Algebraic Semantics for Propositional Awareness Logics
AAlgebraic Semantics for PropositionalAwareness Logics.
Evan Piermont ∗ October 17, 2019
Abstract
This paper puts forth a class of algebraic structures, relativized Booleanalgebras (RBAs), that provide semantics for propositional logic in whichtruth/validity is only defined relative to a local domain. In particular, thejoin of an event and its complement need not be the top element. Nonethe-less, behavior is locally governed by the laws of propositional logic. By fur-ther endowing these structures with operators—akin to the theory of modalAlgebras—RBAs serve as models of modal logics in which truth is relative. Inparticular, modal RBAs provide semantics for various well known awarenesslogics.
Key words: Relativized Boolean Algebras; Awareness Frames; Awareness Log-ics.
In many logical settings, it is desirable that truth and validity are not defined glob-ally, but relative to some local domain. For example, one may not want to discuss theproperties of objects when they do not exist, or the necessity, knowledge, or obligationof statements when they are not defined. However, we may still want the logic tobehave in a classical manor when examined locally, that is, when fixing the domain.This relative definition of truth has become commonplace in the epistemic formaliza-tion of (un)awareness, where an agent’s reasoning is restricted by her awareness but isotherwise rational.This paper puts forth a class of algebraic structures, relativized Boolean algebras (RBAs), that provide semantics for propositional logic in which truth is only definedrelative to a local domain, but within a given domain behavior is classical. By furtherendowing these structures with operators—akin to the theory of modal algebras or ∗ Royal Holloway, University of London, Department of Economics, [email protected] a r X i v : . [ c s . L O ] O c t R Y R X R R X B ∨ Y B X B Y B Z B B X B ∨ Z B Y B ∨ Z B B Figure 1: The RBA from Example 1. The arrows indicate the partial ordering ≥ . Theblue elements compose B , and the red elements, R .Boolean algebras with operators—RBAs serve as models of modal logics in which truthis relative. In particular, RBAs with operators provide semantics for various well knownawareness logics.Like a Boolean algebra and RBA is a set endowed with meet, join, and negationoperations, and bottom and top elements: RB = (cid:104) RB, ∧ , ∨ , ¬ , , (cid:105) . These operationssatisfy the axioms of Boolean algebras except X ∨ ¬ X , which we denote by 1 X , neednot be the top element, and 0 need not be the identify for ∨ . The elements of RBAscan be ordered via the usual condition Y ≥ X iff X ∧ Y = X .In place of the Boolean axioms are the weakened versions X ∨ X and X ∧ π ( X ) = { Z | Z = 1 X } is itself aBoolean algebra. Hence, if we think of π ( X ) as the domain on which the truth of X is defined, then within a domain, truth behaves classically. Additionally, RBAs satisfyone additional property: 1 Y ≥ X implies ¬ ( Y ∧ X ) = ¬ Y ∧ X . This final propertyensures that if Y ≥ X then Z (cid:55)→ Z ∧ X is a Boolean homomorphism from π ( Y ) to π ( X ). Hence, RBAs are naturally equipped with an ordering on domains and a senseof projection between them. Example . Let RB consist of the union of the elements of Boolean Algebras, B (for2lue) and R (for red), generated by the sets { X B , Y B , Z B } and { X R , Y R } , respectively.Moreover, define the Boolean homomorphism π R : B → R defined by X B (cid:55)→ X R , Y B (cid:55)→ Y R and Z B (cid:55)→ R . The operations on RB , when restricted to either Booleanalgebra, coincide with the Boolean operations thereon. For W B ∈ B and W R ∈ R , set W B ∧ W R = π R ( W B ) ∧ W R , and W B ∨ W R = π R ( W B ) ∨ W R . The top element is 1 B and the bottom is 0 R . This algebra is visualized by Figure 1. Consider the proposition p representing “cryptographic protocol x is insecure” and q representing “there is a quantum algorithm breaking protocol x .” Associate p to theevent X B and q to the event X R . Then RB models the situation in which p is alwayseither true or false (since ¬ X B = Y B ∨ Z B so that X B ∨ ¬ X B = 1) but q is true orfalse only on the local domain where quantum computers exist (since ¬ X R = Y R sothat X B ∨ ¬ X B = 1 R (cid:54) = 1).Then, although q → p is defined only when both q and p are (hence on R ), it isstill valid in the particular sense that it is true whenever it is defined.Just as powersets serve as concrete examples of Boolean algebras, given a set W ,we can defined a concrete RBA over { ( A, B ) | B ⊆ W, A ⊆ B } . and with operations defined by neg . ¬ ( A, B ) = ( B \ A, B ) meet . ( A, B ) ∧ ( A (cid:48) , B (cid:48) ) = ( A ∩ A (cid:48) , B ∩ B (cid:48) ) join . ( A, B ) ∨ ( A (cid:48) , B (cid:48) ) = (( A ∪ A (cid:48) ) ∩ ( B ∩ B (cid:48) ) , B ∩ B (cid:48) )Theorem 2 is a Stone-like representation theorem, showing that every RBA can beembedded into a concrete RBA. This inclusion, for the RBA considered in Example 1is shown in Figure 2.As hinted at in Example 1, RBAs serve as models of propositional logic in whichtruth and validity are relative by considering a homomorphism, h : L → RB , betweena propositional language, L , and an RBA (i.e., a map such that h ( ¬ ϕ ) = ¬ h ( ϕ ), h ( ϕ ∧ ψ ) = h ( ϕ ) ∧ h ( ψ ), etc.) In concrete RBAs, the association between the formula ϕ and the event ( A, B ) is intended to be thought of as specifying that ϕ is defined at B Although every RBA can be constructed as the disjoint union of Boolean algebras and homomor-phisms between them, as above, these Boolean algebras do not need to be ordered (in the sense thatthe homomorphisms are surjective) as in Example 1. A . Hence the complement of ( A, B ) is not ( A c , B c ) but rather ( B \ A, B )—the event where ϕ is defined but not true. A similar interpretation holds for the meetsand joins. ϕ is valid in RB if h ( ϕ ) = ( B, B ); if ϕ is true wherever it is defined. { x, y }{ x } { y } { z }∅{ x, z } { y, z }{ x, y, z } Figure 2: The RBA from Example 1 as embedded into the CRBA generated by W = { x, y, z } . The blue arrows are the elements of B and the red the elements of R . There are two interrelated methods of capturing awareness within a formal episte-mology. First are the models that capture awareness semantically, where knowledge andawareness is understood in terms of the subsets of a set called a state-space [2; 8; 11].Second are models that capture awareness syntacticly, where knowledge and awarenessare understood in terms of statements about the world [3; 12; 1; 5]. In state-space models, knowledge and awareness are represented by operators, f K and f A that map events (subsets of the state-space) to events. The event that an agentknows E is f K ( E ); and that she is aware of event E is f A ( E ). Dekel et al. [2] showedthat under mild and intuitive conditions on these operators, the only possibility wasbeing aware of everything or nothing.To circumvent this impossibility result, Heifetz, Meier, and Schipper [8] (HMS) There have also been several papers examining the connection/equivalence between extant modelsof the two approaches [7; 9]. project onto the lower spaces, in the sense that they are strictly more expressive. Then,roughly, an agent in state ω is aware of those events which are in state-spaces lowerin the ordering than the space containing ω . By considering multiple state-spaces,the definition of truth becomes inherently relative: there are states in which neither anevent nor its complement obtain. Nonetheless, when restricted to events in a particularstate-space, behavior is classical.Syntactic models of awareness, conversely, are necessarily contingent on a logicallanguage, L , with two modalities A and K , respectively. The truth of formulas is thenmodeled via Kripke frames/models where at each possible world, ω ∈ W , the DM isaware of a subset of the underlying logical language, A ( ω ) ⊆ L , and considers a subsetof the worlds R( ω ) ⊆ W , possible. Often each state ω in endowed with only a subsetof the full language, L ( ω ), and ϕ ∈ L is modeled as true of false only at those stateswhere ϕ ∈ L ( ω ) [13; 6; 4]. Again, truth is relatively defined: the worlds where ¬ ϕ istrue is the relative complement of those worlds where ϕ is true—relative to the worldswhere it is defined. Validity is likewise relative; ϕ is considered valid if it is true in allstates where it is defined .To accommodate reasoning about knowledge (and awareness), we can enrich anRBA, RB , with an operator, a function f K : RB → RB . The interpretation is asin the state-space models, or indeed in the more general theory of modal algebras: f K ( X ) is the element representing knowledge of the element X . We require that f K respects meets, maps the top element to itself, and that f K ( X ) ∈ π ( X ). The firsttwo conditions represent the standard properties of modalities in normal modal logics:the K axiom and necessitation, respectively. The final requirement constitutes thatknowledge of an element must be defined in the same domain as the element itself. From f K we can define f A , representing awareness, as f A ( X ) = f K (1 X ). The definition of f A , in addition to ensuring that awareness is domain specific, also embodies a weakenedform of necessitation: the agent knows all (and only) tautologies she is aware of. Example . Take the RBA from Example 1 and consider the map, f K : RB → RB as given by f K ( X R ) = f K ( Y R ) = 0 R , f K ( X B ) = f K ( Y B ) = 0 B , f K ( X B ∨ Z B ) = f K ( Y B ∨ Z B ) = Z B and which coincides with the identity map everywhere else. Thisis visualized by the left side of Figure 3. Then f A is simply the map W (cid:55)→ W .Going back to our propositions p and q , from example 1, f K represents the stateof affairs in which an agent who is aware of quantum computers (i.e., is aware of q ) isnecessarily uncertain about the security of the protocol (i.e., does not know p ). Thisis because the two elements resolving the truth of p and representing awareness of q ,5 R Y R X R R X B ∨ Y B X B Y B Z B B X B ∨ Z B Y B ∨ Z B B ω x ω y ω z p, q ¬ p, ¬ q ¬ p Figure 3: On the left, the RBA from Example 1 endowed with the operator f K fromExample 2, as represented by the arrows. The right side shows a Kripke frame wherethe awareness sets are the languages generated by the propositions modeled and theaccessibility relation, R , is partitional and given by the lines above the worlds. The(Boolean) algebra generated by the red worlds is R , and by the blue, B . The associationof ϕ (cid:55)→ { ω | ω | = ϕ } , produces the same model as in Example 1, and f K , from Example2, then corresponds to X (cid:55)→ { ω | R ( ω ) ⊆ X } .namely, X R and Y R , are known only at the bottom element. Conversely, an agent whois unaware of q may be certain of ¬ p ; this is represented by Z B .Propositions 6 and 7 show that modal RBAs are equivalent to awareness frames(Kripke semantics for awareness logics) in exactly the same manner that modal algebrasand Kripke frames are equivalent. For every modal RBA there is an awareness framethat models the same theories and that constructed out of its ultrafilters. Conversely,for every awareness frame there is a modal RBA constructed from the powerset of itsworlds and that models the same theories. For example, the concrete RBA that embedsthe RBA from Example 1, itself visualized in Figure 2, models the same theories as theKripke frame shown on the right side of Figure 3.Stringing these constructions together, we obtain a version of the celebrated repre-sentation theorem of Jonnson and Tarski [10], that every modal RBA can be isomor-phically embedded into the (modal) concrete RBA constructed from its correspondingultrafilter frame. Along the way this shows that the (propositionally generated) aware-ness logics considered by Fagin and Halpern [3] are complete and sound with respectto the class of modal RBAs. 6 Relativized Boolean Algebras
