If You're Happy, Then You Know It: The Logic of Happiness... and Sadness
IIf You’re Happy, Then You Know It: The Logic ofHappiness . . . and Sadness
Sanaz Azimipour
Berlin University of Applied Sciences, Berlin, Germany
Pavel Naumov
King’s College, Pennsylvania, USA
Abstract
The article proposes a formal semantics of happiness and sadness modalities inimperfect information setting. It shows that these modalities are not definablethrough each other and gives a sound and complete axiomatization of theirproperties.
1. Introduction
Emotions provide motivation for many aspects of human behavior. To beable to understand and predict human actions, artificial agents must be ableto identify, comprehend, and reason about human emotions. Different formalmodels of human emotions have been studied in AI literature. Doyle, Shoham,and Wellman propose a logic of relative desire [1]. Lang, Van Der Torre, andWeydert introduce utilitarian desires [2]. Meyer states logical principles aim-ing at capturing anger and fear [3]. Steunebrink, Dastani, and Meyer expandthis work to hope [4]. Adam, Herzig, and Longin propose formal definitionsof hope, fear, relief, disappointment, resentment, gloating, pride, shame, ad-miration, reproach, gratification, remorse, gratitude, and anger [5]. Lorini andSchwarzentruber define regret and elation [6].The focus of this article is on happiness and sadness. These notions havelong been studied in literature on philosophy [7, 8, 9, 10], psychology [11, 12],and economics [13, 14]. Note that happiness/sadness are vague terms that havemultiple meanings that overlap with several other terms such as joy/distressand elation/disappointment . Email addresses:
[email protected] (Sanaz Azimipour), [email protected] (PavelNaumov) For example, Merriam-Webster dictionary lists joy as synonym for happiness and happi-ness as synonym for joy. At the same time, thesaurus.com lists elation as synonym for joyand joy as synonym for elation.
Preprint submitted to Elsevier January 5, 2021 a r X i v : . [ c s . A I] J a n wo approaches to capturing happiness and sadness in formal logical systemshave been proposed. The first approach is based on Ortony, Clore, and Collins’definitions of joy and distress (the capitalization is original):. . . we shall often use the terms “joy” and “distress” as convenientshorthands for the reactions of being PLEASED ABOUT A DESIR-ABLE EVENT and DISPLEASED ABOUT AN UNDESIRABLEEVENT, respectively. [15, p.88]Adam, Herzig, and Longin formalized these definitions. An agent feels joy about ϕ if she believes that ϕ is true and she desires ϕ . An agent feels distress about ϕ if she believes ϕ is true, but she desires ϕ to be false [5]. Similarly, Loriniand Schwarzentruber define that an agent is elated/disappointed about ϕ if ϕ is desirable/undesirable to the agent, agent knows that ϕ is true, and she alsoknows that the others could have prevented ϕ from being true [6]. AlthoughAdam, Herzig, and Longin use beliefs while Lorini and Schwarzentruber useknowledge and the latter authors also add “could have prevented” part, bothdefinitions could be viewed as a variation of Ortony, Clore, and Collins’ defini-tions of joy/distress.Meyer suggested a very different approach to defining these notions. Hewrites “an agent that is happy observes that its subgoals (towards certain goals)are being achieved, and is ‘happy’ with it”. He acknowledges, however, thatthis definition might be capturing only one of the forms of what people meanby happiness [3].Note, for example, that if Pavel, the second author of this article, receivesan unexpected gift from Sanaz, the first author, then he will experience “joy”as defined by Ortony, Clore, and Collins. However, he will not be “happy” asdefined by Meyer because receiving such a gift has never been among Pavel’sgoals .In this article we adopt Ortony, Clore, and Collins’ definitions, but we useterms happiness/sadness instead of joy/distress. While the cited above workssuggest formal semantics and list formal properties of happiness and sadness,none of them gives an axiomatization of these properties. In this article wepropose such an axiomatization and prove its completeness. We also show thatnotions of happiness and sadness in our formal system are, in some sense, dualbut are not definable through each other.The rest of the article is structured as follows. First, we formally defineepistemic models with preferences that serve as the foundation of our semanticsof happiness and sadness. Then, we define formal syntax and semantics of oursystem and illustrate them with several examples. In Section 7, we show thatsadness cannot be defined through happiness. In spite of this, as we show inSection 8 there is a certain duality between the properties of the happiness andsadness. We use this duality to observe that sadness can not be defined through Ortony, Clore, and Collins give a similar example with an unexpected inheritance froman unknown relative.
2. Epistemic Models with Preferences
Throughout the article we assume a fixed countable set of agents A and acountable set of propositional variables. The semantics of our logical systemis defined in terms of epistemic models with preferences. These models extendstandard Kripke models for epistemic logic with a preference relation for eachagent in set A . Definition 1.
A tuple ( W, {∼ a } a ∈A , {≺ a } a ∈A , π ) is called an epistemic modelwith preferences if W is a set of epistemic worlds, ∼ a is an “indistinguishability” equivalence relation on set W for eachagent a ∈ A , ≺ a is a strict partial order preference relation on set W for each agent a ∈ A , π ( p ) is a subset of W for each propositional variable p . For any two sets of epistemic worlds
U, V ⊆ W , we write U ≺ a V if u ≺ a v foreach world u ∈ U and each world v ∈ V . gift not sent gift lost in mailgift received, note lost in mailgift and note received pp ss,ps,ps,p s,p s,ps,pw uvt Figure 1: Gift Scenario.
3n example of an epistemic model with preferences is depicted in Figure 1.It captures mentioned in the introduction scenario in which the first author ofthis article, Sanaz, is sending a gift to the second author, Pavel. If Pavel receivesthe gift, he will acknowledge it by sending back a thank you note. We assumethat either the gift or the note can be lost in the mail and that there is noadditional communication between the authors.This epistemic model with preferences in Figure 1 has four worlds corre-sponding to four different scenarios. In world t Sanaz did not send the gift. Inworld v , she sent the gift, but it was lost in the mail. In world u , Pavel receivedthe gift, but his note is lost in the mail. Finally, in world w Pavel received thegift and Sanaz received his note. Sanaz cannot distinguish the world v , in whichthe gift is lost, from world u , in which the note is lost. Pavel cannot distinguishthe world t , in which gift is not sent, from world v , in which the gift is lost. Healso cannot distinguish the world u , in which the note is lost, from world w , inwhich the note is received. In the diagram, the indistinguishability relations ofSanaz and Pavel are denoted by dashed lines labeled by s and p respectively.Although in general, according to Definition 1, different agents might have dif-ferent preference relations between world, in this scenario we assume that Sanazand Pavel have the same preferences. These preferences are shown in the di-agram using directed edges. For example, the directed edge from world v toworld t labeled with s, p means that they both would prefer if Sanaz does notsend the gift at all to the scenario when the gift is lost in the mail.
3. Syntax and Semantics
In this section we introduce the formal syntax and semantics of our logicalsystem. Throughout the article we assume a fixed countable set of agents A and a fixed nonempty countable set of propositional variables. The language Φof our logical system is defined by grammar: ϕ := p | ¬ ϕ | ϕ → ϕ | N ϕ | K a ϕ | H a ϕ | S a ϕ, where p is a propositional variable and a ∈ A is an agent. We read formula N ϕ as “statement ϕ is true in each world”, formula K a ϕ as “agent a knows ϕ ”,formula H a ϕ as “agent a is happy about ϕ ”, and formula S a ϕ as “agent a is sadabout ϕ ”. We assume that Boolean connectives conjunction ∧ , disjunction ∨ ,and biconditional ↔ are defined through negation ¬ and implication → in thestandard way. By N ϕ we denote formula ¬ N ¬ ϕ . We read N ϕ as “statement ϕ is true in at least one of the worlds”. Definition 2.
For any world w ∈ W of an epistemic model with preferences ( W, {∼ a } a ∈A , {≺ a } a ∈A , π ) and any formula ϕ ∈ Φ , satisfaction relation w (cid:13) ϕ is defined as follows: w (cid:13) p , if w ∈ π ( p ) , w (cid:13) ¬ ϕ , if w (cid:49) ϕ , w (cid:13) ϕ → ψ , if w (cid:49) ϕ or w (cid:13) ψ , w (cid:13) N ϕ , if u (cid:13) ϕ for each world u ∈ W , w (cid:13) K a ϕ , if u (cid:13) ϕ for each world u ∈ W such that w ∼ a u , w (cid:13) H a ϕ , if the following three conditions are satisfied: (a) u (cid:13) ϕ for each world u ∈ W such that w ∼ a u , (b) for any two worlds u, u (cid:48) ∈ W , if u (cid:49) ϕ and u (cid:48) (cid:13) ϕ , then u ≺ a u (cid:48) , (c) there is a world u ∈ W such that u (cid:49) ϕ , w (cid:13) S a ϕ , if the following three conditions are satisfied: (a) u (cid:13) ϕ for each world u ∈ W such that w ∼ a u , (b) for any two worlds u, u (cid:48) ∈ W , if u (cid:13) ϕ and u (cid:48) (cid:49) ϕ , then u ≺ a u (cid:48) , (c) there is a world u ∈ W such that u (cid:49) ϕ . Items 6 and 7 of Definition 2 capture the notions of happiness and sadnessstudied in this article. Item 6 states that to be happy about a condition ϕ ,the agent must know that ϕ is true, the agent must prefer the worlds in whichcondition ϕ is true to those where it is false, and the condition ϕ must notbe trivial. These three parts are captured by items 6(a), 6(b), and 6(c) of theabove definition. Note that we require condition ϕ to be non-trivial to excludean agent being happy about conditions that always hold in the model. Thus,for example, we believe that an agent cannot be happy that 2 + 2 = 4.Similarly, item 7 states that an agent is sad about condition ϕ if she knowsthat ϕ is true, she prefers worlds in which condition ϕ is false to those inwhich condition ϕ is true, and condition ϕ is not trivial. Note that being sad isdifferent from not being happy. In fact, later in this article we show that neitherof modalities H and S is expressible through the other.We conclude this section with a technical observation that follows from Def-inition 2. Lemma 1.
For any epistemic model with preferences ( W, {∼ a } a ∈A , {≺ a } a ∈A , π ) ,any agent a ∈ A , and any formulae ϕ, ψ ∈ Φ , if w (cid:13) ϕ iff w (cid:13) ψ for each world w ∈ W , then w (cid:13) H a ϕ iff w (cid:13) H a ψ for each world w ∈ W , w (cid:13) S a ϕ iff w (cid:13) S a ψ for each world w ∈ W . (cid:2)
4. Gift Scenario
In this section we illustrate Definition 2 using the gift scenario depicted in thediagram in Figure 1. For the benefit of the reader, we reproduce this diagramin Figure 2.
Proposition 1. z (cid:13) H p ( “Pavel received a gift from Sanaz” ) iff z ∈ { w, u } .Proof. ( ⇒ ) : Suppose z / ∈ { w, u } . Thus, z ∈ { t, v } . Hence, see Figure 2, z (cid:49) “Pavel received a gift from Sanaz” . Therefore, z (cid:49) H p (“Pavel received a gift from Sanaz”) by item 6(a) of Defini-tion 2 because z ∼ p z . 5 ift not sent gift lost in mailgift received, note lost in mailgift and note received pp ss,ps,ps,p s,p s,ps,pw uvt Figure 2: Gift Scenario (repeated from Figure 1). ( ⇐ ) : Suppose z ∈ { w, u } . To show z (cid:13) H p (“Pavel received a gift from Sanaz”),we will verify conditions (a), (b) and (c) of item 6 of Definition 2 separately: Condition a:
Consider any world z (cid:48) such that z ∼ p z (cid:48) . To verify the condition, itsuffices to show that z (cid:48) (cid:13) “Pavel received a gift from Sanaz”. Indeed, assump-tions z ∈ { w, u } and z ∼ p z (cid:48) imply that z (cid:48) ∈ { w, u } , see Figure 2. Therefore, z (cid:48) (cid:13) “Pavel received a gift from Sanaz”, again see Figure 2. Condition b:
Consider any two epistemic worlds x, y such that x (cid:49) “Pavel received a gift from Sanaz” , (1) y (cid:13) “Pavel received a gift from Sanaz” . (2)To verify the condition, it suffices to show that x ≺ p y . Indeed, assump-tions (1) and (2) implies that x ∈ { t, v } and y ∈ { w, u } , see Figure 2. Note that { t, v } ≺ p { w, u } , see Figure 2. Therefore, x ≺ p y . Condition c: t (cid:49) “Pavel received a gift from Sanaz”. (cid:2) Proposition 2. z (cid:13) H s ( “Pavel received a gift from Sanaz” ) iff z ∈ { w } .Proof. ( ⇒ ) : Suppose that z / ∈ { w } . Thus, either z ∈ { t } or z ∈ { u, v } . Weconsider these two cases separately. Case I : z ∈ { t } . Then, z (cid:49) “Pavel received a gift from Sanaz”, see Figure 2.Therefore, z (cid:13) H s (“Pavel received a gift from Sanaz”) by item 6(a) of Defini-tion 2 and because z ∼ s z . Case II : z ∈ { u, v } . Then, z ∼ s v , see Figure 2. Note that, see again Figure 2, v (cid:49) “Pavel received a gift from Sanaz”. Therefore, by item 6(a) of Definition 2, z (cid:49) H s (“Pavel received a gift from Sanaz”).6 ⇐ ) : To show that w (cid:13) H s (“Pavel received a gift from Sanaz”), we verify con-ditions (a), (b), and (c) of item 6 of Definition 2: Condition a:
Consider any world z such that w ∼ s z . It suffices to show that z (cid:13) “Pavel received a gift from Sanaz”. Indeed, assumption w ∼ s z impliesthat z = w , see Figure 2. Note that w (cid:13) “Pavel received a gift from Sanaz”,see again Figure 2. Condition b:
The proof is similar to the proof of
Condition b in Proposition 1.
