1) and k¬(A ⇒ B) ∨ ¬(B ⇒ C)k = {n∗ } (2)
For an arbitrary state n ∈ ω we have that M, n |= (A ⇒ B) ∧ (B ⇒ C) ⇒ (A ⇒ C) iff one of
the following holds:
(a) there exists S ⊆ k¬(A ⇒ B) ∨ ¬(B ⇒ C)k such that S ∈ f (n, k(A ⇒ B) ∧ (B ⇒ C)k)
(b) there exists T ⊆ k(A ⇒ B) ∧ (B ⇒ C) ∧ (A ⇒ C)k such that T ∈ f (n, k(A ⇒ B) ∧ (B ⇒
C)k)
Case (a): By (2) and (iii) there is no S ⊆ k¬(A ⇒ B) ∨ ¬(B ⇒ C)k such that S ∈ f (n, k(A ⇒
B) ∧ (B ⇒ C)k).
Case (b): By (1) and (iii) there is no T ⊆ k(A ⇒ B) ∧ (B ⇒ C) ∧ (A ⇒ C)k such that T ∈
f (n, k(A ⇒ B) ∧ (B ⇒ C)k).
It follows then that M, n 6|= (A ⇒ B) ∧ (B ⇒ C) ⇒ (A ⇒ C) and the proof is complete.
Weak Monotonicity: We have to show that
F 6|= (A ⇒ B) ⇒ (A ∧ C ⇒ B)
Let M = (ω, f, V ) be a model of F such that V (A) = ω − {n1 }, V (B) = V (C) = ω and for
n∗ ∈ ω let the following hold:
(i) (∀n ∈ ω − {n∗ }) f (n, kAk) = {ω − {n1 }}, while f (n∗ , kAk) = {ω}
(ii) (∀n ∈ ω − {n∗ }) f (n, kA ∧ Ck) = {ω − {n1 }}, while f (n∗ , kA ∧ Ck) = {ω}
40
(iii) (∀n ∈ ω) f (n, k(A ⇒ B)k) = {ω}
By definition then, we have that
kA ⇒ Bk = kA ∧ C ⇒ Bk = ω − {n∗ }
This means that
k(A ⇒ B) ∧ (A ∧ C ⇒ B)k = ω − {n∗ } (1) and k¬(A ⇒ B)k = {n∗ } (2)
For an arbitrary state n ∈ ω we have that M, n |= (A ⇒ B) ⇒ (A ∧ C ⇒ B) iff one of the
following holds:
(a) there exists S ⊆ k¬(A ⇒ B)k such that S ∈ f (n, kA ⇒ Bk)
(b) there exists T ⊆ k(A ⇒ B) ∧ (A ∧ C ⇒ B)k such that T ∈ f (n, kA ⇒ Bk)
Case (a): By (2) and (iii) there is no S ⊆ k¬(A ⇒ B)k such that S ∈ f (n, kA ⇒ Bk).
Case (b): By (1) and (iii) there is no T ⊆ k(A ⇒ B) ∧ (A ∧ C ⇒ B)k such that
T ∈ f (n, kA ⇒ Bk).
It follows then that M, n 6|= (A ⇒ B) ⇒ (A ∧ C ⇒ B) and the proof is complete.
Modus Ponens: We have to show that
F 6|= A ∧ (A ⇒ B) → B
Let M = (ω, f, V ) be a model of F such that V (A) = ω, V (B) = ω − {n} and for the world n
let f (n, kAk) = {ω − {n}}. Then M, n |= A and by definition M, n |= (A ⇒ B), which gives
M, n |= A ∧ (A ⇒ B).
But M, n 6|= B by construction.
It follows then that M, n 6|= A ∧ (A ⇒ B) → B and the proof is complete.
Weak Modus Ponens: We have to show that
F 6|= A ∧ (A ⇒ B) ⇒ B
Let M = (ω, f, V ) be a model of F such that V (A) = ω − {n1 } and V (B) = ω and for n∗ ∈ ω
let the following hold:
(i) (∀n ∈ ω − {n∗ }) f (n, kAk) = {ω − {n1 }}, while f (n∗ , kAk) = {ω}
(ii) (∀n ∈ ω) f (n, kA ∧ (A ⇒ B)k) = {ω}
By definition then, we have that
kA ⇒ Bk = ω − {n∗ }
This means that
kA ∧ (A ⇒ B) ∧ Bk = ω − {n1 , n∗ } (1) and k¬A ∨ ¬(A ⇒ B)k = {n1 , n∗ } (2)
For an arbitrary state n ∈ ω we have that M, n |= A ∧ (A ⇒ B) ⇒ B iff one of the following
holds:
41
(a) there exists S ⊆ k¬A ∨ ¬(A ⇒ B)k such that S ∈ f (n, kA ∧ (A ⇒ B)k)
(b) there exists T ⊆ kA ∧ (A ⇒ B) ∧ Bk such that T ∈ f (n, kA ∧ (A ⇒ B)k)
Case (a): By (2) and (ii) there is no S ⊆ k¬A ∨ ¬(A ⇒ B)k such that S ∈ f (n, kA∧(A ⇒ B)k).