Call A = (cid:104) A, ∧ , ∨ , ¬ , , (cid:105) an algebra of Boolean similarity type when A is a set,0 , ∈ A , ∧ and ∨ are binary operations taking A × A → A , referred to as the meet and join , receptively, and ¬ is a unary operation taking A → A referred to as the complement . A homomorphism h : A → A (cid:48) , is a function h : A → A (cid:48) that maps h (1) = 1 (cid:48) and that respects the operations (i.e., h ( X ∧ Y ) = h ( X ) ∧ h ( Y ), etc).If A is an algebra whose elements are partially ordered by ≥ , then a filter , u , onthe algebra A is a subset of A such that (i) 1 ∈ u , (ii) u is ≥ -upwards closed and (iii)if X, Y ∈ u then X ∧ Y ∈ u . A filter is called proper if u is a proper subset of A and strongly proper it does not contain X ∧ ¬ X for any X ∈ A .An ultrafilter is a filter that (v) is strongly proper and there is no strongly properfilter, v on A such that u is a proper subset of v . Let F ( A ) and U ( A ) denote the setof filters and ultrafilters on A . Of special importance is the class of Boolean algebras (whose elements are gener-ically referred to as B ) that satisfy the axioms of Boolean algebras (see for example[14]), written here for connivence: ba1 . ∧ and ∨ are associative, communicative, and distributive. ba2 . X ∨ ¬ X = 1 ba3 . X ∧ ¬ X = 0. ba4 . X ∨ X ∧ X .Let BA denote the class of Boolean Algebras. The operations induce a partial orderingon B via Y ≥ X iff X ∨ Y = Y iff X ∧ Y = X . It is will known that condition (v)in the definition of an ultrafilter is, for Boolean algebras, equivalent to: for all X ∈ B either X ∈ u or ¬ X ∈ u , but not both. An algebra of Boolean similarity type, RB = (cid:104) RB, ∧ , ∨ , ¬ , , (cid:105) , is a relativizedBoolean algebra if it satisfies the laws below. To expedite their description, set thefollowing notation 1 X ≡ X ∨ ¬ X , 0 X ≡ X ∧ ¬ X , and Y ≥ X iff X ∧ Y = X . Let π ( RB ) = { X | X ∈ RB } . For any X ∈ RB let π ( X ) = { Z ∈ RB | Z = 1 X } . In Boolean algebras or other structures where X ∧¬ X = 0 for all X , there is no distinction betweenproper and strongly proper filters. b1 . ∧ and ∨ are associative, communicative, and distributive and with ¬ satisfyDeMorgan’s laws. rb2 . X ∧ X = X ∨ X = X ∧ ¬¬ X = X . rb3 . X ∨ X . rb4 . X ∧ rb5 . 1 Y ≥ X implies ¬ ( Y ∧ X ) = ¬ Y ∧ X .Let RBA denote the class of relativized Boolean algebras. It is immediate from( rb2 ) and ( rb4 ) that 1 ≥ X ≥ X . Notice, however, that unlike in BooleanAlgebras, X ∧ Y = X need not be equivalent to X ∨ Y = Y . Remark . If B ∈ BA then B ∈ RBA .For a Boolean algebra, X ∨ ¬ X = 1 irrespective of X , so the join is global , in thesense that the join of an element and its complement is greater than any other element.Likewise X ∧ ¬ X = 0 for all X . As the name betrays, for a relativized Boolean algebra,the meet and join operations are relative , in the sense that their identities, i.e., 1 X and0 X , depend on the elements on which they are acting. Theorem 1, below, establishesthat the operations of a RBA will obey the laws of Boolean algebras locally, within theset of elements which have the same operational identities. That is: π ( X ) forms aBoolean algebra. Moreover, the set these local Boolean structures themselves form asemi-lattice on which the projections maps are homomorphisms. Lemma 1.
The following are true for all RB ∈ RBA .(i) ≥ is a weak order.(ii) If X ≥ Y and X (cid:48) ≥ Y then X ∧ X (cid:48) ≥ Y and X ∨ X (cid:48) ≥ Y .(iii) If X ≥ Y then X ≥ Y and X ∧ Y = 0 Y .(iv) If X ≥ Y then X ∧ Y ∈ π ( Y ) .(v) X ∧ Y = 1 X ∧ Y = 1 X ∨ Y = 1 X ∨ Y . Theorem 1.
For each X , π ( X ) = { Z ∈ RB | Z = 1 X } is a Boolean Algebra (with X and X as the top and bottom elements, and the inherited operations). Moreover,the map h X : Y (cid:55)→ Y ∧ X is a homomorphism from { Y | Y ≥ X } to π ( X ) (where theformer is an algebra with and X as the top and bottom elements, and the inheritedoperations). roof. That π ( X ) is closed under ¬ , ∧ , and ∨ is immediate. That these relationssatisfy ( ba1 ) follows from ( rb1 ). Let Y ∈ π ( X ). Then Y ∨ ¬ Y = Y ∨ X ∨ X ∨ ¬ X = 1 X . Likewise, Y ∧ ¬ Y = ¬ ( Y ∨ ¬ Y ) = ¬ ( X ∨ ¬ X ) = X ∧ ¬ X = 0 X .So ( ba2 ) and ( ba3 ) hold. Finally, consider Y ∧ X and Y ∨ X . We have Y ∧ X = Y ∧ ( Y ∨
1) = ( Y ∧ Y ) ∨ ( Y ∧
1) = Y ∨ Y = Y and also Y ∨ X = Y ∨ ( Y ∧ ¬ Y ) =( Y ∨ Y ) ∧ ( Y ∨ ¬ Y ) = Y ∧ X = Y , indicating ( ba4 ).Next, consider the map h X . Lemma 1(iv) states that the image of h X is indeed π ( X ). Now let 1 Y ≥ X and 1 Z ≥ X . We have ¬ h X ( Y ) = ¬ ( Y ∧ X ) = ¬ Y ∧ X = h X ( ¬ Y ) by ( rb5 ). We have h X ( Y ∧ Z ) = ( Y ∧ Z ) ∧ X = ( Y ∧ X ) ∧ ( Z ∧ X ) = h X ( Y ) ∧ h X ( Z ). The case for ∨ is identical. Just as the powerset of a set forms the prototypical example of a BA , RBA s can alsobe given concrete instantiations as (a subset of) a powerset. A concrete