Condition c: t (cid:49) “Pavel received a gift from Sanaz”. (cid:2) The next proposition shows that Sanaz is happy that Pavel is happy only ifshe gets the thank you card and, thus, she knows that he received the the gift.
Proposition 3. z (cid:13) H s H p ( “Pavel received a gift from Sanaz” ) iff z ∈ { w } .Proof. Note that x (cid:13) “Pavel received a gift from Sanaz” iff x ∈ { w, u } , seeFigure 2. Thus, by Proposition 1, for any world x ∈ W , x (cid:13) “Pavel received a gift from Sanaz”iff x (cid:13) H p (“Pavel received a gift from Sanaz”) . Hence, by Lemma 1, for any world x ∈ W , x (cid:13) H s (“Pavel received a gift from Sanaz”)iff x (cid:13) H s H p (“Pavel received a gift from Sanaz”) . Therefore, z (cid:13) H s H p (“Pavel received a gift from Sanaz”) iff z ∈ { w } and Propo-sition 2. (cid:2) Note that because Sanaz never acknowledges the thank you card, Pavel doesnot know that Sanaz is happy. Hence, he cannot be happy that she is happy.This is captured in the next proposition.
Proposition 4. z (cid:49) H p H s ( “Pavel received a gift from Sanaz” ) for each world z ∈ { w, u, v, t } .Proof. We consider the following two cases separately:
Case I : z ∈ { w, u } . Then, z ∼ p u , see Figure 2. Note that by Proposition 2, u (cid:49) H s (“Pavel received a gift from Sanaz”) . Therefore, z (cid:49) H p H s (“Pavel received a gift from Sanaz”) by item 6(a) of Defi-nition 2 and the statement z ∼ p u . Case II : z ∈ { v, t } . Then, z (cid:49) H s (“Pavel received a gift from Sanaz”) by Propo-sition 2. Therefore, z (cid:49) H p H s (“Pavel received a gift from Sanaz”) by item 6(a)of Definition 2. (cid:2) The proof of the next statement is similar to the proof of Proposition 4except that it refers to Proposition 3 instead of Proposition 2.7 roposition 5. z (cid:49) H p H s H p ( “Pavel received a gift from Sanaz” ) for each world z ∈ { w, u, v, t } . (cid:2) The next proposition states that Sanaz is sad about Pavel not receiving thegift only if she does not send it. Informally, this proposition is true becauseSanaz cannot distinguish world v in which the gift is lost from world u in whichthe card is lost. Proposition 6. z (cid:13) S s ¬ ( “Pavel received a gift from Sanaz” ) iff z ∈ { t } .Proof. ( ⇒ ) : Suppose that z / ∈ { t } . Thus, either z ∈ { w } or z ∈ { u, v } . Weconsider the these two cases separately: Case I : z ∈ { w } . Then, z (cid:13) “Pavel received a gift from Sanaz”, see Figure 2.Thus, z (cid:49) ¬ (“Pavel received a gift from Sanaz”) by item 2 of Definition 2.Therefore, z (cid:49) S s ¬ (“Pavel received a gift from Sanaz”) by item 7(a) of Defi-nition 2. Case II : z ∈ { u, v } . Then, z ∼ s u , see Figure 2. Note that u (cid:13) “Pavel received a gift from Sanaz” , see Figure 2. Thus, u (cid:49) ¬ (“Pavel received a gift from Sanaz”) by item 2 ofDefinition 2. Therefore, z (cid:49) S s ¬ (“Pavel received a gift from Sanaz”) by item7(a) of Definition 2 and the statement z ∼ s u .( ⇐ ) : To prove that t (cid:13) S s ¬ (“Pavel received a gift from Sanaz”), we verifyconditions (a), (b), and (c) of item 7 in Definition 2 separately: Condition a:
Consider any world z (cid:48) such that t ∼ s z (cid:48) . It suffices to show that z (cid:48) (cid:13) ¬ (“Pavel received a gift from Sanaz”). Indeed, note that t (cid:49) (“Pavel received a gift from Sanaz”) , see Figure 2. Then, t (cid:13) ¬ (“Pavel received a gift from Sanaz”) by item 2 ofDefinition 2. Also note that the assumption t ∼ s z (cid:48) implies that t = z (cid:48) , seeFigure 2. Therefore, z (cid:48) (cid:13) ¬ (“Pavel received a gift from Sanaz”). Condition b:
Consider any two epistemic worlds x, y such that x (cid:13) ¬ (“Pavel received a gift from Sanaz”) ,y (cid:49) ¬ (“Pavel received a gift from Sanaz”) . To verify the condition, it suffices to show that x ≺ s y . Indeed, by item 2 ofDefinition 2, x (cid:49) “Pavel received a gift from Sanaz” ,y (cid:13) “Pavel received a gift from Sanaz” . Thus, x ∈ { t, v } and y ∈ { w, u } , see Figure 2. Note that { t, v } ≺ s { w, u } , seealso Figure 2. Therefore, x ≺ s y . 8 ondition c: Note that w (cid:13) “Pavel received a gift from Sanaz”. Therefore, w (cid:49) ¬ (“Pavel received a gift from Sanaz”) by item 2 of Definition 2. (cid:2) By Proposition 6, Sanaz is sad about Pavel not receiving the gift only if shedoes not send it. Since Pavel cannot distinguish world t in which the gift is sentfrom world v in which it is lost, Pavel cannot know that Sanaz is sad. This isformally captured in the next proposition. Proposition 7. z (cid:49) K p S s ¬ ( “Pavel received a gift from Sanaz” ) for each world z ∈ { w, u, t, v } .Proof. We consider the following two cases separately:
Case I: z ∈ { w, u } . Then, z (cid:49) S s ¬ (“Pavel received a gift from Sanaz”) byProposition 6. Therefore, z (cid:49) K p S s ¬ (“Pavel received a gift from Sanaz”) byitem 5 of Definition 2. Case II: z ∈ { t, v } . Then, z ∼ p v , see Figure 2. Note that v (cid:49) S s ¬ (“Pavel received a gift from Sanaz”)by Proposition 6. Therefore, z (cid:49) K p S s ¬ (“Pavel received a gift from Sanaz”) byitem 5 of Definition 2 and the statement z ∼ p v . (cid:2) Proposition 8. z (cid:13) S p ¬ ( “Pavel received a gift from Sanaz” ) iff z ∈ { v, t } .Proof. ( ⇒ ) : Suppose that z / ∈ { v, t } . Thus, z ∈ { w, u } . Hence, see Figure 2, z (cid:13) “Pavel received a gift from Sanaz”. Then, by item 2 of Definition 2, z (cid:49) ¬ (“Pavel received a gift from Sanaz”) . Therefore, z (cid:49) S p ¬ (“Pavel received a gift from Sanaz”) by item 7(a) of Defini-tion 2.( ⇐ ) : Suppose that z ∈ { v, t } . We verify conditions (a), (b), and (c) from item7 of Definition 2 to prove that z (cid:13) S p ¬ (“Pavel received a gift from Sanaz”): Condition a:
Consider any world z (cid:48) such that z ∼ p z (cid:48) . It suffices to show that z (cid:48) (cid:13) ¬ (“Pavel received a gift from Sanaz”). Indeed, the assumptions z ∈ { v, t } and z ∼ p z (cid:48) imply that z (cid:48) ∈ { v, t } , see Figure 2. Thus, see again Figure 2, z (cid:48) (cid:49) “Pavel received a gift from Sanaz”. Therefore, by item 2 of Definition 2, z (cid:48) (cid:13) ¬ (“Pavel received a gift from Sanaz”). Condition b:
The proof is similar to the proof of
Condition b in Proposition 6.
Condition c:
Note that w (cid:13) “Pavel received a gift from Sanaz”. Therefore, w (cid:49) ¬ (“Pavel received a gift from Sanaz”) by item 2 of Definition 2. (cid:2) Proposition 9. z (cid:13) S s S p ¬ ( “Pavel received a gift from Sanaz” ) iff z ∈ { t } . roof. Note that x (cid:49) “Pavel received a gift from Sanaz” iff x ∈ { t, v } , seeFigure 2. Thus, x (cid:13) ¬ (“Pavel received a gift from Sanaz”) iff x ∈ { t, v } byitem 2 of Definition 2. Thus, by Proposition 8, for any world x ∈ W , x (cid:13) ¬ (“Pavel received a gift from Sanaz”)iff x (cid:13) S p ¬ (“Pavel received a gift from Sanaz”) . Hence, by Lemma 1, for any world z ∈ W , z (cid:13) S s ¬ (“Pavel received a gift from Sanaz”)iff z (cid:13) S s S p ¬ (“Pavel received a gift from Sanaz”) . Therefore, z (cid:13) S s S p ¬ (“Pavel received a gift from Sanaz”) iff z ∈ { t } byProposition 6. (cid:2) Proposition 10. z (cid:49) K p S s S p ¬ ( “Pavel received a gift from Sanaz” ) for eachworld z ∈ { w, u, v, t } .Proof. We consider the following two cases separately:
Case I : z ∈ { w, u } . Then, z (cid:49) S s S p ¬ (“Pavel received a gift from Sanaz”) byProposition 9. Therefore, z (cid:49) K p S s S p ¬ (“Pavel received a gift from Sanaz”) byitem 5 of Definition 2. Case II : z ∈ { t, v } . Then, z ∼ p v , see Figure 2. Note that, by Proposition 9, v (cid:49) S s S p ¬ (“Pavel received a gift from Sanaz”). Therefore, by item 5 of Defini-tion 2, z (cid:49) K p S s S p ¬ (“Pavel received a gift from Sanaz”). (cid:2)
5. The Battle of Cuisines Scenario
In this section we illustrate Definition 2 using a scenario based on a classicalstrategic game. Suppose that the two co-authors finally met and are decidingIranian RussianIranian 1,3 0,0Russian 0,0 3,1
Table 1: The Battle of Cuisines. Sanaz is the first player, Pavel is the second. on a restaurant to have dinner. Sanaz, being Iranian, wants to explore Russiancuisine, while Pavel would prefer to have dinner in an Iranian restaurant. Theepistemic worlds in this model are pairs ( r s , r p ) of restaurants choices madeby Sanaz and Pavel respectively, where r s , r p ∈ { Iranian , Russian } . We willconsider the situation after they both arrived to a restaurant and thus eachof them already knows the choice made by the other. Hence, both of them10an distinguish all epistemic worlds. In other words, this is a perfect informa-tion scenario. We specify the preference relations of Sanaz and Pavel throughtheir respective utility functions u s and u p captured in Table 1. For exam-ple, (Russian , Iranian) ≺ s (Iranian , Iranian) because the value of Sanaz’s utilityfunction in world (Iranian , Iranian) is larger than in world (Russian , Iranian): u s (Iranian , Iranian) = 1 ,u s (Russian , Iranian) = 0 . Proposition 11. (Russian , Russian) (cid:49) H p ( “Pavel is in the Russian restaurant” ) . Proof.