Case (b): By (1) and (ii) there is no T ⊆ kA ∧ (A ⇒ B) ∧ Bk such that T ∈ f (n, kA ∧ (A ⇒
B)k).
It follows then that M, n 6|= A ∧ (A ⇒ B) ⇒ B and the proof is complete.
AC, OR, CV, MOD, CA, SDA, Monotonicity: For all these axioms the counterexample
is obvious. For an arbitrary world n ∈ ω just let f (n, kAk), f (n, k¬Ak), f (n, kBk), f (n, kCk),
f (n, kA ∧ Bk), f (n, kA ∨ Bk) and f (n, kA ∧ Ck) be different from one another, according to
the respective axiom, and all are invalid in n.
Now for another variant. While the previous semantics asked that the proposition of interest be entailed by a proposition ‘necessary’ in (a member of) the appropriate neighborhood
(see the comments before Definition 3.13), we now require that it is explicitly participating in
the neighborhood.
⇒
⇒
Definition 3.16 [Conditional Logic m2 ]. Let m2 be the logic consisting of all A ∈ L⇒
valid in F = (ω, f ), where a conditional is evaluated as follows:
For a model M over F, M, n |= (A ⇒ B) iff either
(i) k¬Ak ∈ f (n, kAk), or
(ii) kA ∧ Bk ∈ f (n, kAk)
⇒
Theorem 3.17 The logic m2 is closed under the rules RCEA and RCEC
Proof. RCEA: Let M |= (A ≡ B). This means that kAk = kBk, k¬Ak = k¬Bk and
kA ∧ Ck = kB ∧ Ck. We have to show that
M |= (A ⇒ C) ≡ (B ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ C). Then, either:
(i) k¬Ak ∈ f (n, kAk), or
(ii) kA ∧ Ck ∈ f (n, kAk)
42
Case (i): By kAk = kBk and k¬Ak = k¬Bk we also have that k¬Bk ∈ f (n, kBk) and by
definition, M, n |= (B ⇒ C) follows.
Case (ii): Similarly, by kAk = kBk and kA ∧ Ck = kB ∧ Ck we obtain kB ∧ Ck ∈ f (n, kBk)
and thus M, n |= (B ⇒ C).
So, if M, n |= (A ⇒ C), then M, n |= (B ⇒ C), which gives that
M, n |= (A ⇒ C) → (B ⇒ C)
Similarly, M, n |= (B ⇒ C) → (A ⇒ C). Since the world n was arbitrarily chosen, the proof
is complete.
RCEC: Let M |= (A ≡ B). This means that kC ∧ Ak = kC ∧ Bk. We have to show that
M |= (C ⇒ A) ≡ (C ⇒ B)
Assume an arbitrary state n ∈ ω, such that M, n |= (C ⇒ A). Then, either:
(i) k¬Ck ∈ f (n, kCk), or
(ii) kC ∧ Ak ∈ f (n, kCk)
Case (i): By definition, we also have that M, n |= (C ⇒ B).
Case (ii): By kC ∧ Ak = kC ∧ Bk we obtain kC ∧ Bk ∈ f (n, kCk) and thus M, n |= (C ⇒ B).
So, if M, n |= (C ⇒ A), then M, n |= (C ⇒ B), which gives that
M, n |= (C ⇒ A) → (C ⇒ B)
Similarly, M, n |= (C ⇒ B) → (C ⇒ A). Since the world n was arbitrarily chosen, the proof
is complete.
⇒
Theorem 3.18 The logic m2 :
1. is not closed under the rules RCK and RCE
2. does not contain any of the axioms ID, CUT, AC, CC, Loop, OR, CV, CSO,
CM, MP, MOD, CA, CS, CEM, SDA, Transitivity, Weak Transitivity,
Monotonicity, Weak Monotonicity, Modus Ponens and Weak Modus Ponens
Proof. RCK, RCE, ID, CUT, AC, CC, Loop, OR, CV, CSO, MP, MOD, CA, CS,
CEM, SDA, Transitivity, Weak Transitivity, Monotonicity, Weak Monotonicity,
Modus Ponens, Weak Modus Ponens: The counterexamples of all these rules and axioms
⇒
are identical to the respective cases of the logic m1 . The counterexample of the axiom CM
follows.