RBA based ona set W has elements which are of the form ( A, B ) where both A and B are subsetsof W and A ⊆ B . The operations are relative versions of the usual powerset Booleanalgebra operations: for example ¬ ( A, B ) = ( B \ A, B ).In this section we will establish a version of Stone’s representation theorem for
RBA s,showing that each
RBA can be embedded into a concrete relative Boolean algebra.If W is a set, let denote (cid:104){ ( A, B ) | B ⊆ W, A ⊆ B } , ∧ , ∨ , ¬ , ( ∅ , ∅ ) , ( W, W ) (cid:105) withthe operations being defined as follows: neg . ¬ ( A, B ) = ( B \ A, B ) meet . ( A, B ) ∧ ( A (cid:48) , B (cid:48) ) = ( A ∩ A (cid:48) , B ∩ B (cid:48) ) join . ( A, B ) ∨ ( A (cid:48) , B (cid:48) ) = (( A ∪ A (cid:48) ) ∩ ( B ∩ B (cid:48) ) , B ∩ B (cid:48) )Let CRBA be the class of all such algebras. It is easy to check that isa relativized Boolean algebra, with π ( ) = { ( B, B ) | B ⊆ W } ∼ = P ( W ) and π ( A, B ) = { ( A (cid:48) , B ) | A (cid:48) ⊆ B } ∼ = P ( B ) both being Boolean algebras. Notice herethat the ordering, ≥ , is simply the product ordering on P ( W ): ( A, B ) ≥ ( A (cid:48) , B (cid:48) ) ifand only if A ⊇ A (cid:48) and B ⊇ B (cid:48) . Also notice that 1 ( A,B ) = ( B, B ) and 0 ( A,B ) = ( ∅ , B ).If u is a filter of RB , then let π ( u ) = u ∩ π ( RB ) and π ( u, X ) = u ∩ π ( X ).With these definitions, we can can define the following filters on RB , which may notbe ultrafilters themselves but whose projections are either empty or are ultrafilters. F RB = { u ∈ F ( RB ) | π ( u, X ) ∈ U ( π ( X )) , for all X ∈ π ( u ) } . If B = ∅ then we have the trivial algebra where 0 = 1, but this is no problem. F RB . Using this construction, Theorem 2 shows that every RB ∈ RBA can beembedded into the concrete
RBA based on the set F RB . This clearly mirrors thestandard results for representing a Boolean algebra in the powerset of its ultrafilters.Here, however, the filters we must work with are not ultrafilters, since π ( RB ) neednot be a Boolean algebra (for example, for a given 1 X , there may be no 1 Y such that1 X ∧ Y = 0). Lemma 2.
Let F ∈ F ( RB ) be strongly proper and let X / ∈ F . Then F can be extendedto u ∈ F RB such that π ( F ) = π ( u ) and X / ∈ u . Theorem 2.
The map h : RB → P ( F RB ) defined by X (cid:55)→ ( { u ∈ F RB | X ∈ u } , { u ∈ F RB | X ∈ u } ) is an injective homomorphism.Proof. Clearly h (1) = ( F RB , F RB ). Take h ( X ) = ( A, B ) and h ( Y ) = ( A (cid:48) , B (cid:48) ). By( rb3 ), 1 X = 1 ¬ X ; we have { u ∈ F RB | ¬ X ∈ u } = B . If u / ∈ B , then ¬ X / ∈ u , and if u ∈ B then u ∩ π ( X ) is an ultrafilter on π ( X ). Hence for all u ∈ B , either X ∈ u or ¬ X ∈ u . This indicates that h ( X ) = ( B \ A, B ) as desired.Since a filter contains X ∧ Y if and only if it contains X and it contains Y , { u ∈ F RB | X ∧ Y ∈ u } = A ∩ A (cid:48) . Further, by Lemma 1(v), 1 X ∧ Y = 1 X ∧ Y , so { u ∈ F RB | X ∧ Y ∈ u } = B ∩ B (cid:48) : h ( X ∧ Y ) = ( A ∩ A (cid:48) , B ∩ B (cid:48) ).Next, by Lemma 1(v), 1 X ∨ Y = 1 X ∧ Y , so { u ∈ F RB | X ∨ Y ∈ u } = B ∩ B (cid:48) ,as well. Now: If u / ∈ B ∩ B (cid:48) , then X ∨ Y / ∈ u , and if u ∈ B then u ∩ π ( X ∨ Y ) isan ultrafilter on π ( X ∨ Y ). It is well known that an ultrafilter on a Boolean algebracontains X ∨ Y if and only if it contains X or it contains Y . This indicates that h ( X ∨ Y ) = (( A ∪ A (cid:48) ) ∩ ( B ∩ B (cid:48) ) , B ∩ B (cid:48) ), as desired.Finally to see that h is injective, assume X (cid:54) = Y . If 1 X (cid:54) = 1 Y then assume withoutloss of generality that 1 Y (cid:54)≥ X . So { Z | Z ≥ X , Z ∈ RB } is a strongly proper filteron RB and can, by Lemma 2, be extended to an element of F RB that does not include1 Y . Thus B (cid:54) = B (cid:48) .If 1 X = 1 Y , and (without loss if generality Y (cid:54)≥ X ), then { Z | Z ≥ X } is a stronglyproper filter on RB not containing Y . By Lemma 2 again, we can extend to an elementof F RB that does not include Y . Thus A (cid:54) = A (cid:48) .10 Models of Propositional Logic
For P , a set of propositional variables, let L ( P ) be the language defined by thegrammar ϕ ::= p | | ¬ ϕ | ( ϕ ∧ ϕ )where p ∈ P . We employ the standard logical abbreviations: 0 ≡ ¬
1, ( ϕ ∨ ψ ) ≡¬ ( ¬ ϕ ∧ ¬ ψ ) and ( ϕ → ψ ) ≡ ( ¬ ϕ ∨ ψ ). For ϕ ∈ L ( P ) let P ( ϕ ) collect those propositionalvariables which are subformula of ϕ .In an abuse of notation, we let L ( P ) also denote the algebra of Boolean similaritytype in which the base set is L ( P ) itself and the meet, join, and complement operationsand the top and bottom elements being denoted by their grammatical counterparts.This is the free algebra of Boolean similarity type generated by P .Say that ϕ ∈ L ( P ) is valid in RBA , denoted
RBA | = ϕ , if for all RB ∈ RBA and allhomomorphisms h : L ( P ) → RB we have that h ( ϕ ) = 1 X for some X ∈ B . Theorem 3. ϕ is a theorem of classical propositional logic if and only if RBA | = ϕ .Proof. For RB ∈ RBA , let h : L ( P ) → RB and let h ( ϕ ) = X for some classicalvalidity, ϕ . Note that for all subformula, ψ , of ϕ , it must be that 1 h ( ψ ) ≥ X (this isthe consequence of Lemma 1). Now, define the homomorphism h (cid:48) : L ( P ) → π ( X ) via: h (cid:48) : p (cid:55)→ ( h ( p ) ∧ X ) if 1 h ( p ) ≥ X X otherwise.By Theorem 1, it must be that for all subformula, ψ , of ϕ , that h (cid:48) ( ψ ) = ( h ( ψ ) ∧ X ) = h X ( h ( ψ )) ∈ π ( X ) . Since ϕ is a theorem of classical logic and h (cid:48) is a homomorphism to π ( X ) ∈ BA , itmust be that h (cid:48) ( ϕ ) = 1 X , and hence h ( ϕ ) = 1 X as desired.Completeness follows from Remark 1. Let L A,K ( P ) denote the extension of L ( P ) to include the modalities A and K : L A,K ( P ) is defined by ϕ ::= p | | ¬ ϕ | ( ϕ ∧ ϕ ) | Aϕ | Kϕ.