Note that(Iranian , Iranian) (cid:49) “Pavel is in the Russian restaurant” , (Russian , Russian) (cid:13) “Pavel is in the Russian restaurant” . At the same time, see Table 1, u p (Russian , Russian) = 1 < u p (Iranian , Iranian) . Hence, (Iranian , Iranian) ⊀ p (Russian , Russian). Therefore, by item 6(b) of Def-inition 2, (Russian , Russian) (cid:49) H p (“Pavel is in the Russian restaurant”). (cid:2) Note that Sanaz prefers world (Russian , Russian) to any other world. This,however, does not mean that she is happy about everything in this world. Thenext two propositions illustrate this.
Proposition 12. (Russian , Russian) (cid:49) H s ( “Sanaz is in the Russian restaurant” ) . Proof.
Note that(Iranian , Iranian) (cid:49) “Sanaz is in the Russian restaurant” , (Russian , Iranian) (cid:13) “Sanaz is in the Russian restaurant” . At the same time, see Table 1, u s (Russian , Iranian) = 0 < u s (Iranian , Iranian) . Hence, (Iranian , Iranian) ⊀ s (Russian , Iranian). Therefore, by item 6(b) of Def-inition 2, (Russian , Russian) (cid:49) H s (“Sanaz is in the Russian restaurant”). (cid:2) Proposition 13. ( x, y ) (cid:13) H s ( “Sanaz and Pavel are in the same restaurant” ) iff x = y . roof. ( ⇒ ) : Suppose that x (cid:54) = y . Then,( x, y ) (cid:49) “Sanaz and Pavel are in the same restaurant” , ( x, x ) (cid:13) “Sanaz and Pavel are in the same restaurant” . Note also that u s ( x, y ) = 0 < ≤ u s ( x, x ) because x (cid:54) = y , see Table 1. Hence( x, x ) ⊀ s ( x, y ). Therefore,( x, y ) (cid:49) H s (“Sanaz and Pavel are in the same restaurant”)by item 6(b) of Definition 2.( ⇐ ) : Suppose x = y . We verify conditions (a), (b), and (c) from item 6 of Defini-tion 2 to prove that ( x, y ) (cid:13) H s (“Sanaz and Pavel are in the same restaurant”): Condition a:
Since this is a model with perfect information, it suffices to showthat ( x, y ) (cid:13) “Sanaz and Pavel are in the same restaurant”, which is true dueto the assumption x = y . Condition b:
Consider any two worlds ( x , y ) , ( x , y ) ∈ W such that( x , y ) (cid:49) “Sanaz and Pavel are in the same restaurant” , (3)( x , y ) (cid:13) “Sanaz and Pavel are in the same restaurant” . (4)It suffices to show that ( x , y ) ≺ s ( x , y ). Indeed, statements (3) and (4) implythat x (cid:54) = y and x = y , respectively. Thus, u s ( x , y ) = 0 < ≤ u s ( x , y ),see Table 1. Therefore, ( x , y ) ≺ s ( x , y ). Condition c:
Note that(Russian , Iranian) (cid:49) “Sanaz and Pavel are in the same restaurant” . (cid:2) The proof of the next proposition is similar to the proof of the one above.
Proposition 14. ( x, y ) (cid:13) H p ( “Sanaz and Pavel are in the same restaurant” ) iff x = y . (cid:2) Proposition 15. ( x, y ) (cid:13) H p H s ( “Sanaz and Pavel are in the same restaurant” ) iff x = y .Proof. Note that ( x, y ) (cid:13) “Sanaz and Pavel are in the same restaurant” iff x = y . Thus, by Proposition 13, for any world ( x, y ) ∈ W ,( x, y ) (cid:13) “Sanaz and Pavel are in the same restaurant”iff ( x, y ) (cid:13) H s (“Sanaz and Pavel are in the same restaurant”) . x, y ) ∈ W ,( x, y ) (cid:13) H p (“Sanaz and Pavel are in the same restaurant”)iff ( x, y ) (cid:13) H p H s (“Sanaz and Pavel are in the same restaurant”) . Therefore, ( x, y ) (cid:13) H p H s (“Sanaz and Pavel are in the same restaurant”) iff x = y by Proposition 14. (cid:2) Proposition 16. ( x, y ) (cid:13) H s H p H s ( “Sanaz and Pavel are in the same restaurant” ) iff x = y .Proof. Note that ( x, y ) (cid:13) “Sanaz and Pavel are in the same restaurant” iff x = y . Thus, by Proposition 15, for any world ( x, y ) ∈ W ,( x, y ) (cid:13) “Sanaz and Pavel are in the same restaurant”iff ( x, y ) (cid:13) H p H s (“Sanaz and Pavel are in the same restaurant”) . Hence, by Lemma 1, for any world ( x, y ) ∈ W ,( x, y ) (cid:13) H s (“Sanaz and Pavel are in the same restaurant”)iff ( x, y ) (cid:13) H s H p H s (“Sanaz and Pavel are in the same restaurant”) . Therefore, ( x, y ) (cid:13) H s H p H s (“Sanaz and Pavel are in the same restaurant”) iff x = y by Proposition 13. (cid:2) Proposition 17. ( x, y ) (cid:13) H s ( “Sanaz and Pavel are in the Russian restaurant” ) iff x = y = Russian .Proof. ( ⇒ ) : By item 6(a) of Definition 2, the assumption of the proposi-tion ( x, y ) (cid:13) H s (“Sanaz and Pavel are in the Russian restaurant”) implies that( x, y ) (cid:13) “Sanaz and Pavel are in the Russian restaurant”. Therefore, x = y =Russian.( ⇐ ) : Suppose x = y = Russian. To prove that( x, y ) (cid:13) H s (“Sanaz and Pavel are in the Russian restaurant”) , we verify conditions (a), (b), and (c) from item 6 of Definition 2: Condition a:
Since this is a model with perfect information, it suffices to showthat ( x, y ) (cid:13) “Sanaz and Pavel are in the Russian restaurant”, which is truedue to the assumption x = y = Russian.13 ondition b: Consider any two worlds ( x , y ) , ( x , y ) ∈ W such that( x , y ) (cid:49) “Sanaz and Pavel are in the Russian restaurant” , (5)( x , y ) (cid:13) “Sanaz and Pavel are in the Russian restaurant” . (6)It suffices to show that ( x , y ) ≺ s ( x , y ). Indeed, statement (5) implies that u s ( x , y ) ≤
1, see Table 1. Similarly, statement (6) implies that u s ( x , y ) = 3.Thus, u s ( x , y ) ≤ < u s ( x , y ). Therefore, ( x , y ) ≺ s ( x , y ). Condition c: (Russian , Iranian) (cid:49) “Sanaz and Pavel are in the Russian restaurant”. (cid:2)
Proposition 18. (Russian , Russian) (cid:49) H p ( “Sanaz and Pavel are in the Russian restaurant” ) . Proof.
Note that u p (Iranian , Iranian) = 3 > u p (Russian , Russian). Thus,(Iranian , Iranian) ⊀ p (Russian , Russian). Therefore, the proposition is true byitem 6(b) of Definition 2. (cid:2)
Proposition 19. (Russian , Russian) (cid:49) H p H s ( “Sanaz and Pavel are in the Russian restaurant” ) . Proof.
Note that(Iranian , Iranian) (cid:49) H s (“Sanaz and Pavel are in the Russian restaurant”) , (Russian , Russian) (cid:13) H s (“Sanaz and Pavel are in the Russian restaurant”) . by Proposition 17. At the same time, u p (Iranian , Iranian) = 3 > u p (Russian , Russian) . Thus, (Iranian , Iranian) ⊀ p (Russian , Russian). Therefore, the proposition istrue by item 6(b) of Definition 2. (cid:2)
Proposition 20. ( x, y ) (cid:13) S s ( “Sanaz and Pavel are in different restaurants” ) iff x (cid:54) = y. Proof. ( ⇒ ) : Suppose that x = y . Thus,( x, y ) (cid:49) “Sanaz and Pavel are in different restaurants” . Thus, by item 7(a) of Definition 2,( x, y ) (cid:49) S s (“Sanaz and Pavel are in different restaurants”) . ⇐ ) : Suppose that x (cid:54) = y . To prove that( x, y ) (cid:13) S s (“Sanaz and Pavel are in different restaurants”) , we verify conditions (a), (b), and (c) from item 7 of Definition 2: Condition a:
Since this is a model with perfect information, it suffices to showthat ( x, y ) (cid:13) “Sanaz and Pavel are in different restaurants”, which is true dueto the assumption x (cid:54) = y . Condition b:
The proof of this condition is similar to the proof of
Condition b in the proof of Proposition 13.
Condition c: (Iranian , Iranian) (cid:49) “Sanaz and Pavel are in different restaurants”. (cid:2)
The proof of the following statement is similar to the proof of the one above.
Proposition 21. ( x, y ) (cid:13) S p ( “Sanaz and Pavel are in different restaurants” ) iff x (cid:54) = y. (cid:2) Proposition 22. ( x, y ) (cid:13) S p S s ( “Sanaz and Pavel are in different restaurants” ) iff x (cid:54) = y. Proof.
Note that ( x, y ) (cid:13) “Sanaz and Pavel are in different restaurants” iff x (cid:54) = y . Thus, by Proposition 20, for any world ( x, y ) ∈ W ,( x, y ) (cid:13) “Sanaz and Pavel are in different restaurants”iff ( x, y ) (cid:13) S s (“Sanaz and Pavel are in different restaurants”) . Hence, by Lemma 1, for any world ( x, y ) ∈ W ,( x, y ) (cid:13) S p (“Sanaz and Pavel are in different restaurants”)iff ( x, y ) (cid:13) S p S s (“Sanaz and Pavel are in different restaurants”) . Therefore, ( x, y ) (cid:13) S p S s (“Sanaz and Pavel are in different restaurants”) iff x = y by Proposition 21. (cid:2)
6. Lottery Example
Be happy for this moment. Thismoment is your life.The Rub´aiy´at of Omar Khayy´am
As our next example, consider a hypothetical situation when Sanaz andPavel play lottery with Omar Khayy´am, an Iranian mathematician, astronomer,15hilosopher, and poet. Each of them gets a lottery ticket and it is known thatexactly one out of three tickets is winning. We consider the moment when eachof the players has already seen her or his own ticket, but does not know yetwhat are the tickets of the other players. We assume that each of the threeplayers prefers the outcome when she or he wins the lottery.
Sanaz wins Omar winsPavel winspo soss p o pwu v
Figure 3: Lottery Epistemic Model with Preferences.