43
CM: We have to show that
F 6|= (A ⇒ B ∧ C) → (A ⇒ B) ∧ (A ⇒ C)
Let M = (ω, f, V ) be a model of F such that V (A) = ω, V (B) = ω − {n1 }, V (C) = ω − {n2 }
and for an arbitrary world n ∈ ω let f (n, kAk) = {ω − {n1 , n2 }}. By definition then, we have
that
M, n |= (A ⇒ B ∧ C)
But M, n 6|= (A ⇒ B), because
k¬Ak 6∈ f (n, kAk) and kA ∧ Bk 6∈ f (n, kAk)
It follows then that M, n 6|= (A ⇒ B ∧ C) → (A ⇒ B) ∧ (A ⇒ C) and the proof is complete.
Our last logic will require that the neighborhoods are filters over ω. Keeping in
mind that in F = (ω, f ), where
ω
f : ω × 2ω → 22
maps worlds and propositions, to neighborhoods of cofinite subsets of ω, we further assume
that:
(i) If S ∈ f (n, X) and S ⊆ T then T ∈ f (n, X) (upwards closure)
(ii) If S, T ∈ f (n, X) then S ∩ T ∈ f (n, X) (closure under intersections)
Note that f is well defined: if S is cofinite and S ⊆ T then T is also cofinite and if S, T are
cofinite then S ∩ T is also cofinite.
⇒
⇒
Definition 3.19 [Conditional Logic m3 ]. Let m3 be the logic consisting of all A ∈ L⇒
valid in F = (ω, f ), where a conditional is evaluated as follows:
For a model M over F, M, n |= (A ⇒ B) iff either
(i) k¬Ak ∈ f (n, kAk), or
(ii) kA ∧ Bk ∈ f (n, kAk)
Fact 3.20 If we replaced (i) and (ii) in the above definition with:
(i) there exists S ⊆ k¬Ak such that S ∈ f (n, kAk)
(ii) there exists T ⊆ kA ∧ Bk such that T ∈ f (n, kAk)
then the definition of (A ⇒ B) would be equivalent, because of the condition (i) in the
definition of f .
44
⇒
Theorem 3.21 The logic m3 :
1. is closed under the rules RCEA, RCK and RCEC
2. contains the axioms CC and CM
Proof. RCEA: Let M |= (A ≡ B). This means that kAk = kBk, k¬Ak = k¬Bk and
kA ∧ Ck = kB ∧ Ck. We have to show that
M |= (A ⇒ C) ≡ (B ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ C). Then, either:
(i) k¬Ak ∈ f (n, kAk), or
(ii) kA ∧ Ck ∈ f (n, kAk)
Case (i): By kAk = kBk and k¬Ak = k¬Bk we also have that k¬Bk ∈ f (n, kBk) and by
definition, M, n |= (B ⇒ C) follows.
Case (ii): Similarly, by kAk = kBk and kA ∧ Ck = kB ∧ Ck we obtain kB ∧ Ck ∈ f (n, kBk)
and thus M, n |= (B ⇒ C).
So, if M, n |= (A ⇒ C), then M, n |= (B ⇒ C), which gives that
M, n |= (A ⇒ C) → (B ⇒ C)
Similarly, M, n |= (B ⇒ C) → (A ⇒ C). Since the world n was arbitrarily chosen, the proof
is complete.
RCK: Let M |= (A1 ∧ . . . ∧ An ) → B. This means that kC ∧ A1 ∧ . . . ∧ An k ⊆ kC ∧ Bk. We
have to show that
M |= (C ⇒ A1 ∧ . . . ∧ C ⇒ An ) → (C ⇒ B)
Assume an arbitrary state n ∈ ω, such that M, n |= (C ⇒ A1 ∧ . . . ∧ C ⇒ An ). Obviously,
M, n |= (C ⇒ A1 ) and . . . and M, n |= (C ⇒ An ). Then, either:
(i) k¬Ck ∈ f (n, kCk), or
(ii) kC ∧ A1 k ∈ f (n, kCk) and . . . and kC ∧ An k ∈ f (n, kCk). By the condition (ii) in the
definition of f , this gives kC ∧ A1 ∧ . . . ∧ An k ∈ f (n, kCk)
Case (i): By definition, we also have that M, n |= (C ⇒ B).
Case (ii): By kC ∧ A1 ∧ . . . ∧ An k ⊆ kC ∧ Bk and the condition (i) in the definition of f , we
obtain kC ∧ Bk ∈ f (n, kCk) and thus M, n |= (C ⇒ B).
So, if M, n |= (C ⇒ A1 ∧ . . . ∧ C ⇒ An ), then M, n |= (C ⇒ B), which gives that
M, n |= (C ⇒ A1 ∧ . . . ∧ C ⇒ An ) → (C ⇒ B)
45
Since the world n was arbitrarily chosen, the proof is complete.