Consider the following axioms, all of which are standard, and whose merits andinterpretations are discussed in the literature cited in the introduction.11 xioms:K. ( Kϕ ∧ K ( ϕ → ψ )) → Kψ . D. ¬ K T. Kϕ → ϕ . Kϕ → KKϕ . ( ¬ Kϕ ∧ Aϕ ) → K ¬ Kϕ . AGP. Aϕ → Aψ , for all ψ ∈ L A,K ( P ( ϕ )). A0. Kϕ → Aϕ . KA. Aϕ ↔ KAϕ . Rules of Inference:MP.
From ϕ and ϕ → ψ infer ψ (modusponens). Sub.
From ϕ infer all of its substitutioninstances. Nec K . From ϕ infer Kϕ . Nec AK . From ϕ infer Aϕ → Kϕ .Let AX denote the smallest logic containing the tautologies of propositional logicand AGP ∪ K ∪ D ∪ A0 and which is closed under MP , Sub , and
Nec AK . AX is theaxiom system considered in [3] when awareness is generated by primitive propositionsand when the accessibility relation is serial. A awareness frame is a pre-ordered set ( W, ≥ ) endowed with a serial binary relation,R. We will set R( ω ) = { ω (cid:48) ∈ Ω | ω R ω (cid:48) } . Although we refer to the elements of W asworlds or states, note they will not have the standard interpretation of specifying thetruth of all formulas, but will rather model only a subset of the language.A awareness model for the language L A,K ( P ) is a awareness frame, ( W, ≥ , R) alongwith two functions, L : P → P ( W ) and V : P → P ( W ) such (i) L ( p ) is ≥ upwardsclosed, and (ii) V ( p ) ⊆ L ( p ) for all p ∈ P . Abusing notation let L A,K ( ω ) = L A,K ( L ( ω ))specify the language at world ω . It is the content of (i) that if ω ≥ ω (cid:48) then L A,K ( ω ) ⊇L A,K ( ω (cid:48) ).An awareness model M = ( W, ≥ , R , L, V ) defines, at every ω ∈ W the truth of allformula in L A,K ( ω ). Truth is defined recursively via the operator | = as • (cid:104) M, ω (cid:105) | = p iff ω ∈ V ( p ), Our inclusion of D will, as usual, specify those models where the accessibility relation is serial.There is no intrinsic problem considering a weaker logic without D (and therefore without any restric-tion on the accessibility relation), but to obtain a complete and sound axiomatization, we must replaceit with a novel axiom: K → Aϕ . (cid:104) M, ω (cid:105) | = ¬ ϕ iff (cid:104) M, ω (cid:105) (cid:54)| = ϕ , • (cid:104) M, ω (cid:105) | = ϕ ∧ ψ iff (cid:104) M, ω (cid:105) | = ϕ and (cid:104) M, ω (cid:105) | = ψ , • (cid:104) M, ω (cid:105) | = Aϕ iff for all ω (cid:48) ∈ R( ω ), ϕ ∈ L A,K ( ω (cid:48) ), • (cid:104) M, ω (cid:105) | = Kϕ iff for all ω (cid:48) ∈ R( ω ), (cid:104) M, ω (cid:48) (cid:105) | = ϕ .For a model M , let V ( ϕ ) = { ω ∈ W | (cid:104) M, ω (cid:105) | = ϕ } collect the worlds in which ϕ holds, and L ( ϕ ) = { ω ∈ W | ϕ ∈ L A,K ( ω ) } the worlds where ϕ is defined. The reuse of V and L is desired, as V and L are extensions of the functions in the definition of M .Call ϕ valid in M if it is true everywhere it is defined: if V ( ϕ ) = L ( ϕ ). Call ϕ validin the class of awareness models, denoted AM | = ϕ , if it is valid in all M . Theorem 4. ϕ is a theorem of AX iff AM | = ϕ . The proof relies on a slight variant of the canonical model argument and thusrelegated to the appendix.The roles of T , , and are as in standard Kripke frames, axiomatizing theclass of frames wherein R is reflexive, transitive, and Euclidean, respectively. KA corresponds to the frames such that ω (cid:48) ∈ R( ω ) and ω (cid:48)(cid:48) ∈ R( ω (cid:48) ) implies that ω (cid:48)(cid:48) ≥ ω (cid:48)(cid:48)(cid:48) for some ω (cid:48)(cid:48)(cid:48) ∈ R( ω ). In particular, it is true in the class of transitive frames (i.e.,under AX ∪ ).Notice that Necessitation (from ϕ infer Kϕ ) is not sound in AM . Indeed, considera model M and some valid ϕ which is not in L A,K ( ω (cid:48) ) for some ω (cid:48) . Then if ω (cid:48) ∈ R( ω )with ϕ ∈ L A,K ( ω ) we have that (cid:104) M, ω (cid:105) | = ϕ (since ϕ was valid and defined at ω , butnot (cid:104) M, ω (cid:105) (cid:54)| = Kϕ (since ϕ is not defined at ω (cid:48) , hence (cid:104) M, ω (cid:48) (cid:105) (cid:54)| = ϕ ). Necessitation issound and (along with the other axioms) complete within the class of frames whereR( ω ) ⊆ { ω (cid:48) ∈ W | ω (cid:48) ≥ ω } . However, this class of models is remarkably boring asevidenced by the validity of ϕ → Aϕ . Let RB be a relativized Boolean algebra. Then ( RB , f K ) is a modal relativizedBoolean algebra, or MRBA, if f K : RB → RB such that the following conditions hold f1 . 