Figure 3 depicts the epistemic model with preferences capturing the abovescenario. It has three epistemic worlds, u , v , and w in which the winner isSanaz, Omar, and Pavel respectively. Dashed lines represent indistinguishabilityrelations. For example, the dashed line between worlds w and v labeled with s shows that Sanaz cannot distinguish the world in which Pavel wins from theone in which Omar wins. This is true because we consider the knowledge at themoment when neither of the players knows yet what are the tickets of the otherplayers. The directed edges between worlds represent preference relations. Forexample, the directed edge from world w to world v labeled with o captures thefact that Omar would prefer to win the lottery rather than to lose it. Proposition 23. x (cid:13) H s ( “Sanaz won the lottery” ) iff x = u .Proof. ( ⇒ ) : Suppose that x (cid:54) = u . Thus, x (cid:49) “Sanaz won the lottery”, see Fig-ure 3. Therefore, x (cid:49) H s (“Sanaz won the lottery”) by item 6(a) of Definition 2.( ⇐ ) : Suppose that x = u . To prove the required, it suffices to verify conditions(a), (b), and (c) from item 6 of Definition 2: Condition a:
Consider any world y such that u ∼ s y . We will show that y (cid:13) “Sanaz won the lottery”. Indeed, assumption u ∼ s y implies that u = y ,see Figure 3. Therefore, y (cid:13) “Sanaz won the lottery”, see again Figure 3. Condition b:
Consider any worlds y, z such that y (cid:49) “Sanaz won the lottery”and z (cid:13) “Sanaz won the lottery”. We will show that y ≺ s z . Indeed, assump-16ion y (cid:49) “Sanaz won the lottery” implies that y ∈ { w, v } , see Figure 3. Simi-larly, assumption z (cid:13) “Sanaz won the lottery” implies that z = u . Statements y ∈ { w, v } and z = u imply that y ≺ s z , see again Figure 3. Condition c: w (cid:49) “Sanaz won the lottery”. (cid:2) Proposition 24. u (cid:49) H s ( “Pavel lost the lottery” ) .Proof. Note that w (cid:49) “Pavel lost the lottery”, v (cid:13) “Pavel lost the lottery”, and w ⊀ s v , see Figure 3. Therefore, u (cid:49) H s (“Pavel lost the lottery”) by item 6(b)of Definition 2. (cid:2) Proposition 25. u (cid:49) K p H s ( “Sanaz won the lottery” ) .Proof. Note that u ∼ p v , see Figure 3. Also, v (cid:49) H s (“Sanaz won the lottery”)by Proposition 23. Therefore, u (cid:49) K p H s (“Sanaz won the lottery”) by item 5 ofDefinition 2. (cid:2) Proposition 26. u (cid:13) S p ( “Pavel lost the lottery” ) .Proof. It suffices to verify conditions (a), (b), and (c) from item 7 of Definition 2:
Condition a:
Consider any world y such that u ∼ p y . We will show that y (cid:13) “Pavel lost the lottery”. Indeed, assumption u ∼ p y implies that y ∈ { u, v } ,see Figure 3. Therefore, y (cid:13) “Pavel lost the lottery”, see again Figure 3. Condition b:
Consider any worlds y, z such that y (cid:13) “Pavel lost the lottery” and z (cid:49) “Pavel lost the lottery”. We will show that y ≺ p z . Indeed, assumption y (cid:13) “Pavel lost the lottery” implies that y ∈ { u, v } , see Figure 3. Similarly,assumption z (cid:49) “Pavel lost the lottery” implies that z = w . Statements y ∈{ u, v } and z = w imply that y ≺ s z , see again Figure 3. Condition c: w (cid:49) “Pavel lost the lottery”. (cid:2) Proposition 27. u (cid:13) K s S p ( “Pavel lost the lottery” ) .Proof. Consider any world y such that u ∼ s y . By item 5 of Definition 2, it suf-fices to show that y (cid:13) S p (“Pavel lost the lottery”). Indeed, assumption u ∼ s y implies that u = y , see Figure 3. Therefore, y (cid:13) S p (“Pavel lost the lottery”) byProposition 26. (cid:2)
7. Undefinability of Sadness through Happiness
In this section we prove that sadness is not definable through happiness.More formally, we show that formula S a p is not equivalent to any formula inthe language Φ − S : ϕ := p | ¬ ϕ | ϕ → ϕ | N ϕ | K a ϕ | H a ϕ, S from the full language Φ of ourlogical system. In the next section we will use a duality principle to claim thathappiness is not definable through sadness either.Without loss of generality, in this section we assume that the set of agents A contains a single agent a and the set of propositional variables contains a singlevariable p . To prove undefinability of sadness through happiness we considertwo epistemic models with preferences depicted in Figure 4. By (cid:13) l and (cid:13) r wemean the satisfaction relations for the left and and the right model respectively.In Lemma 4, we show that these two models are indistinguishable in languagethe Φ − S . In Lemma 5 and Lemma 6, we prove that w (cid:13) l S a p and w (cid:49) r S a p respectively. Together, these three statements imply undefinability of modality S in the language Φ − S , which is stated in the end of this section as Theorem 1.We start with two auxiliary lemmas used in the proof of Lemma 4. W W W p p a W W W p p aa a Figure 4: Two Models.
Lemma 2. w (cid:49) l H a ϕ for each world w ∈ { w , w , w } and each formula ϕ ∈ Φ .Proof. Suppose that w (cid:13) l H a ϕ . Thus, by item 6 of Definition 2, u (cid:13) l ϕ for each world u ∈ W such that w ∼ a u, (7) u ≺ a u (cid:48) for all worlds u, u (cid:48) ∈ W such that u (cid:49) ϕ and u (cid:48) (cid:13) ϕ, (8)and there is a world (cid:98) w ∈ { w , w , w } such that (cid:98) w (cid:49) ϕ. (9)Since relation ∼ a is reflexive, statement (7) implies that w (cid:13) ϕ. (10)Thus, using statements (8) and (9), (cid:98) w ≺ a w. (11)Hence, see Figure 4 (left), (cid:98) w ∈ { w , w } , (12) w = w . (13)18ote that w ∼ a w , see Figure 4 (left). Thus, w ∼ a w by statement (13).Then, by statement (7), w (cid:13) ϕ. (14)Hence, (cid:98) w (cid:54) = w because of statement (9). Thus, (cid:98) w = w due to statement (12).Then, w (cid:49) ϕ because of statement (9). Therefore, w ≺ a w by statements (8)and (14), which is a contradiction, see Figure 4 (left). (cid:2) Lemma 3. w (cid:49) r H a ϕ for each world w ∈ { w , w , w } and each formula ϕ ∈ Φ .Proof. Suppose w (cid:13) r H a ϕ . Thus, w (cid:13) r ϕ by item 6(a) of Definition 2 andthe reflexivity of relation ∼ a . At the same time, by item 6(c) of Definition 2,there must exist a world u ∈ W such that u (cid:49) r ϕ . By item 6(b) of the sameDefinition 2, the assumption w (cid:13) r H a ϕ and the statements u (cid:49) r ϕ and w (cid:13) r ϕ imply that u ≺ a w , which is a contradiction because relation ≺ a in the rightmodel is empty, see Figure 4. (cid:2) Lemma 4. w (cid:13) l ϕ iff w (cid:13) r ϕ for each world w and each formula ϕ ∈ Φ − S .Proof. We prove the statement by structural induction on formula ϕ . For thecase when formula ϕ is propositional variable p , observe that π l ( p ) = { w , w } = π r ( p ) by the choice of the models, see Figure 4. Thus, w (cid:13) l p iff w (cid:13) r p forany world w by item 1 of Definition 2. The case when formula ϕ is a negationor an implication follows from the induction hypothesis and items 2 and 3 ofDefinition 2 in the straightforward way.Suppose formula ϕ has the form N ψ . If w (cid:49) l N ψ , then by item 4 of Defini-tion 2, there must exists a world u ∈ { w , w , w } such that u (cid:49) l ψ . Hence, bythe induction hypothesis, u (cid:49) r ψ . Therefore, w (cid:49) r N ψ , by item 4 of Definition 2.The proof in the other direction is similar.Assume that formula ϕ has the form K a ψ . If w (cid:49) l K a ψ , then, by item 5 ofDefinition 2, there is a world u ∈ { w , w , w } such that w ∼ la u and u (cid:49) l ψ .Then, w ∼ ra u because relations ∼ l and ∼ r are equal, see Figure 4 and, by theinduction hypothesis, u (cid:49) r ψ . Therefore, w (cid:49) r K a ψ by item 5 of Definition 2.The proof in the other direction is similar.Finally, suppose formula ϕ has the form H a ψ . Therefore, w (cid:49) l ϕ and w (cid:49) r ϕ by Lemma 2 and Lemma 3 respectively. (cid:2) Lemma 5. w (cid:13) l S a p .Proof. We verify the three conditions from item 7 of Definition 2 separately:
Condition a:
Observe that w ∈ { w , w } = π l ( p ), see Figure 4 (left). Then, w (cid:13) l p by item 1 of Definition 2. Note also that there is only one world u ∈ { w , w , w } such that w ∼ la u (namely, the world w itself), see Figure 4(left). Therefore, u (cid:13) p for each world u ∈ { w , w , w } such that w ∼ a u .19 ondition b: Note that π l ( p ) = { w , w } , see Figure 4 (left). Then, w (cid:13) l p , w (cid:13) l p , and w (cid:49) l p by item 1 of Definition 2. Also, observe that w ≺ la w and w ≺ la w , see Figure 4 (left). Thus, for any world u, u (cid:48) ∈ { w , w , w } , if u (cid:13) p and u (cid:48) (cid:49) p , then u ≺ a u (cid:48) . Condition c: w (cid:49) l p by item 1 of Definition 2 and because w / ∈ { w , w } = π l ( p ), see Figure 4 (left).This concludes the proof of the lemma. (cid:2) Lemma 6. w (cid:49) r S a p .Proof. Note that π r ( p ) = { w , w } , see Figure 4 (right). Thus, w (cid:13) r p and w (cid:49) r p by item 1 of Definition 2. Observe also, that w ⊀ a w , Figure 4 (right).Therefore, w (cid:49) r S a p by item 7(b) of Definition 2. (cid:2) The next theorem follows from the three lemmas above.
Theorem 1.
Modality S is not definable in language Φ − S . (cid:2)
8. Duality of Happiness and Sadness
As we have shown in the previous section, sadness modality is not definablethrough happiness modality. In spite of this, there still is a connection betweenthese two modalities captured below in Theorem 2. To understand this con-nection, we need to introduce the notion of a converse model and the notion of τ -translation. As usual, for any binary relation R ⊆ X × Y , by converse relation R c we mean the set of pairs { ( y, x ) ∈ Y × X | ( x, y ) ∈ R } . Definition 3.
By the converse model M c of an epistemic model with preference M = ( W, {∼ a } a ∈A , {≺ a } a ∈A , π ) , we mean model ( W, {∼ a } a ∈A , {≺ c a } a ∈A , π ) . For any epistemic model with preference, by (cid:13) we denote the satisfaction rela-tion for this model and by (cid:13) c the satisfaction relation for the converse model. Definition 4.
For any formula ϕ ∈ Φ , formula τ ( ϕ ) ∈ Φ is defined recursivelyas follows: τ ( p ) ≡ p , where p is a propositional variable, τ ( ¬ ϕ ) ≡ ¬ τ ( ϕ ) , τ ( ϕ → ψ ) ≡ τ ( ϕ ) → τ ( ψ ) , τ ( N ϕ ) ≡ N τ ( ϕ ) , τ ( K a ϕ ) ≡ K a τ ( ϕ ) , τ ( H a ϕ ) ≡ S a τ ( ϕ ) , τ ( S a ϕ ) ≡ H a τ ( ϕ ) . We are now ready to state and prove the “duality principle” that connectsmodalities H and S . 20 heorem 2. w (cid:13) ϕ iff w (cid:13) c τ ( ϕ ) , for each world w of an epistemic model withpreferences.Proof. We prove the theorem by induction on structural complexity of formula ϕ . If ϕ is a propositional variable, then the statement of the theorem holdsbecause the model and the converse model have the same valuation function π . If ϕ is a negation, an implication, an N -formula, or an K -formula, then thestatement of the theorem follows from the induction hypothesis and items 2, 3,4, and 5 of Definition 2 respectively.Suppose that formula ϕ has the form H a ψ . First, assume that w (cid:13) H a ψ .Thus, by item 6 of Definition 2, the following three conditions are satisfied:(a) u (cid:13) ψ for each world u ∈ W such that w ∼ a u ,(b) for any two worlds u, u (cid:48) ∈ W , if u (cid:49) ψ and u (cid:48) (cid:13) ψ , then u ≺ a u (cid:48) ,(c) there is a world u ∈ W such that u (cid:49) ψ .Hence, by the induction hypothesis,(d) u (cid:13) c τ ( ψ ) for each world u ∈ W such that w ∼ a u ,(e) for any two worlds u, u (cid:48) ∈ W , if u (cid:49) c τ ( ψ ) and u (cid:48) (cid:13) c τ ( ψ ), then u ≺ a u (cid:48) ,(f) there is a world u ∈ W such that u (cid:49) c τ ( ψ ).Note that statement (e) is logically equivalent to(g) for any two worlds u, u (cid:48) ∈ W , if u (cid:13) c τ ( ψ ) and u (cid:48) (cid:49) c τ ( ψ ), then u (cid:48) ≺ a u .By the definition of converse partial order, statement (g) is equivalent to(h) for any two worlds u, u (cid:48) ∈ W , if u (cid:13) c τ ( ψ ) and u (cid:48) (cid:49) c τ ( ψ ), then u ≺ c a u (cid:48) .Thus, w (cid:13) c S a τ ( ψ ) by item 7 of Definition 2 using statements (d), (h), and (f).Therefore, w (cid:13) c τ ( H a ψ ). The proof in the other direction and the proof for thecase when formula ϕ has the form S a ψ are similar. (cid:2) The next theorem follows from Theorem 1 and Theorem 2.
Theorem 3.