RCEC: Let M |= (A ≡ B). This means that kC ∧ Ak = kC ∧ Bk. We have to show that
M |= (C ⇒ A) ≡ (C ⇒ B)
Assume an arbitrary state n ∈ ω, such that M, n |= (C ⇒ A). Then, either:
(i) k¬Ck ∈ f (n, kCk), or
(ii) kC ∧ Ak ∈ f (n, kCk)
Case (i): By definition, we also have that M, n |= (C ⇒ B).
Case (ii): By kC ∧ Ak = kC ∧ Bk we obtain kC ∧ Bk ∈ f (n, kCk) and thus M, n |= (C ⇒ B).
So, if M, n |= (C ⇒ A), then M, n |= (C ⇒ B), which gives that
M, n |= (C ⇒ A) → (C ⇒ B)
Similarly, M, n |= (C ⇒ B) → (C ⇒ A). Since the world n was arbitrarily chosen, the proof
is complete.
CC: We have to show that
F |= (A ⇒ B) ∧ (A ⇒ C) → (A ⇒ B ∧ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ B) ∧ (A ⇒ C), where M is a model
of F. Obviously, M, n |= (A ⇒ B) and M, n |= (A ⇒ C). Then, either:
(i) k¬Ak ∈ f (n, kAk), or
(ii) kA ∧ Bk ∈ f (n, kAk) and kA ∧ Ck ∈ f (n, kAk)
Case (i): By definition, M, n |= (A ⇒ B ∧ C) follows.
Case (ii): By the condition (ii) in the definition of f , we also have that kA ∧ B ∧ Ck ∈
f (n, kAk), and thus M, n |= (A ⇒ B ∧ C).
So, if M, n |= (A ⇒ B) ∧ (A ⇒ C), then M, n |= (A ⇒ B ∧ C), which gives that
M, n |= (A ⇒ B) ∧ (A ⇒ C) → (A ⇒ B ∧ C)
Since the world n and model M were arbitrarily chosen, the proof is complete.
CM: We have to show that
F |= (A ⇒ B ∧ C) → (A ⇒ B) ∧ (A ⇒ C)
Assume an arbitrary state n ∈ ω, such that M, n |= (A ⇒ B ∧ C), where M is a model of F.
Then, either:
46
(i) k¬Ak ∈ f (n, kAk), or
(ii) kA ∧ B ∧ Ck ∈ f (n, kAk)
Case (i): By definition, M, n |= (A ⇒ B) and M, n |= (A ⇒ C), so M, n |= (A ⇒ B)∧(A ⇒ C)
follows.
Case (ii): We also have that kA ∧ B ∧ Ck ⊆ kA ∧ Bk and kA ∧ B ∧ Ck ⊆ kA ∧ Ck. By
condition (i) in the definition of f , this gives
kA ∧ Bk ∈ f (n, kAk) and kA ∧ Ck ∈ f (n, kAk)
By definition then, M, n |= (A ⇒ B) and M, n |= (A ⇒ C), and thus M, n |= (A ⇒ B) ∧ (A ⇒
C).
So, if M, n |= (A ⇒ B ∧ C), then M, n |= (A ⇒ B) ∧ (A ⇒ C), which gives that
M, n |= (A ⇒ B ∧ C) → (A ⇒ B) ∧ (A ⇒ C)
Since the world n and model M were arbitrarily chosen, the proof is complete.
⇒
Theorem 3.22 The logic m3 :
1. is not closed under the rule RCE
2. does not contain the axioms ID, CUT, AC, Loop, OR, CV, CSO, MP,
MOD, CA, CS, CEM, SDA, Transitivity, Weak Transitivity, Monotonicity,
Weak Monotonicity, Modus Ponens and Weak Modus Ponens
Proof. RCE: We have to show that
M |= (A → B) and M 6|= (A ⇒ B)
for some model M of F.
Let M = (ω, f, V ) be a model of F such that V (A) = {n | n is even} and V (B) = ω. Then
(∀n ∈ ω) M, n |= (A → B), because (∀n ∈ ω) M, n |= B. Thus M |= (A → B).
But we have that both k¬Ak and kA ∧ Bk are not co-finite, so for an arbitrary world n ∈ ω
k¬Ak 6∈ f (n, kAk) and kA ∧ Bk 6∈ f (n, kAk)
It follows then that M, n 6|= (A ⇒ B) and consequently M 6|= (A ⇒ B).
ID: We have to show that
F 6|= (A ⇒ A)
Let M = (ω, f, V ) be a model of F such that V (A) = {n | n is even}. Then we have that both
kAk and k¬Ak are not co-finite, so for an arbitrary world n ∈ ω
47
k¬Ak 6∈ f (n, kAk) and kAk 6∈ f (n, kAk)
It follows then that M, n 6|= (A ⇒ A) and the proof is complete.