1 f K ( X ) = 1 X , 13 . f K (1) = 1 f3 . f K ( X ∧ Y ) = f K ( X ) ∧ f K ( Y ). fD . f K (0 X ) = 0 X .Let MRBA denote the class of modal RBAs. From f K we can define the additionalmap f A : RB → RB via f A : X (cid:55)→ f K (1 X ).These conditions reflect the properties of knowledge and awareness in relation tothe elements where they are defined: ( f1 ) states that knowledge (and awareness) of anelement is defined exactly when the event itself is defined; ( f2 ) reflects our weakenedform of necessitation: something which is tautological and always defined is alwaysknown; ( f3 ) encodes the distributive property of knowledge. As always, we have that( f3 ) implies that f K is monotone. Finally, ( fD ) instantiates our restriction to non-trivial knowledge as D did in frame semantics.If ( RB , f K ) is a MRBA and h : L ( P ) → RB is a homomorphism we can extend h to h + : L A,K ( P ) → RB via (inductively) h + ( Aϕ ) = f A ( h + ( ϕ )) and h + ( Kϕ ) = f K ( h + ( ϕ )). Then say that ϕ ∈ L A,K ( P ) is valid in MRBA , denoted
MRBA | = ϕ , iffor all ( RB , f K ) ∈ MRBA and all homomorphisms from h : L ( P ) → RB we have that h + ( ϕ ) = 1 X for some X ∈ RB . Theorem 5.
MRBA | = ϕ iff AM | = ϕ . The proof of Theorem 5 is the conjunction of the following two propositions. Propo-sition 6 constructs, for each awareness model, a corresponding MRBA (and homomor-phism) such that for each formula, ( V ( ϕ ) , L ( ϕ )) = h + ( ϕ ). Then, in converse fash-ion, Proposition 7 constructs an awareness model, for each ( RB , f K , h ), such that h + ( ϕ ) = ( V ( ϕ ) , L ( ϕ )). If F = ( W, ≥ , R) is an awareness frame, define the concrete MRBA, ( , f K,R )and f K,R : (
A, B ) (cid:55)→ ( { ω | R K ( ω ) ⊆ A ) ∩ B, B )Verifying that f K,R here defined satisfies ( f1 )-( f3 ) is straight forward. ( fD ) followsfrom the assumption that R is serial. Proposition 6.
Let F = ( W, ≥ , R) and M = ( F, L, V ) be an awareness model andtake h M : L ( P ) → to be the homomorphism defined by h ( p ) = ( V ( p ) , L ( p )) then h M + ( ϕ ) = ( V ( ϕ ) , L ( ϕ )) . roof. This is done by induction of the structure of formula. We show the inductivestep for Kϕ : h M + ( Kϕ ) = f K,R ( h M + ( ϕ ))= f K,R ( V ( ϕ ) , L ( ϕ ))= ( { ω | R ( ω ) ⊂ V ( ϕ ) } ∩ L ( ϕ ) , L ( Kϕ ))= ( { ω | ω (cid:48) ∈ R ( ω ) = ⇒ (cid:104) M, ω (cid:48) (cid:105) | = ϕ, ϕ ∈ L A,K ( ω ) } , L ( Kϕ ))= ( V ( Kϕ ) , L ( Kϕ )) . The second equality is our inductive hypothesis. I promise, the other steps are evensimpler.Proposition 6 proves that
MRBA | = ϕ implies AM | = ϕ . To see this, notice that if MRBA | = ϕ then for every ( RB , f K ), and for every homomorphism h : L ( P ) → RB , wehave h + ( ϕ ) = 1 X for some X ∈ RB . In particular, for each model ( F, L, V ) this is truefor and h M : p (cid:55)→ ( V ( p ) , L ( p )). Thus, Proposition 6 requires that V ( ϕ ) = L ( ϕ ).Since this holds for all models, we have that ϕ is valid in AM . In dual fashion, the next Proposition shows that AM | = ϕ implies MRBA | = ϕ byconstructing an awareness model for each ( RB , f K , h ) that yield the same validities.As usual, the worlds will be sets of ultrafilter like objects. By the reasoning outlinedin Section 2, we consider the filters F RB = { u ∈ F ( RB ) | π ( u, X ) ∈ U ( π ( X )) , for all X ∈ π ( u ) } . Then, if ( RB , f K ) ∈ MRBA , define the ultrafilter frame as ( P ( F RB ) , ≥ RB , R RB )where u ≥ RB v iff π ( u ) ⊇ π ( v ) and u R RB v iff f K ( X ) ∈ u implies X ∈ v . Proposition 7.