Modality H is not definable in language Φ − H . (cid:2)
9. Axioms of Emotions
In addition to propositional tautologies in language Φ, our logical systemcontains the following axioms, where E ∈ { H , S } .1. Truth: N ϕ → ϕ , K a ϕ → ϕ , and E a ϕ → ϕ ,2. Distributivity: N ( ϕ → ψ ) → ( N ϕ → N ψ ), K a ( ϕ → ψ ) → ( K a ϕ → K a ψ ),3. Negative Introspection: ¬ N ϕ → N ¬ N ϕ , and ¬ K a ϕ → K a ¬ K a ϕ ,4. Knowledge of Necessity: N ϕ → K a ϕ ,5. Emotional Introspection: E a ϕ → K a E a ϕ ,21. Emotional Consistency: H a ϕ → ¬ S a ϕ ,7. Coherence of Potential Emotions: NE a ϕ ∧ NE a ψ → N ( ϕ → ψ ) ∨ N ( ψ → ϕ ), NH a ϕ ∧ NS a ψ → N ( ϕ → ¬ ψ ) ∨ N ( ¬ ψ → ϕ ),8. Counterfactual: E a ϕ → ¬ N ϕ ,9. Emotional Predictability: NH a ϕ ∨ NS a ¬ ϕ → ( K a ϕ → H a ϕ ), NH a ¬ ϕ ∨ NS a ϕ → ( K a ϕ → S a ϕ ),10. Substitution: N ( ϕ ↔ ψ ) → ( E a ϕ → E a ψ ).The Truth, the Distributivity, and the Negative Introspection axioms for modal-ities N and K are well-known properties from modal logic S5. The Truth axiomfor modality E states that if an agent is either happy or sad about ϕ , thenstatement ϕ must be true. This axiom reflects the fact that our emotions aredefined through agent’s knowledge. We will mention belief-based emotions inthe conclusion.The Knowledge of Necessity axioms states that each agent knows all state-ments that are true in all worlds of the model. The Emotional Introspectionaxiom captures one of the two possible interpretations of the title of this article:each agent knows her emotions. The other interpretation of the title is statedbelow as Lemma 8. The Emotional Consistency axiom states that an agentcannot be simultaneously happy and sad about the same thing.Recall that N ϕ stands for formula ¬ N ¬ ϕ . Thus, formula NE a ϕ means thatagent a might have the emotion E about statement ϕ . In other words, formula NE a ϕ states that agent a has a potential emotion E about ϕ . The Coherenceof Potential Emotions axioms expresses the fact that potential emotions of anyagent are not independent. The first of these axioms says that if an agent has thesame potential emotions about statements ϕ and ψ , then one of these statementsmust imply the other in each world of our model. The second of these axiomssays that if an agent has opposite potential emotions about statements ϕ and ψ , then these statements cannot be both true in any world of the model. Weprove soundness of these axioms along with the soundness of the other axiomsof our system in Section 10.The Counterfactual axiom states that an agent cannot have an emotionabout something which is universally true in the model. This axiom reflectsitems 6(c) and 7(c) of Definition 2.Because the assumptions of both Emotional Predictability axioms containdisjunctions, each of these axioms could be split into two statements. The firststatement of the first Emotional Predictability axiom says that if an agent ispotentially happy about ϕ , then she must be happy about ϕ each time when sheknows that ϕ is true. The second statement of the same axiom says that if anagent is potentially sad about ¬ ϕ , then she must be happy about ϕ each timewhen she knows that ϕ is true. The second Emotional Predictability axioms isthe dual form of the first axiom.Finally, the Substitution axiom states that if two statements are equivalentin each world of the model and an agent has an emotion about one of them,22hen she must have the same emotion about the other statement.We write (cid:96) ϕ and say that statement ϕ is a theorem of our logical sys-tem if ϕ is derivable from the above axioms using the Modus Ponens and theNecessitation inference rules: ϕ, ϕ → ψψ ϕ N ϕ . For any set of statements X ⊆ Φ, we write X (cid:96) ϕ if formula ϕ is derivablefrom the theorems of our system and the set X using only the Modus Ponensinference rule. We say that set X is inconsistent if there is a formula ϕ ∈ Φsuch that X (cid:96) ϕ and X (cid:96) ¬ ϕ . Lemma 7.
Inference rule ϕ K a ϕ is derivable.Proof. This rule is a combination of the Necessitation inference rule, the Knowl-edge of Necessity axiom, and the Modus Ponens inference rule. (cid:2)
Lemma 8. (cid:96) E a ϕ → K a ϕ .Proof. Note that (cid:96) E a ϕ → ϕ by the Truth axiom. Thus, (cid:96) K a ( E a ϕ → ϕ ) byLemma 7. Hence, (cid:96) K a E a ϕ → K a ϕ by the Distributivity axiom and the ModusPonens inference rule. Therefore, (cid:96) E a ϕ → K a ϕ by the Emotional Introspectionaxiom and propositional reasoning. (cid:2) The next three lemmas are well-known in model logic. We reproduce theirproofs here to keep the article self-contained.
Lemma 9 (deduction). If X, ϕ (cid:96) ψ , then X (cid:96) ϕ → ψ .Proof. Suppose that sequence ψ , . . . , ψ n is a proof from set X ∪ { ϕ } and thetheorems of our logical system that uses the Modus Ponens inference rule only.In other words, for each k ≤ n , either1. (cid:96) ψ k , or2. ψ k ∈ X , or3. ψ k is equal to ϕ , or4. there are i, j < k such that formula ψ j is equal to ψ i → ψ k .It suffices to show that X (cid:96) ϕ → ψ k for each k ≤ n . We prove this by inductionon k through considering the four cases above separately. Case I : (cid:96) ψ k . Note that ψ k → ( ϕ → ψ k ) is a propositional tautology, and thus,is an axiom of our logical system. Hence, (cid:96) ϕ → ψ k by the Modus Ponensinference rule. Therefore, X (cid:96) ϕ → ψ k . Case II : ψ k ∈ X . Then, X (cid:96) ψ k . Case III : formula ψ k is equal to ϕ . Thus, ϕ → ψ k is a propositional tautology.Therefore, X (cid:96) ϕ → ψ k . 23 ase IV : formula ψ j is equal to ψ i → ψ k for some i, j < k . Thus, by the in-duction hypothesis, X (cid:96) ϕ → ψ i and X (cid:96) ϕ → ( ψ i → ψ k ). Note that formula( ϕ → ψ i ) → (( ϕ → ( ψ i → ψ k )) → ( ϕ → ψ k )) is a propositional tautology.Therefore, X (cid:96) ϕ → ψ k by applying the Modus Ponens inference rule twice. (cid:2) Lemma 10. If ϕ , . . . , ϕ n (cid:96) ψ , then (cid:50) ϕ , . . . , (cid:50) ϕ n (cid:96) (cid:50) ψ , where (cid:50) is eithermodality N or modality K a .Proof. Lemma 9 applied n times to assumption ϕ , . . . , ϕ n (cid:96) ψ implies that (cid:96) ϕ → ( · · · → ( ϕ n → ψ ) . . . ). Thus, (cid:96) (cid:50) ( ϕ → ( · · · → ( ϕ n → ψ ) . . . )),by either the Necessitation inference rule (if (cid:50) is modality N ) or Lemma 7 (if (cid:50) is modality K ). Hence, by the Distributivity axiom and the Modus Ponensinference rule, (cid:96) (cid:50) ϕ → (cid:50) ( ϕ · · · → ( ϕ n → ψ ) . . . ) . Then, (cid:50) ϕ (cid:96) (cid:50) ( ϕ · · · → ( ϕ n → ψ ) . . . ) by the Modus Ponens inference rule.Thus, again by the Distributivity axiom and the Modus Ponens inference rule, (cid:50) ϕ (cid:96) (cid:50) ϕ → (cid:50) ( ϕ · · · → ( ϕ n → ψ ) . . . ). Therefore, (cid:50) ϕ , . . . , (cid:50) ϕ n (cid:96) (cid:50) ψ , byrepeating the last two steps n − (cid:2) Lemma 11 (positive introspection). (cid:96) (cid:50) ϕ → (cid:50)(cid:50) ϕ , where (cid:50) is either modal-ity N or modality K a .Proof. Formula (cid:50) ¬ (cid:50) ϕ → ¬ (cid:50) ϕ is an instance of the Truth axiom. Thus, bycontraposition, (cid:96) (cid:50) ϕ → ¬ (cid:50) ¬ (cid:50) ϕ . Hence, taking into account the followinginstance of the Negative Introspection axiom: ¬ (cid:50) ¬ (cid:50) ϕ → (cid:50) ¬ (cid:50) ¬ (cid:50) ϕ , we have (cid:96) (cid:50) ϕ → (cid:50) ¬ (cid:50) ¬ (cid:50) ϕ. (15)At the same time, ¬ (cid:50) ϕ → (cid:50) ¬ (cid:50) ϕ is an instance of the Negative Introspectionaxiom. Thus, (cid:96) ¬ (cid:50) ¬ (cid:50) ϕ → (cid:50) ϕ by the law of contrapositive in the propositionallogic. Hence, (cid:96) (cid:50) ( ¬ (cid:50) ¬ (cid:50) ϕ → (cid:50) ϕ ) by either the Necessitation inference rule (if (cid:50) is modality N ) or Lemma 7 (if (cid:50) is modality K ). Thus, by the Distributivityaxiom and the Modus Ponens inference rule, (cid:96) (cid:50) ¬ (cid:50) ¬ (cid:50) ϕ → (cid:50)(cid:50) ϕ. The latter,together with statement (15), implies the statement of the lemma by proposi-tional reasoning. (cid:2)
Lemma 12. N ( ϕ → ψ ) , N ( ¬ ϕ → ¬ ψ ) (cid:96) N ( ϕ ↔ ψ ) , N ( ϕ → ¬ ψ ) , N ( ¬ ϕ → ψ ) (cid:96) N ( ϕ ↔ ¬ ψ ) .Proof. It is provable in the propositional logic that ϕ → ψ, ¬ ϕ → ¬ ψ (cid:96) ϕ ↔ ψ .Thus, N ( ϕ → ψ ) , N ( ¬ ϕ → ¬ ψ ) (cid:96) N ( ϕ ↔ ψ ) by Lemma 10. The proof of thesecond part of the lemma is similar. (cid:2) emma 13. N ( ϕ ↔ ψ ) , NH a ϕ (cid:96) NH a ψ , N ( ϕ ↔ ψ ) , NS a ¬ ϕ (cid:96) NS a ¬ ψ , N ( ϕ ↔ ¬ ψ ) , NS a ¬ ϕ (cid:96) NS a ψ .Proof. Formula N ( ϕ ↔ ψ ) → ( H a ϕ → H a ψ ) is an instance of the Substitutionaxiom. Thus, (cid:96) N ( ϕ ↔ ψ ) → ( ¬ H a ψ → ¬ H a ϕ ) by the laws of propositionalreasoning. Hence, N ( ϕ ↔ ψ ) , ¬ H a ψ (cid:96) ¬ H a ϕ by the Modus Ponens rule appliedtwice. Then, NN ( ϕ ↔ ψ ) , N ¬ H a ψ (cid:96) N ¬ H a ϕ by Lemma 10. Thus, by Lemma 11and the Modus Ponens inference rule, N ( ϕ ↔ ψ ) , N ¬ H a ψ (cid:96) N ¬ H a ϕ . Hence, N ( ϕ ↔ ψ ) (cid:96) N ¬ H a ψ → N ¬ H a ϕ by Lemma 9. Then, by the laws of propositionalreasoning, N ( ϕ ↔ ψ ) (cid:96) ¬ N ¬ H a ϕ → ¬ N ¬ H a ψ . Thus, by the definition ofmodality N , we have N ( ϕ ↔ ψ ) (cid:96) NH a ϕ → NH a ψ . Therefore, by the ModusPonens inference rule, N ( ϕ ↔ ψ ) , NH a ϕ (cid:96) NH a ψ .To prove the second statement, observe that ( ϕ ↔ ψ ) → ( ¬ ϕ ↔ ¬ ψ ) isa propositional tautology. Thus, ϕ ↔ ψ (cid:96) ¬ ϕ ↔ ¬ ψ by the Modus Ponensinference rule. Hence, N ( ϕ ↔ ψ ) (cid:96) N ( ¬ ϕ ↔ ¬ ψ ) by Lemma 10. Then, to proveof the second statement, it suffices to show that N ( ¬ ϕ ↔ ¬ ψ ) , NS a ¬ ϕ (cid:96) NS a ¬ ψ .The proof of this is the same as of the first statement.The proof of the third statement is similar to the proof of the second, but itstarts with the tautology ( ϕ ↔ ¬ ψ ) → ( ¬ ϕ ↔ ψ ). (cid:2) Lemma 14 (Lindenbaum).
Any consistent set of formulae can be extendedto a maximal consistent set of formulae.Proof.