CUT: We have to show that
F 6|= (A ∧ B ⇒ C) ∧ (A ⇒ B) → (A ⇒ C)
Let M = (ω, f, V ) be a model of F such that V (A) = V (B) = ω, V (C) = ω − {n1 } and for an
arbitrary world n ∈ ω let f (n, kAk) = {ω} and f (n, kA ∧ Bk) = {ω, ω − {n1 }}. By definition
then, we have that
M, n |= (A ∧ B ⇒ C) and M, n |= (A ⇒ B)
and thus M, n |= (A ∧ B ⇒ C) ∧ (A ⇒ B).
But M, n 6|= (A ⇒ C), because
k¬Ak 6∈ f (n, kAk) and kA ∧ Ck 6∈ f (n, kAk)
It follows then that M, n 6|= (A ∧ B ⇒ C) ∧ (A ⇒ B) → (A ⇒ C) and the proof is complete.
Loop: For k = 2: We have to show that
F 6|= (A0 ⇒ A1 ) ∧ (A1 ⇒ A2 ) ∧ (A2 ⇒ A0 ) → (A0 ⇒ A2 )
Let M = (ω, f, V ) be a model of F such that V (A0 ) = ω, V (A1 ) = ω − {n1 }, V (A2 ) = ω − {n2 }
and for an arbitrary world n ∈ ω let f (n, kA0 k) = {ω, ω−{n1 }}, f (n, kA1 k) = {ω, ω−{n1 }, ω−
{n2 }, ω − {n1 , n2 }} and f (n, kA2 k) = {ω, ω − {n2 }}. By definition then, we have that
M, n |= (A0 ⇒ A1 ), M, n |= (A1 ⇒ A2 ) and M, n |= (A2 ⇒ A0 )
and thus M, n |= (A0 ⇒ A1 ) ∧ (A1 ⇒ A2 ) ∧ (A2 ⇒ A0 ).
But M, n 6|= (A0 ⇒ A2 ), because
k¬A0 k 6∈ f (n, kA0 k) and kA0 ∧ A2 k 6∈ f (n, kA0 k)
It follows then that M, n 6|= (A0 ⇒ A1 ) ∧ (A1 ⇒ A2 ) ∧ (A2 ⇒ A0 ) → (A0 ⇒ A2 ) and the proof
is complete.
CSO: We have to show that
F 6|= (A ⇒ B) ∧ (B ⇒ A) → ((A ⇒ C) ≡ (B ⇒ C))
Let M = (ω, f, V ) be a model of F such that V (A) = ω, V (B) = ω − {n1 }, V (C) = ω − {n2 }
and for an arbitrary world n ∈ ω let f (n, kAk) = {ω, ω − {n1 }, ω − {n2 }, ω − {n1 , n2 }} and
f (n, kBk) = {ω, ω − {n1 }}. By definition then, we have that
M, n |= (A ⇒ B), M, n |= (B ⇒ A) and M, n |= (A ⇒ C)
and thus M, n |= (A ⇒ B) ∧ (B ⇒ A) and M, n |= (A ⇒ C).
But M, n 6|= (B ⇒ C), because
48
k¬Bk 6∈ f (n, kBk) and kB ∧ Ck 6∈ f (n, kBk)
It follows then that M, n 6|= (A ⇒ B) ∧ (B ⇒ A) → ((A ⇒ C) → (B ⇒ C)) and the proof is
complete.
MP: We have to show that
F 6|= (A ⇒ B) → (A → B)
Let M = (ω, f, V ) be a model of F such that V (A) = ω, V (B) = ω − {n} and for the world n
let f (n, kAk) = {ω, ω − {n}}. By definition then, we have that M, n |= (A ⇒ B).
But M, n 6|= (A → B), because M, n |= A ∧ ¬B.
It follows then that M, n 6|= (A ⇒ B) → (A → B) and the proof is complete.
CS: We have to show that
F 6|= (A ∧ B) → (A ⇒ B)
Let M = (ω, f, V ) be a model of F such that V (A) = V (B) = {n} and for the world n let
f (n, kAk) = {ω}. By definition then, we have that M, n |= (A ∧ B).
But M, n 6|= (A ⇒ B), because
k¬Ak 6∈ f (n, kAk) and kA ∧ Bk 6∈ f (n, kAk)
It follows then that M, n 6|= (A ∧ B) → (A ⇒ B) and the proof is complete.
CEM: We have to show that
F 6|= (A ⇒ B) ∨ (A ⇒ ¬B)
Let M = (ω, f, V ) be a model of F such that V (A) = V (B) = {n | n is even} and for an
arbitrary state n ∈ ω let f (n, kAk) = {ω}. Then we have that M, n 6|= (A ⇒ B) and
M, n 6|= (A ⇒ ¬B), because
k¬Ak 6∈ f (n, kAk), kA ∧ Bk 6∈ f (n, kAk) and kA ∧ ¬Bk 6∈ f (n, kAk)
It follows then that M, n 6|= (A ⇒ B) ∨ (A ⇒ ¬B) and the proof is complete.