Let h : L ( P ) → RB be a homomorphism and h + its extension to L A,K ( P ) → . Let M = ( P ( F RB ) , ≥ RB , R RB , L h , V h ) where L h ( p ) = { u ∈ F RB | h ( p ) ∈ u } and V h ( p ) = { u ∈ F RB | h ( p ) ∈ u } . Then for all ϕ ∈ L A,K ( P ) , L h ( ϕ ) = { u ∈ F RB | h + ( ϕ ) ∈ u } , and V h ( ϕ ) = { u ∈ F RB | h + ( ϕ ) ∈ u } . Proof.
As always, the proof is by induction on the structure of formula. This is straight-forward with the help of the following lemma:
Lemma 3.
Let u ∈ F RB . Then R( u ) ⊆ { v ∈ F RB | X ∈ v } iff f K ( X ) ∈ u . roof. The if direction is immediate given the definition of R. We show the only if viaits contrapositive: if f K ( X ) / ∈ u then there exists some v ∈ F RB such that f K ( X (cid:48) ) ∈ u implies X (cid:48) ∈ v and X / ∈ v .Define v − = { X (cid:48) ∈ RB | f K ( X (cid:48) ) ∈ u } . By assumption X / ∈ v − . By ( f2 ), 1 ∈ v − ,by ( f3 ) v − is an upset and is closed under meets, hence v − ∈ F ( RB ). By ( fD )0 Y / ∈ v − for any Y ∈ RB ; so v − is strongly proper. Lemma 2 allows us to extend v − to v ∈ F RB such that π ( v ) = π ( v − ) and with X / ∈ v . (cid:70) We first show that for all ϕ , L h ( ϕ ) = { v ∈ F RB | h + ( ϕ ) ∈ v } . This is by inductionof the complexity of ϕ . The base case is the definition of L . We will show the cases for ∧ and K (negation is trivial and the argument for A is exactly the argument for K ).Let u ∈ L h ( ϕ ∧ ψ ). So, ϕ ∧ ψ ∈ L ( u ) iff ϕ ∈ L ( u ) and ψ ∈ L ( u ). By theinductive hypothesis, this is iff 1 h + ( ϕ ) ∈ u and 1 h + ( ψ ) ∈ u . Since u ∈ F RB , this is iff1 h + ( ϕ ) ∧ h + ( ψ ) = 1 h + ( ϕ ) ∧ h + ( ψ ) = 1 h + ( ϕ ∧ ψ ) ∈ u .Let u ∈ L h ( Kϕ ). So, Kϕ ∈ L ( u ) iff ϕ ∈ L ( u ). By the inductive hypothesis, this isiff 1 h + ( ψ ) = 1 f K ( h + ( ψ )) = 1 h + ( Kψ ) ∈ u . Where the second equality is via ( f1 ).Next, we show that for all ϕ , V h ( ϕ ) = { v ∈ F RB | h + ( ϕ ) ∈ v } . Again, thisis by induction of the complexity of ϕ , and again, we will just show the interestingsteps: Let u ∈ V h ( Kϕ ). So, (cid:104) M, u (cid:105) | = Kϕ ; iff for all v ∈ R( u ), (cid:104) M, v (cid:105) | = ϕ . By theinductive hypothesis, this is iff R( u ) ⊆ { v ∈ F RB | h + ( ϕ ) ∈ v } , which, by Lemma 3 isiff f K ( h + ( ϕ )) = h + ( Kϕ ) ∈ u .Let u ∈ V h ( Aϕ ). So, (cid:104) M, u (cid:105) | = Aϕ ; iff for all v ∈ R( u ), v ∈ L h ( ϕ ). By the previouspart of the proof (concerning L h ), this is iff R( u ) ⊆ { v ∈ F RB | h + ( ϕ ) ∈ v } , which, byLemma 3 is iff f K (1 h + ( ϕ ) ) = f A ( h + ( ϕ )) = h + ( Aϕ ) ∈ u .To see that Proposition 7 shows that AM | = ϕ implies MRBA | = ϕ (and thuscompletes the proof of Theorem 5), let ϕ be valid in AM and pick your favorite MRBA ,( RB , f K ) and homomorphism h : L ( P ) → RB . Then in particular, ϕ is valid in M = ( P ( F RB ) , ≥ RB , R RB , L h , V h ), meaning V h ( ϕ ) = L h ( ϕ ). Proposition 7 thenindicates that { u ∈ F RB | h + ( ϕ ) ∈ u } = { u ∈ F RB | h + ( ϕ ) ∈ u } , which can only be true if h + ( ϕ ) = 1 h + ( ϕ ) , indicating validity in ( RB , h ). A Example 3
Let RB consist of the union of the elements of Boolean Algebras, B (for blue) and R (for red), generated by the sets { Y B , ¬ Y B } and { X R , ¬ X R } , respectively. Moreover,16 R ¬ X R X R R Y B B ¬ Y B B Figure 4: The RBA from Example 3. The arrows indicate the partial ordering ≥ . Theblue elements compose B , and the red elements, R .define the Boolean homomorphism π R : B → R defined by Y B (cid:55)→ R . The operationson RB , when restricted to either Boolean algebra, coincide with the Boolean operationsthereon. For W B ∈ B and W R ∈ R , set W B ∧ W R = π R ( W B ) ∧ W R , and W B ∨ W R = π R ( W B ) ∨ W R . The top element is 1 B and the bottom is 0 R . This algebra is visualizedby Figure 4.Notice that π R ( B ) (cid:54) = R but rather is the trivial 0 − B Proof of Theorem 4
Proof.