The standard proof of Lindenbaum’s lemma applies here [16, Proposition2.14]. (cid:2)
10. Soundness
The Truth, the Distributivity, and the Negative Introspection axioms formodalities K and N are well-known principles of S5 logic. The soundness ofthe Knowledge of Necessity axiom follows from Definition 2. Below we provethe soundness of each of the remaining axioms as a separate lemma. We statestrong soundness as Theorem 4 in the end of the section. In the lemmas below weassume that w ∈ W is an arbitrary world of an epistemic model with preferences M = ( W, {∼ a } a ∈A , {≺ a } a ∈A , π ). Lemma 15. If w (cid:13) E a ϕ , then w (cid:13) ϕ .Proof. First, we consider the case E = H . Note that w ∼ a w because ∼ a is anequivalence relation. Thus, the assumption w (cid:13) H a ϕ implies w (cid:13) ϕ by item6(a) of Definition 2. The proof for the case E = S is similar, but it uses item7(a) of Definition 2 instead of item 6(a). (cid:2) emma 16. If w (cid:13) E a ϕ , then w (cid:13) K a E a ϕ .Proof. First, we consider the case E = H . Consider any world w (cid:48) ∈ W such that w ∼ a w (cid:48) . By item 5 of Definition 2, it suffices to show that w (cid:48) (cid:13) H a ϕ . Indeed,by item 6 of Definition 2, the assumption w (cid:13) H a ϕ of the lemma implies that(a) u (cid:13) ϕ for each world u ∈ W such that w ∼ a u ,(b) for any two worlds u, u (cid:48) ∈ W , if u (cid:49) ϕ and u (cid:48) (cid:13) ϕ , then u ≺ a u (cid:48) ,(c) there is a world u ∈ W such that u (cid:49) ϕ ,By the assumption w ∼ a w (cid:48) , statement (a) implies that(a’) u (cid:13) ϕ for each world u ∈ W such that w (cid:48) ∼ a u .Finally, statements (a’), (b), and (c) imply that w (cid:48) (cid:13) H a ϕ by item 6 of Defini-tion 2. The proof for the case E = S is similar, but it uses item 7 of Definition 2instead of item 6. (cid:2) Lemma 17. If w (cid:13) H a ϕ , then w (cid:49) S a ϕ .Proof. By item 6(a) of Definition 2, the assumption w (cid:13) H a ϕ implies that w (cid:13) ϕ . By item 6(c) of Definition 2, the same assumption implies that there isa world w (cid:48) ∈ W such that w (cid:48) (cid:49) ϕ . By item 6(b) of Definition 2 the assumption w (cid:13) H a ϕ and statements w (cid:48) (cid:49) ϕ and w (cid:13) ϕ imply that w (cid:48) ≺ a w . Thus, w ⊀ a w (cid:48) because relation ≺ a is a strict partial order. Therefore, w (cid:49) S a ϕ by item 7(b)of Definition 2 and statements w (cid:13) ϕ and w (cid:48) (cid:49) ϕ . (cid:2) Lemma 18. If w (cid:13) NE a ϕ and w (cid:13) NE a ψ , then either w (cid:13) N ( ϕ → ψ ) or w (cid:13) N ( ψ → ϕ ) .Proof. First, we consider the case E = H . Suppose that w (cid:49) N ( ϕ → ψ ) and w (cid:49) N ( ψ → ϕ ). Thus, by item 4 of Definition 2, there are epistemic worlds w , w ∈ W , such that w (cid:49) ϕ → ψ and w (cid:49) ψ → ϕ . Hence, by item 3 ofDefinition 2, w (cid:13) ϕ, w (cid:49) ψ, w (cid:13) ψ, w (cid:49) ϕ. (16)At the same time, by the definition of modality N and items 2 and 4 of Defini-tion 2, the assumption w (cid:13) NH a ϕ of the lemma implies that there is a world w (cid:48) such that w (cid:48) (cid:13) H a ϕ . Hence, w ≺ a w by item 6(b) of Definition 2 and parts w (cid:49) ϕ and w (cid:13) ϕ of statement (16).Similarly, the assumption w (cid:13) NH a ψ of the lemma and parts w (cid:49) ψ and w (cid:13) ψ of statement (16) imply that w ≺ a w . Note that statements w ≺ a w and w ≺ a w are inconsistent because relation ≺ a is a strict partial order.The proof in the case E = S is similar, but it uses item 7(b) of Definition 2instead of item 6(b). (cid:2) emma 19. If w (cid:13) NH a ϕ and w (cid:13) NS a ψ , then either w (cid:13) N ( ϕ → ¬ ψ ) or w (cid:13) N ( ¬ ψ → ϕ ) .Proof. Suppose that w (cid:49) N ( ϕ → ¬ ψ ) and w (cid:49) N ( ¬ ψ → ϕ ). Thus, by item 4 ofDefinition 2, there are epistemic worlds w , w ∈ W , such that w (cid:49) ϕ → ¬ ψ and w (cid:49) ¬ ψ → ϕ . Hence, by item 3 and item 2 of Definition 2, w (cid:13) ϕ, w (cid:13) ψ, w (cid:49) ψ, w (cid:49) ϕ. (17)At the same time, by the definition of modality N and items 2 and 4 of Defini-tion 2, the assumption w (cid:13) NH a ϕ of the lemma implies that there is a world w (cid:48) such that w (cid:48) (cid:13) H a ϕ . Hence, w ≺ a w by item 6(b) of Definition 2 and parts w (cid:49) ϕ and w (cid:13) ϕ of statement (17).Also, by item 7(b) of Definition 2, the assumption w (cid:13) NS a ψ of the lemmaand parts w (cid:13) ψ and w (cid:49) ψ of statement (17) imply that w ≺ a w . Notethat statements w ≺ a w and w ≺ a w are inconsistent because relation ≺ a is a strict partial order. (cid:2) Lemma 20. If w (cid:13) E a ϕ , then w (cid:49) N ϕ .Proof. First, suppose that E = H . Then, by item 6(c) of Definition 2, theassumption w (cid:13) E a ϕ implies that there is an epistemic world u ∈ W such that u (cid:49) ϕ . Therefore, w (cid:49) N ϕ by item 4 of Definition 2.The proof in the case E = S is similar, but it uses item 7(c) of Definition 2instead of item 6(c). (cid:2) Lemma 21.
If either w (cid:13) NH a ϕ or w (cid:13) NS a ¬ ϕ , then statement w (cid:13) K a ϕ implies w (cid:13) H a ϕ .Proof. First, suppose that w (cid:13) NH a ϕ . Thus, by the definition of modality N and items 2 and 4 of Definition 2, there is an epistemic world w (cid:48) ∈ W such that w (cid:48) (cid:13) H a ϕ . Hence, by item 6 of Definition 2,(a) u (cid:13) ϕ for each world u ∈ W such that w (cid:48) ∼ a u ,(b) for any two worlds u, u (cid:48) ∈ W , if u (cid:49) ϕ and u (cid:48) (cid:13) ϕ , then u ≺ a u (cid:48) ,(c) there is a world u ∈ W such that u (cid:49) ϕ .Also, by item 5 of Definition 2, the assumption w (cid:13) K a ϕ implies that(a’) u (cid:13) ϕ for each world u ∈ W such that w ∼ a u .By item 6 of Definition 2, statements (a’), (b), and (c) imply w (cid:13) H a ϕ .Next, suppose that w (cid:13) NS a ¬ ϕ . Thus, by the definition of modality N anditems 2 and 4 of Definition 2, there is an epistemic world w (cid:48) ∈ W such that w (cid:48) (cid:13) S a ¬ ϕ . Hence, by item 7 of Definition 2,(a) u (cid:13) ¬ ϕ for each world u ∈ W such that w (cid:48) ∼ a u ,27b) for any two worlds u, u (cid:48) ∈ W , if u (cid:13) ¬ ϕ and u (cid:48) (cid:49) ¬ ϕ , then u ≺ a u (cid:48) ,(c) there is a world u ∈ W such that u (cid:49) ¬ ϕ .Also, by item 5 of Definition 2, the assumption w (cid:13) K a ϕ implies that(a’) u (cid:13) ϕ for each world u ∈ W such that w ∼ a u .Note that by item 2 of Definition 2, statement (b) implies that(b’) for any two worlds u, u (cid:48) ∈ W , if u (cid:49) ϕ and u (cid:48) (cid:13) ϕ , then u ≺ a u (cid:48) .And, by item 2 of Definition 2, statement (a) implies that(c’) w (cid:48) (cid:49) ϕ because relation ∼ a is reflexive. Finally, note that by item 6 of Definition 2,statements (a’), (b’), and (c’) imply w (cid:13) H a ϕ . (cid:2) The proof of the next lemma is using the converse models and translation τ that have been introduced in Definition 3 and Definition 4 respectively. Lemma 22.
If either w (cid:13) NH a ¬ ϕ or w (cid:13) NS a ϕ , then statement w (cid:13) K a ϕ implies w (cid:13) S a ϕ .Proof. Let M c be the converse model of the epistemic model with preferences M and (cid:13) c is the satisfaction relation for the model M c . By Lemma 21, if ei-ther w (cid:13) c NH a τ ( ϕ ) or w (cid:13) c NS a ¬ τ ( ϕ ), then statement w (cid:13) c K a τ ( ϕ ) implies w (cid:13) c H a τ ( ϕ ). Thus, by Definition 4, if either w (cid:13) c τ ( NS a ϕ ) or w (cid:13) c τ ( NH a ¬ ϕ ),then statement w (cid:13) c τ ( K a ϕ ) implies w (cid:13) c τ ( S a ϕ ). Therefore, by Theorem 2, ifeither w (cid:13) NS a ϕ or w (cid:13) NH a ¬ ϕ , then statement w (cid:13) K a ϕ implies w (cid:13) S a ϕ . (cid:2) Lemma 23. If w (cid:13) N ( ϕ ↔ ψ ) and w (cid:13) E a ϕ , then w (cid:13) E a ψ .Proof. First, we consider the case E = H . By item 6 of Definition 2, theassumption w (cid:13) H a ϕ implies that(a) u (cid:13) ϕ for each world u ∈ W such that w ∼ a u ,(b) for any two worlds u, u (cid:48) ∈ W , if u (cid:49) ϕ and u (cid:48) (cid:13) ϕ , then u ≺ a u (cid:48) ,(c) there is a world u ∈ W such that u (cid:49) ϕ .Thus, by the assumption w (cid:13) N ( ϕ ↔ ψ ) and item 4 of Definition 2,(a’) u (cid:13) ψ for each world u ∈ W such that w ∼ a u ,(b’) for any two worlds u, u (cid:48) ∈ W , if u (cid:49) ψ and u (cid:48) (cid:13) ψ , then u ≺ a u (cid:48) ,(c’) there is a world u ∈ W such that u (cid:49) ψ .By item 6 of Definition 2, statements (a’), (b’), and (c’) imply w (cid:13) H a ψ .The proof in the case E = S is similar, but it uses item 7 of Definition 2instead of item 6. (cid:2) The strong soundness theorem below follows from the lemmas proven above.
Theorem 4.
For any epistemic world w of an epistemic model with preferences,any set of formulae X ⊆ Φ , and any formula ϕ ∈ Φ , if w (cid:13) χ for each formula χ ∈ X and X (cid:96) ϕ , then w (cid:13) ϕ . (cid:2)
1. Utilitarian Emotions
Lang, van er Torre, and Weydert introduced a notion of utilitarian desirewhich is based on a utility function rather than a preference relation [2]. Al-though desire, as an emotion, is different from the happiness and sadness emo-tions that we study in this article, their approach could be adopted to happinessand sadness as well. To do this, one needs to modify Definition 1 to includeagent-specific utility functions instead of agent-specific preference relations:
Definition 5.
A tuple ( W, {∼ a } a ∈A , { u a } a ∈A , π ) is called an epistemic modelwith utilities if W is a set of epistemic worlds, ∼ a is an “indistinguishability” equivalence relation on set W for eachagent a ∈ A , u a is a “utility” function from set W to real numbers for each agent a ∈ A , π ( p ) is a subset of W for each propositional variable p . Below is the definition of the satisfaction relation for epistemic model withutilities. Its parts 6(b) and 7(b) are similar to Lang, van er Torre, and Weydertutilitarian desire definition in [2]. Unlike the current article, [2] does not proveany completeness results. In the definition below we assume that language Φ ismodified to incorporate a no-negative real “degree” parameter into modalities H da and S da . We read statement H da ϕ as “agent a is happy about ϕ with degree d ”. Similarly, we read S da ϕ as “agent a is sad about ϕ with degree d ”. Definition 6.