Transitivity: We have to show that
F 6|= (A ⇒ B) ∧ (B ⇒ C) → (A ⇒ C)
Let M = (ω, f, V ) be a model of F such that V (A) = ω, V (B) = ω − {n1 }, V (C) = ω − {n2 }
and for an arbitrary world n ∈ ω let f (n, kAk) = {ω, ω − {n1 }} and f (n, kBk) = {ω, ω −
{n1 }, ω − {n2 }, ω − {n1 , n2 }}. By definition then, we have that
M, n |= (A ⇒ B) and M, n |= (B ⇒ C)
and thus M, n |= (A ⇒ B) ∧ (B ⇒ C).
But M, n 6|= (A ⇒ C), because
49
k¬Ak 6∈ f (n, kAk) and kA ∧ Ck 6∈ f (n, kAk)
It follows then that M, n 6|= (A ⇒ B) ∧ (B ⇒ C) → (A ⇒ C) and the proof is complete.
Weak Transitivity: We have to show that
F 6|= (A ⇒ B) ∧ (B ⇒ C) ⇒ (A ⇒ C)
Let M = (ω, f, V ) be a model of F such that V (A) = ω − {n1 }, V (B) = ω − {n2 }, V (C) = ω
and for n∗ ∈ ω let the following hold:
(i) (∀n ∈ ω−{n∗ }) f (n, kAk) = {ω, ω−{n1 }, ω−{n2 }, ω−{n1 , n2 }}, while f (n∗ , kAk) = {ω}
(ii) (∀n ∈ ω − {n∗ }) f (n, kBk) = {ω, ω − {n2 }}, while f (n∗ , kBk) = {ω}
(iii) (∀n ∈ ω) f (n, k(A ⇒ B) ∧ (B ⇒ C)k) = {ω}
By definition then, we have that
kA ⇒ Bk = kB ⇒ Ck = kA ⇒ Ck = ω − {n∗ }
This means that
k(A ⇒ B) ∧ (B ⇒ C) ∧ (A ⇒ C)k = ω − {n∗ } (1) and k¬(A ⇒ B) ∨ ¬(B ⇒ C)k = {n∗ } (2)
For an arbitrary state n ∈ ω we have that M, n |= (A ⇒ B) ∧ (B ⇒ C) ⇒ (A ⇒ C) iff one of
the following holds:
(a) k¬(A ⇒ B) ∨ ¬(B ⇒ C)k ∈ f (n, k(A ⇒ B) ∧ (B ⇒ C)k)
(b) k(A ⇒ B) ∧ (B ⇒ C) ∧ (A ⇒ C)k ∈ f (n, k(A ⇒ B) ∧ (B ⇒ C)k)
Case (a): By (2) and (iii) we have k¬(A ⇒ B) ∨ ¬(B ⇒ C)k 6∈ f (n, k(A ⇒ B) ∧ (B ⇒ C)k).
Case (b): By (1) and (iii) we have k(A ⇒ B) ∧ (B ⇒ C) ∧ (A ⇒ C)k 6∈ f (n, k(A ⇒ B) ∧ (B ⇒
C)k).
It follows then that M, n 6|= (A ⇒ B) ∧ (B ⇒ C) ⇒ (A ⇒ C) and the proof is complete.
Weak Monotonicity: We have to show that
F 6|= (A ⇒ B) ⇒ (A ∧ C ⇒ B)
Let M = (ω, f, V ) be a model of F such that V (A) = ω − {n1 }, V (B) = V (C) = ω and for
n∗ ∈ ω let the following hold:
(i) (∀n ∈ ω − {n∗ }) f (n, kAk) = {ω, ω − {n1 }}, while f (n∗ , kAk) = {ω}
(ii) (∀n ∈ ω − {n∗ }) f (n, kA ∧ Ck) = {ω, ω − {n1 }}, while f (n∗ , kA ∧ Ck) = {ω}
(iii) (∀n ∈ ω) f (n, kA ⇒ Bk) = {ω}
50
By definition then, we have that
kA ⇒ Bk = kA ∧ C ⇒ Bk = ω − {n∗ }
This means that
k(A ⇒ B) ∧ (A ∧ C ⇒ B)k = ω − {n∗ } (1) and k¬(A ⇒ B)k = {n∗ } (2)
For an arbitrary state n ∈ ω we have that M, n |= (A ⇒ B) ⇒ (A ∧ C ⇒ B) iff one of the
following holds:
(a) k¬(A ⇒ B)k ∈ f (n, kA ⇒ Bk)
(b) k(A ⇒ B) ∧ (A ∧ C ⇒ B)k ∈ f (n, kA ⇒ Bk)
Case (a): By (2) and (iii) we have k¬(A ⇒ B)k 6∈ f (n, kA ⇒ Bk).