Soundness is straightforward. To show completeness, we follows the usual con-ical construction with a slight caveat. For the frame, let W c denote the set of allpairs (Γ , Q ) where Γ is a maximally consistent set of formula containing AX in thelanguage L A,K ( Q ) and Q ⊆ P . Order W c via (Γ , Q ) ≥ c (Γ (cid:48) , Q (cid:48) ) iff Q ⊇ Q (cid:48) . Con-struct the relations according to (Γ , Q )R c (Γ (cid:48) , Q (cid:48) ) iff { ϕ ∈ L A,K ( P ) | Kϕ ∈ Γ } ⊆ Γ (cid:48) and { ϕ ∈ L A,K ( P ) | Aϕ ∈ Γ } ⊆ L A,K ( Q (cid:48) ). Then to construct the canonical model, set L c ( p ) = { (Γ , Q ) ∈ Ω c | p ∈ Q } and V c ( p ) = { (Γ , Q ) ∈ Ω c | p ∈ Γ } . An induction onthe complexity of ϕ shows that, for all (Γ , Q ), ϕ ∈ Γ iff (cid:104) M c , (Γ , Q ) (cid:105) | = ϕ . The onlynontrivial steps, for Aϕ and Kϕ , are direct consequences of the following lemmas:17 emma 4. Fix (Γ , Q ) . If Aϕ / ∈ Γ then there exists a (Γ (cid:48) , Q (cid:48) ) such that (Γ , Q )R(Γ (cid:48) , Q (cid:48) ) and ϕ / ∈ Q (cid:48) .Proof. Since Γ contains
AGP , we have that { ψ | Aψ ∈ Γ } = L A,K ( Q (cid:48) ) for some Q (cid:48) ⊂ P . Since Aϕ / ∈ Γ, ϕ / ∈ L A,K ( Q (cid:48) ). Set Γ − = { ψ | Kψ ∈ Γ } . By A0 , Γ − ⊆L A,K ( Q (cid:48) ), and by D , Γ − (cid:54) = L A,K ( Q (cid:48) ). Notice also that, by Nec AK , we have that Γ − contains all tautologies in L A,K ( Q ). This allows for the standard argument that Γ − isa consistent set of formulas and can therefore be extended to a maximally consistentset, Γ (cid:48) ⊂ L A,K ( Q (cid:48) ). (Γ (cid:48) , Q (cid:48) ) is the desired world. (cid:70) Lemma 5.
Fix (Γ , Q ) . If Kϕ / ∈ Γ then there exists a (Γ (cid:48) , Q (cid:48) ) such that (Γ , Q )R(Γ (cid:48) , Q (cid:48) ) and ϕ / ∈ Γ (cid:48) .Proof. There are two cases to consider. First, if
Aϕ / ∈ Γ, then by Lemma 4, there is anassessable world, (Γ (cid:48) , Q (cid:48) ), such that ϕ / ∈ L A,K ( Q (cid:48) ), and hence clearly, ϕ / ∈ Γ (cid:48) .So assume that Aϕ ∈ Γ. Since Γ contains
AGP , we have that { ψ | Aψ ∈ Γ } = L A,K ( Q (cid:48) ) for some Q (cid:48) ⊆ P . Then consider the set Γ − = {¬ ϕ } ∪ { ψ | Kψ ∈ Γ } . SinceΓ contains A0 , we have that Γ − ⊆ L A,K ( Q (cid:48) ). As usual, Γ − can be extended to amaximally consistent set, Γ (cid:48) in L A,K ( Q (cid:48) ). Again, (Γ (cid:48) , Q (cid:48) ) is the desired world. (cid:70) C Other Proofs Omitted from the Main Text
Proof of Lemma 1.
Parts (i-iii) are immediate from definitions.(iv) ( X ∧ Y ) ∨ X ∧ (( Y ∨ ∨
1) = 1 X ∧ ( Y ∨
1) = 1 X ∧ Y = 1 Y .(v) 1 X ∧ Y = ( X ∨ ∧ ( Y ∨
1) = ( X ∧ Y ) ∨ X ∧ Y . Further 1 X ∨ Y =( X ∨ ∨ ( Y ∨
1) = ( X ∨ Y ) ∨ ¬ ( ¬ X ∧ ¬ Y ) ∨ ¬ X ∧ ¬ Y ) ∨ ¬ X ∨ ∧ ( ¬ Y ∨
1) = ( X ∨ ∧ ( Y ∨
1) = 1 X ∧ Y , where the elimination ofnegations comes from the fact that X ∨ X ∨ ¬ X = ¬ X ∨ rb3 ). Proof of Lemma 2.
We will show that if F ∈ F ( RB ) is strongly proper, then for all1 X ∈ π ( F ), ¬ X / ∈ F then F (cid:48) = { Z ∧ Y | Z ≥ X, Y ∈ F } is in F ( RB ) and is stronglyproper and π ( F ) = π ( F (cid:48) ) and ¬ X / ∈ F (cid:48) . This suffices, since we can then appeal tothe usual Zornesque arguments, to choose a maximal element of the partial order of allextensions. 18hat F (cid:48) is upwards closed, contains 1, and is closed under intersections is immediate.Thus, we need only show that F (cid:48) is strongly proper. Assume to the contrary that Z ∧ Y = 0 W for some W ∈ RB with Z ≥ X and Y ∈ F . Then { Z , Y } ∈ F and so tois 1 V = 1 Y ∧ X and Y ∧ V . Since Z ∧ Y = 0 W we have also that ( Z ∧ V ) ∧ ( Y ∧ V ) = 0 W ∧ V = 0 V . (1)Now, ( Z ∧ V ) ≥ ( X ∧ V ), the fact that π ( V ) ∈ BA and (1) requires that ¬ ( X ∧ V ) ≥ ( Y ∧ V ). But, since F was upwards closed, this requires that ¬ X ∈ F , since ¬ X ≥ ¬ X ∧ V = ¬ ( X ∧ V ) (by ( rb5 )). This contradicts our assumption.Clearly, π ( F ) ⊆ π ( F (cid:48) ), so to see the other direction, let Z ∧ Y = 1 W for some W ∈ RB with Z ≥ X and Y ∈ F . Then by Lemma 1(v), 1 W = 1 Z ∧ Y ≥ X ∧ Y and so 1 W ∈ π ( F ). That 0 W ∧ V = 0 V follows from the fact that h V given by Lemma 1 is a homomorphism. eferences [1] Oliver Board and Kim-Sau Chung. Object-based unawareness . University of Min-nesota, Department of Economics, 2007.[2] Eddie Dekel, Barton L Lipman, and Aldo Rustichini. Standard state-space modelspreclude unawareness.
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