For any epistemic model with utilities ( W, {∼ a } a ∈A , { u a } a ∈A , π ) ,any world w ∈ W , and any formula ϕ ∈ Φ , satisfaction relation w (cid:13) ϕ is definedas follows: w (cid:13) p , if w ∈ π ( p ) , w (cid:13) ¬ ϕ , if w (cid:49) ϕ , w (cid:13) ϕ → ψ , if w (cid:49) ϕ or w (cid:13) ψ , w (cid:13) N ϕ , if v (cid:13) ϕ for each world v ∈ W , w (cid:13) K a ϕ , if v (cid:13) ϕ for each world v ∈ W such that w ∼ a v , w (cid:13) H da ϕ , if the following three conditions are satisfied: (a) v (cid:13) ϕ for each world v ∈ W such that w ∼ a v , (b) for any v, v (cid:48) ∈ W , if v (cid:49) ϕ and v (cid:48) (cid:13) ϕ , then u a ( v ) + d ≤ u a ( v (cid:48) ) , (c) there is a world v ∈ W such that v (cid:49) ϕ , w (cid:13) S da ϕ , if the following three conditions are satisfied: (a) v (cid:13) ϕ for each world v ∈ W such that w ∼ a u , (b) for any worlds v, v (cid:48) ∈ W , if v (cid:13) ϕ and v (cid:48) (cid:49) ϕ , then u a ( v )+ d ≤ u a ( v (cid:48) ) , (c) there is a world u ∈ W such that u (cid:49) ϕ . We have already defined utility functions for our Battle of Cuisines scenario,see Section 5. In the two propositions below we use this scenario to illustrateutilitarian happiness modality. Note how Sanaz is much happier to be withPavel in a Russian restaurant than she is to be with him in a restaurant.29 roposition 28. ( x, x ) (cid:13) H s ( “Sanaz and Pavel are in the same restaurant” ) ,where x ∈ { Iranian , Russian } .Proof. We verify conditions (a), (b), and (c) from item 6 of Definition 6:
Condition a:
Consider any epistemic world ( y, z ) such that ( x, x ) ∼ s ( y, z ).It suffices to show that ( y, z ) (cid:13) “Sanaz and Pavel are in the same restaurant”.The latter is true because assumption ( x, x ) ∼ s ( y, z ) implies that x = y and x = z in the perfect information setting of the Battle of Cuisines scenario. Condition b:
Consider any worlds ( y, z ) and ( y (cid:48) , z (cid:48) ) such that( y, z ) (cid:49) “Sanaz and Pavel are in the same restaurant” , (18)( y (cid:48) , z (cid:48) ) (cid:13) “Sanaz and Pavel are in the same restaurant” . (19)Statement (18) implies that u s ( y, z ) = 0, see Table 1. Similarly, statement (19)implies that u s ( y (cid:48) , z (cid:48) ) ≥
1. Therefore, u s ( y, z ) + 1 = 1 ≤ u s ( y (cid:48) , z (cid:48) ). Condition c: (Russian , Iranian) (cid:49) “Sanaz and Pavel are in the same restaurant”. (cid:2)
Proposition 29. (Russian , Russian) (cid:13) H s ( “Sanaz and Pavel are in the Russian restaurant” ) . Proof.
Conditions (a) and (c) from item 6 of Definition 6 could be verifiedsimilarly to the proof of Proposition 28. Below we verify condition (b).Consider any worlds ( y, z ) and ( y (cid:48) , z (cid:48) ) such that( y, z ) (cid:49) “Sanaz and Pavel are in the Russian restaurant” , (20)( y (cid:48) , z (cid:48) ) (cid:13) “Sanaz and Pavel are in the Russian restaurant” . (21)Statement (20) implies that u s ( y, z ) ≤
1, see Table 1. Similarly, statement (21)implies that u s ( y (cid:48) , z (cid:48) ) = 3. Therefore, u s ( y, z ) + 2 ≤ u s ( y (cid:48) , z (cid:48) ). (cid:2)
12. Goodness-Based Emotions
Lorini and Schwarzentruber proposed a different framework for defining emo-tions [6]. Instead of specifying preference relations on the epistemic worlds theylabel some of worlds as desirable or “good”. In such a setting they definemodalities “rejoice” and “disappointment” that are similar to our modalities“happiness” and “sadness”. In this section we compare their approach to ours.Although their framework endows agents with actions, it appears that actionsare essential for defining regret and are less important for capturing rejoice anddisappointment. In the definition below, we simplify Lorini and Schwarzen-truber’s framework to action-less models that we call epistemic model withgoodness. 30 efinition 7.
A tuple ( W, {∼ a } a ∈A , { G a } a ∈A , π ) is called an epistemic modelwith goodness if W is a set of epistemic worlds, ∼ a is an “indistinguishability” equivalence relation on set W for eachagent a ∈ A , G a ⊆ W is a nonempty set of “good” epistemic worlds for agent a ∈ A , π ( p ) is a subset of W for each propositional variable p . To represent the gift example from Figure 1 as an epistemic model withgoodness, we need to specify the sets of good epistemic worlds G s and G p ofSanaz and Pavel. A natural way to do this is to assume that G s = G p = { w } .In other words, the ideal outcome for both of them would be if the gift and thecard arrives to the recipients.In the lottery example, the desirable outcome for each agent is when theagents wins the lottery. In other words, G s = { u } , G p = { w } , and G o = { v } ,see Figure 3.In the Battle of Cuisines example captured in Table 1, the choice of goodepistemic worlds is not obvious. On one hand, we can assume that good worldsfor both Sanaz and Pavel are the ones where they have positive pay-offs. Inthis case, G s = G p = { (Iranian , Iranian) , (Russian , Russian) } . Alternatively, wecan choose the good worlds to be those where they get maximal pay-off. Inthat case, G s = { (Russian , Russian) } and G p = { (Iranian , Iranian) } . Note thatour epistemic models with preferences approach provides a more fine-grainedsemantics that does not force the choice between these two alternatives.In the definition below, we rephrase Lorini and Schwarzentruber’s formaldefinitions of “rejoice” and “disappointment” in terms of epistemic models withgoodness. We denote modalities “rejoice” and “disappointment” by H and S respectively to be consistent with the notations in the rest of this article. Definition 8.
For any world w ∈ W of an epistemic model with goodness ( W, {∼ a } a ∈A , { G a } a ∈A , π ) and any formula ϕ ∈ Φ , satisfaction relation w (cid:13) ϕ is defined as follows: w (cid:13) p , if w ∈ π ( p ) , w (cid:13) ¬ ϕ , if w (cid:49) ϕ , w (cid:13) ϕ → ψ , if w (cid:49) ϕ or w (cid:13) ψ , w (cid:13) N ϕ , if u (cid:13) ϕ for each world u ∈ W , w (cid:13) K a ϕ , if u (cid:13) ϕ for each world u ∈ W such that w ∼ a u , w (cid:13) H a ϕ , if the following three conditions are satisfied: (a) u (cid:13) ϕ for each world u ∈ W such that w ∼ a u , (b) u (cid:13) ϕ for each world u ∈ G a , (c) there is a world u ∈ W such that u (cid:49) ϕ , w (cid:13) S a ϕ , if the following three conditions are satisfied: (a) u (cid:13) ϕ for each world u ∈ W such that w ∼ a u , (b) u (cid:49) ϕ for each world u ∈ G a , there is a world u ∈ W such that u (cid:49) ϕ . Consider the discussed above epistemic model with goodness for the giftscenario in which G s = G p = { w } . It is relatively easy to see that all proposi-tions that we proved in Section 4 for preference-based semantics hold true undergoodness-based semantics of modalities H and S given in Definition 8.The situation is different for the Battle of Cuisines scenario. If (cid:13) denotesthe satisfaction relation of the epistemic model with goodness where G s = G p = { (Iranian , Iranian) , (Russian , Russian) } , then the following two propositions aretrue just like they are under our definition of happiness (see Proposition 13 andProposition 12): Proposition 30. (Russian , Russian) (cid:13) H s ( “Sanaz and Pavel are in the same restaurant” ) . Proof.
It suffices to verify conditions (a), (b), and (c) of item 6 of Definition 8:
Condition a:
Since the Battle of the Cuisines is a setting with perfect informa-tion, it suffices to show that(Russian , Russian) (cid:13) (“Sanaz and Pavel are in the same restaurant”) . The last statement is true by the definition of the world (Russian , Russian).
Condition b:
Statement “Sanaz and Pavel are in the same restaurant” is satisfiedin both good worlds: (Iranian , Iranian) and (Russian , Russian).
Condition c:
Statement “Sanaz and Pavel are in the same restaurant” is notsatisfied in the world (Russian , Iranian). (cid:2)
Proposition 31. (Russian , Russian) (cid:49) H s ( “Sanaz is in the Russian restaurant” ) . Proof.
Note that (Iranian , Iranian) is a good world, in which statement “Sanazis in the Russian restaurant” is not satisfied. Therefore, the statement of theproposition is true by item 6(b) of Definition 8. (cid:2)
However, for the satisfaction relation (cid:13) of the epistemic model with good-ness where G s = { (Russian , Russian) } and G p = { (Iranian , Iranian) } , the situ-ation is different: Proposition 32. (Russian , Russian) (cid:13) H s ( “Sanaz and Pavel are in the same restaurant” ) . Proof.
The proof is similar to the proof of Proposition 30 except that in
Con-dition b we only need to consider the world (Russian , Russian). (cid:2) roposition 33. (Russian , Russian) (cid:13) H s ( “Sanaz is in the Russian restaurant” ) . Proof.
It suffices to verify conditions (a), (b), and (c) of item 6 of Definition 8:
Condition a:
Since the Battle of the Cuisines is a setting with perfect informa-tion, it suffices to show that(Russian , Russian) (cid:13) (“Sanaz is in the Russian restaurant”) . The last statement is true by the definition of the world (Russian , Russian).
Condition b:
Note that in the current setting set G s contains only element(Russian , Russian) and that(Russian , Russian) (cid:13) (“Sanaz is in the Russian restaurant”) . Condition c: (Iranian , Russian) (cid:49) (“Sanaz is in the Russian restaurant”). (cid:2) We conclude this section by an observation that the Coherence of PotentialEmotions axiom is not universally true under goodness-based semantics. Indeed,note that according to Proposition 32 and Proposition 33, there is an epistemicworlds in which Sanaz is happy that “Sanaz is in the Russian restaurant” andthere is an epistemic world in which she is happy that “Sanaz and Pavel are inthe same restaurant”. If the Coherence of Potential Emotions axiom holds inthis setting, then one of these statements would imply the other, but neither ofthem does.
13. Canonical Model
In the rest of this article we prove strong completeness of our logical systemwith respect to the semantics given in Definition 2. As usual, the proof of thecompleteness is based on a construction of a canonical model. In this section,for any maximal consistent set of formulae X ⊆ Φ, we define canonical epistemicmodel with preferences M ( X ) = ( W, {∼ a } a ∈A , {≺ a } a ∈A , π ).As it is common in modal logic, we define worlds as maximal consistent setsof formulae. Since the meaning of modality N in our system is “for all worlds”,we require all worlds in the canonical model to have the same N -formulae. Weachieve this through the following definition. Definition 9. W is the set of all such maximal consistent sets of formulae Y that { ϕ ∈ Φ | N ϕ ∈ X } ⊆ Y . Note that although the above definition only requires all N -formulae fromset X to be in set Y , it is possible to show that the converse is also true due tothe presence of the Negative Introspection axiom for modality N in our system. Lemma 24. X ∈ W . roof. Consider any formula N ϕ ∈ X . By Definition 9, it suffices to show that ϕ ∈ X . Indeed, the assumption N ϕ ∈ X implies that X (cid:96) ϕ by the Truthaxiom and the Modus Ponens inference rule. Therefore, ϕ ∈ X because set X is maximal. (cid:2) Definition 10.
For any worlds w, u ∈ W , let w ∼ a u if ϕ ∈ u for each formula K a ϕ ∈ w . Alternatively, one can define w ∼ a u if sets w and u have the same K a -formulae. Our approach results in shorter proofs, but it requires to prove thefollowing lemma. Lemma 25.