Case (b): By (1) and (iii) we have k(A ⇒ B) ∧ (A ∧ C ⇒ B)k 6∈ f (n, kA ⇒ Bk).
It follows then that M, n 6|= (A ⇒ B) ⇒ (A ∧ C ⇒ B) and the proof is complete.
Modus Ponens: We have to show that
F 6|= A ∧ (A ⇒ B) → B
Let M = (ω, <, V ) be a model of F such that V (A) = ω and V (B) = ω − {n} and for the
world n let f (n, kAk) = {ω, ω − {n}}. Then M, n |= A and by definition M, n |= (A ⇒ B),
which gives M, n |= A ∧ (A ⇒ B).
But M, n 6|= B by construction.
It follows then that M, n 6|= A ∧ (A ⇒ B) → B and the proof is complete.
Weak Modus Ponens: We have to show that
F 6|= A ∧ (A ⇒ B) ⇒ B
Let M = (ω, f, V ) be a model of F such that V (A) = ω − {n1 } and V (B) = ω and for n∗ ∈ ω
let the following hold:
(i) (∀n ∈ ω − {n∗ }) f (n, kAk) = {ω, ω − {n1 }}, while f (n∗ , kAk) = {ω}
(ii) (∀n ∈ ω) f (n, kA ∧ (A ⇒ B)k) = {ω}
By definition then, we have that
kA ⇒ Bk = ω − {n∗ }
This means that
kA ∧ (A ⇒ B) ∧ Bk = ω − {n1 , n∗ } (1) and k¬A ∨ ¬(A ⇒ B)k = {n1 , n∗ } (2)
For an arbitrary state n ∈ ω we have that M, n |= A ∧ (A ⇒ B) ⇒ B iff one of the following
holds:
51
(a) k¬A ∨ ¬(A ⇒ B)k ∈ f (n, kA ∧ (A ⇒ B)k)
(b) kA ∧ (A ⇒ B) ∧ Bk ∈ f (n, kA ∧ (A ⇒ B)k)
Case (a): By (2) and (ii) we have k¬A ∨ ¬(A ⇒ B)k 6∈ f (n, kA ∧ (A ⇒ B)k).
Case (b): By (1) and (ii) we have kA ∧ (A ⇒ B) ∧ Bk 6∈ f (n, kA ∧ (A ⇒ B)k).
It follows then that M, n 6|= A ∧ (A ⇒ B) ⇒ B and the proof is complete.
AC, OR, CV, MOD, CA, SDA, Monotonicity: For all these axioms the counterexample
is obvious. For an arbitrary world n ∈ ω just let f (n, kAk), f (n, k¬Ak), f (n, kBk), f (n, kCk),
f (n, kA ∧ Bk), f (n, kA ∨ Bk) and f (n, kA ∧ Ck) be different from one another, according to
the respective axiom, and all are invalid in n.
We summarize our results in this paper in the following table which aims to provide a clear
view of the position of the logics we defined here, with respect to known systems of ‘normality’
conditionals but also with respect to the KLM machinery. The table consists of two parts:
the KLM part and the Conditional Logic part. The logics defined in this work occupy
the rightmost column. Throughout the table a 3means ‘yes’, a shaded rectangle means ‘no’
and a white rectangle means ‘not explicitly proved in the literature’. Note that the landscape
of the logics we defined here is completely identified.
The KLM part reminds us of the axiom and the various rules (first column) used to define
the four KLM systems of logic (second column). The relation of KLM systems to Conditional
Logics can be immediately visualized by locating the conditional analog of each KLM rule, in
the same row, on the second part of the table.
The Conditional Logic part of the table comprises the axioms and rules (see also Section 2.2) of Conditional Logic (third column) and a list of known logics from the literature,
indicating which axioms and rules are (in)validated by the logic. For the logics CU, CE and
V, see Section 2.2. The logics CT4 and C4 were also already discussed in 2.4 as the conditional logics of normality defined by Boutilier and Lamarre respectively. The logic NP was
introduced by James P. Delgrande in 1987 [Del87] and was one of the first attempts to define a
normality conditional. The quite weak logics Λ and C are due to Victor Jauregui [Jau08] and
James P. Delgrande [Del03] respectively. Finally, the last column contains the logics defined
in this paper.
52
RM
OR
Loop
AND
CM
CUT
RW
LLE
REF
∼C
C
∼B
B
À→B
C
∼A
∼C
A
∼C
∼¬B
A∧B
∼C
B ∼C
A∨B ∼C
∼A1 ... Ak ∼A0
A0 ∼Ak
∼B
A ∼C
A ∼B∧C
∼C
A∧B
∼C
A
∼B
A ∼C
A
A0
A
A
A∧B ∼C
A ∼B
A ∼C
A
À≡B
A ∼A
Axioms and Rules
KLM
3
3
3
3
3
3
3
3
3
3
3
3
3
CL
C
3
3
3
3
3
3
3
3
P
Systems
3
3
3
3
3
3
3
3
3
R
A⇒A
3
3
3
(¬A⇒A) → (B⇒A)
(A⇒B) ∧ (C⇒B) → (A∧C ⇒ B)
(A ∧ B) → (A⇒B)
(A⇒B) ∨ (A⇒¬B)
(A∨B ⇒ C) → (A⇒C) ∧ (B⇒C)
(A⇒B) ∧ (B⇒C) → (A⇒C)
MOD
CA
CS
CEM
SDA
Transitivity
⇒ ⇒ ⇒
A ∧ (A⇒B) ⇒ B
(A⇒B) ⇒ (A∧C ⇒ B)
(A⇒B) → (A∧C ⇒ B)
3
3
3
3
3
3
Table 1: Logics Ω, ω, mi vs KLM systems and known Conditional Logics.
Modus Ponens
Weak
Monotonicity
Weak
Monotonicity
Transitivity
(A⇒B) ∧ (B⇒C) ⇒ (A⇒C)
(A⇒B) → (A→B)
MP
Weak
(A ⇒ B∧C) → (A⇒B) ∧ (A⇒C)
CM
(A⇒B) ∧ (B⇒A) → ((A⇒C) ≡ (B⇒C))
3
CSO
3
3
A → B
A ⇒ B
RCE
3
3
3
A ≡ B
(C⇒A) ≡ (C⇒B)
3
3
3
3
3
3
3
3
RCEC
3
3
3
3
3
3
3
3
3
V
3
3
3
3
3
3
3
CE
(A⇒B) ∧ ¬(A⇒¬C) → (A∧C ⇒ B)
(A⇒C) ∧ (B⇒C) → (A∨B ⇒ C)
(A0 ⇒A1 ∧ . . . ∧ Ak ⇒A0 ) → (A0 ⇒Ak )
(A⇒B) ∧ (A⇒C) → (A ⇒ B∧C)
(A⇒B) ∧ (A⇒C) → (A∧B ⇒ C)
(A∧B ⇒ C) ∧ (A⇒B) → (A⇒C)
(A1 ∧ ... ∧ An ) → B
(C⇒A1 ∧ ... ∧ C⇒An ) → (C⇒B)
A ≡ B
(A⇒C) ≡ (B⇒C)
CU
CV
OR
Loop
CC
AC
CUT
RCK
RCEA
ID
Axioms and Rules
3
3
3
3
3
3
3
3
3
3
CT4
CONDITIONAL LOGICS
3
3
3
3
3
3
3
3
3
C4
3
3
3
Λ
3
3
C
Systems
3
3
3
3
3
3
3
3
NP
3
3
3
3
3
3
3
3
3
3
3
3
3
3
⇒
Ω
3
3
3
3
3
3
3
3
3
3
3
3
⇒
ω
3
3
3
⇒
m1
3
3
⇒
m2
3
3
3
3
3
⇒
m3
4
Conclusions
In this paper, we have worked on a majority-based account of normality conditionals, based
on the intuition that a cofinite subset of ω is obviously much larger than its complement. The
attempt of defining conditionals modally over the frame (ω, <) has the obvious advantage that
its modal axiomatization directly leads to a (for instance, tableaux-based) decision procedure,
through an obvious translation. The other direction of employing Scott-Montague type semantics with neighborhoods of cofinite subsets, demonstrates the flexibility of the approach,
as even weak logics can be defined by tuning the truth definitions.
The expected difficulty of obtaining a complete axiomatization, is partly due to the fact
that conditional logic lacks the sophisticated model-theoretic machinery of modal logics that
allows to prove the completeness result for the logic of (ω, <) (p-morphisms, bulldozing, cluster
analysis of transitive frames, etc.). The experimentation with cofinite sets as the guiding
principle behind ‘overwhelming majority’ is however very instructive, as it allows to delineate
the core rules of such an approach.
It is interesting to check, as a question that readily emerges form this work, the nonmonotonic consequence relations that emerge from these conditionals. Another direction for further
research is to place them exactly in the universe of conditional logics (e.g [Nut80]), investigating how are they related to the semantics of D. Lewis [Lew73]. The most challenging question
however, is to relate these majority-default conditionals to the probabilistic conditionals introduced by E. Adams [Ada75] and further developed by J. Pearl and H. Geffner [Pea88, GP92].
Acknowledgements. This paper is a revised and extended version of the ECSQARU 2015
paper, under the same title [KR15b]. We wish to thanks the two anonymous ECSQARU
referees for their constructive comments. Also, we wish to thank James Delgrande, Johannes
Marti and Riccardo Pinossio for going through earlier versions of this work and providing
comments on its structure and content, along with suggestions for linking these logics with
other, more ‘mainstream’ approaches to Conditionals.
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