Relation ∼ a is an equivalence relation on set W .Proof. Reflexivity:
Consider any formula ϕ ∈ Φ. Suppose that K a ϕ ∈ w .By Definition 10, it suffices to show that ϕ ∈ w . Indeed, assumption K a ϕ ∈ w implies w (cid:96) ϕ by the Truth axiom and the Modus Ponens inference rule.Therefore, ϕ ∈ w because set w is maximal. Symmetry:
Consider any epistemic worlds w, u ∈ W such that w ∼ a u andany formula K a ϕ ∈ u . By Definition 10, it suffices to show ϕ ∈ w . Suppose theopposite. Then, ϕ / ∈ w . Hence, w (cid:48) ϕ because set w is maximal. Thus, w (cid:48) K a ϕ by the contraposition of the Truth axiom. Then, ¬ K a ϕ ∈ w because set w ismaximal. Thus, w (cid:96) K a ¬ K a ϕ by the Negative Introspection axiom and theModus Ponens inference rule. Hence, K a ¬ K a ϕ ∈ w because set w is maximal.Then, ¬ K a ϕ ∈ u by assumption w ∼ a u and Definition 10. Therefore, K a ϕ / ∈ u because set w is consistent, which contradicts the assumption K a ϕ ∈ u . Transitivity:
Consider any epistemic worlds w, u, v ∈ W such that w ∼ a u and u ∼ a v and any formula K a ϕ ∈ w . By Definition 10, it suffices to show ϕ ∈ v .Assumption K a ϕ ∈ w implies w (cid:96) K a K a ϕ by Lemma 11 and the Modus Ponensinference rule. Thus, K a K a ϕ ∈ w because set w is maximal. Hence, K a ϕ ∈ u bythe assumption w ∼ a u and Definition 10. Therefore, ϕ ∈ v by the assumption u ∼ a v and Definition 10. (cid:2) The next step in specifying the canonical model is to define preference re-lation ≺ a for each agent a ∈ A , which we do in Definition 13. Towards thisdefinition, we first introduce the “emotional base” ∆ a for each agent a . The set∆ a contains a formula δ if agent a could either be potentially happy about δ orpotentially sad about ¬ ϕ . Definition 11. ∆ a = { δ ∈ Φ | NH a δ ∈ X } ∪ { δ ∈ Φ | NS a ¬ δ ∈ X } . The next lemma holds because set Φ is countable.
Lemma 26.
Set ∆ a is countable for each agent a ∈ A . (cid:2) Next, we introduce a total pre-order (cid:118) a on emotional base ∆ a of each agent a ∈ A . Note that this pre-order is different from canonical preference relation ≺ a that we introduce in Definition 13.34 efinition 12. For any agent a ∈ A and any two formulae δ, δ (cid:48) ∈ ∆ a , let δ (cid:118) δ (cid:48) if N ( δ → δ (cid:48) ) ∈ X . Lemma 27. N ( ¬ δ (cid:48) → ¬ δ ) ∈ X for any agent a ∈ A and any two formulae δ, δ (cid:48) ∈ ∆ a such that δ (cid:118) δ (cid:48) .Proof. Formula ( δ → δ (cid:48) ) → ( ¬ δ (cid:48) → ¬ δ ) is a propositional tautology. Thus, (cid:96) N (( δ → δ (cid:48) ) → ( ¬ δ (cid:48) → ¬ δ )) by the Necessitation inference rule. Hence, (cid:96) N ( δ → δ (cid:48) ) → N ( ¬ δ (cid:48) → ¬ δ ) (22)by the Distributivity axiom and the Modus Ponens inference rule.Suppose that δ (cid:118) δ (cid:48) . Thus, N ( δ → δ (cid:48) ) ∈ X by Definition 12. Hence, X (cid:96) N ( ¬ δ (cid:48) → ¬ δ ) by statement (22) and the Modus Ponens inference rule.Therefore, N ( ¬ δ (cid:48) → ¬ δ ) ∈ X because set X is maximal. (cid:2) Lemma 28.
For any agent a ∈ A , relation (cid:118) is a total pre-order on set ∆ a .Proof. We need to show that relation (cid:118) is reflexive, transitive, and total.
Reflexivity:
Consider an arbitrary formula δ ∈ Φ. By Definition 12, it sufficesto show that N ( δ → δ ) ∈ X . Indeed, formula δ → δ is a propositional tautology.Thus, (cid:96) N ( δ → δ ) by the Necessitation inference rule. Therefore, N ( δ → δ ) ∈ X because set X is maximal. Transitivity:
Consider arbitrary δ , δ , δ ∈ Φ such that N ( δ → δ ) ∈ X and N ( δ → δ ) ∈ X . By Definition 12, it suffices to show that N ( δ → δ ) ∈ X . Indeed, note that formula ( δ → δ ) → (( δ → δ ) → ( δ → δ )) is apropositional tautology. Thus, (cid:96) N (( δ → δ ) → (( δ → δ ) → ( δ → δ ))) bythe Necessitation inference rule. Hence, by the Distributivity axiom and theModus Ponens inference rule, (cid:96) N ( δ → δ ) → N (( δ → δ ) → ( δ → δ )).Then, X (cid:96) N (( δ → δ ) → ( δ → δ )) by the assumption N ( δ → δ ) ∈ X andthe Modus Ponens inference rule. Thus, X (cid:96) N ( δ → δ ) → N ( δ → δ ) bythe Distributivity axiom and the Modus Ponens inference rule. Hence, by theassumption N ( δ → δ ) ∈ X and the Modus Ponens rule, X (cid:96) N ( δ → δ ).Therefore, N ( δ → δ ) ∈ X because set X is maximal. Totality:
Consider arbitrary formulae δ , δ ∈ ∆ a . By Definition 12, it sufficesto show that either N ( δ → δ ) ∈ X or N ( δ → δ ) ∈ X . By Definition 11,without loss of generality, we can assume that on the the following three casestakes place: Case I: NH a δ , NH a δ ∈ X . Then, X (cid:96) N ( δ → δ ) ∨ N ( δ → δ ) by the firstCoherence of Potential Emotions axiom and propositional reasoning. Therefore,because set X is maximal, either N ( δ → δ ) ∈ X or N ( δ → δ ) ∈ X . Case II: NS a ¬ δ , NS a ¬ δ ∈ X . Similarly to the previous case, we can show thateither N ( ¬ δ → ¬ δ ) ∈ X or N ( ¬ δ → ¬ δ ) ∈ X . Without loss of generality,suppose that N ( ¬ δ → ¬ δ ) ∈ X . Note that formula ( ¬ δ → ¬ δ ) → ( δ → δ )is a propositional tautology. Thus, (cid:96) N (( ¬ δ → ¬ δ ) → ( δ → δ )) by the35ecessitation inference rule. Hence, (cid:96) N ( ¬ δ → ¬ δ ) → N ( δ → δ ) by theDistributivity axiom and the Modus Ponens rule. Thus, X (cid:96) N ( δ → δ ) bythe assumption N ( ¬ δ → ¬ δ ) ∈ X . Therefore, N ( δ → δ ) ∈ X because set X is maximal. Case III: NH a δ , NS a ¬ δ ∈ X . Thus, by the second Coherence of PotentialEmotions axiom and propositional reasoning, X (cid:96) N ( δ → ¬¬ δ ) ∨ N ( ¬¬ δ → δ ). Hence, either N ( δ → ¬¬ δ ) ∈ X or N ( ¬¬ δ → δ ) ∈ X because set X is consistent. Then using an argument similar to the one in Case II andpropositional tautologies( δ → ¬¬ δ ) → ( δ → δ ) and ( ¬¬ δ → δ ) → ( δ → δ )one can conclude that either N ( δ → δ ) ∈ X or N ( δ → δ ) ∈ X . (cid:2) We now are ready to define preference relation ≺ a on epistemic worlds ofthe canonical model. Definition 13. w ≺ a u if there is a formula δ ∈ ∆ a such that δ / ∈ w and δ ∈ u . Note that the transitivity of the relation ≺ a is not obvious. We prove it asa part of the next lemma. Lemma 29. ≺ a is a strict partial order on W .Proof. Irreflexivity . Suppose that w ≺ a w for some world w ∈ W . Thus, byDefinition 13, there exists formula δ ∈ ∆ a such that δ / ∈ w and δ ∈ w , which isa contradiction. Transitivity . Consider any worlds w, u, v ∈ W such that w ≺ a u and u ≺ a v .It suffices to prove that w ≺ a v . Indeed, by Definition 13, assumptions w ≺ a u and u ≺ a v imply that there are formulae δ , δ in ∆ a such that δ / ∈ w, δ ∈ u, δ / ∈ u, and δ ∈ v. (23)By Lemma 28, either δ (cid:118) δ or δ (cid:118) δ . We consider these two cases separately. Case I: δ (cid:118) δ . Then, N ( δ → δ ) ∈ X by Definition 13. Hence, δ → δ ∈ u by Definition 9. Thus, u (cid:96) δ by the part δ ∈ u of statement (23) and theModus Ponens inference rule. Therefore, δ ∈ u because set u is maximal,which contradicts the part δ / ∈ u of statement (23). Case II: δ (cid:118) δ . Then, N ( δ → δ ) ∈ X by Definition 13. Thus, δ → δ ∈ v by Definition 9. Hence, v (cid:96) δ by the part δ ∈ v of statement (23) and theModus Ponens inference rule. Then, δ ∈ v because set v is maximal. Therefore, w ≺ a v by Definition 13 and the part δ / ∈ w of statement (23). (cid:2) Definition 14. π ( p ) = { w ∈ W | p ∈ w } . This concludes the definition of the canonical epistemic model with prefer-ences M ( X ) = ( W, {∼ a } a ∈A , {≺ a } a ∈A , π ).36
4. Properties of a Canonical Model
As usual the proof of the completeness is centered around an “induction”or “truth” lemma. In our case, this is Lemma 40. We precede this lemmawith several auxiliary lemmas that are used in the induction step of the proof ofLemma 40. For the benefit of the reader, we grouped these auxiliary lemmas intoseveral subsections. Throughout this section up to and including Lemma 40, weassume a fixed canonical model M ( X ). N Lemma 30.
For any worlds w, u ∈ W and any formula ϕ ∈ Φ , if N ϕ ∈ w ,then ϕ ∈ u .Proof. Suppose that ϕ / ∈ u . Thus, u (cid:48) ϕ because set u is maximal. Hence, N ϕ / ∈ u by the Truth axiom. Then, NN ϕ / ∈ X by Definition 9. Thus, X (cid:48) NN ϕ because set X is maximal. Hence, N ϕ / ∈ X by Lemma 11. Then, ¬ N ϕ ∈ X because set X is maximal. Thus, X (cid:96) N ¬ N ϕ by the Negative Introspectionaxiom and the Modus Ponens inference rule. Hence, N ¬ N ϕ ∈ X because set X is maximal. Then, ¬ N ϕ ∈ w by Definition 9. Therefore, N ϕ / ∈ w becauseset w is consistent. (cid:2) Lemma 31.
For any world w ∈ W and any formula ϕ ∈ Φ , if N ϕ / ∈ w , thenthere is a world u ∈ W such that ϕ / ∈ u .Proof. Consider the set X = {¬ ϕ } ∪ { ψ | N ψ ∈ X } . We start by show-ing that set X is consistent. Suppose the opposite. Then, there are formulae N ψ , . . . , N ψ n ∈ X such that ψ , . . . , ψ n (cid:96) ϕ . Thus, N ψ , . . . , N ψ n (cid:96) N ϕ byLemma 10. Hence, X (cid:96) N ϕ because N ψ , . . . , N ψ n ∈ X . Then, X (cid:96) NN ϕ by Lemma 11 and the Modus Ponens inference rule. Thus, NN ϕ ∈ X be-cause set X is maximal. Hence, N ϕ ∈ w by Definition 9 which contradicts theassumption N ϕ / ∈ w of the lemma. Therefore, set X is consistent.Let set u be a maximum consistent extension of set X . Such set exists byLemma 14. Note that u ∈ W by Definition 9 and the choice of sets X and u .Also, ¬ ϕ ∈ X ⊆ u by the choice of sets X and u . Therefore, ϕ / ∈ u because set u is consistent. (cid:2) K Lemma 32.
For any agent a ∈ A , any worlds w, u ∈ W , and any formula ϕ ∈ Φ , if K a ϕ ∈ w and w ∼ a u , then ϕ ∈ u .Proof. Assumptions K a ϕ ∈ w and w ∼ a u imply ϕ ∈ u by Definition 10. (cid:2) Lemma 33.
For any agent a ∈ A , any world w ∈ W , and any formula ϕ ∈ Φ ,if K a ϕ / ∈ w , then there is a world u ∈ W such that w ∼ a u and ϕ / ∈ u . roof. First, we show that the following set of formulae is consistent X = {¬ ϕ } ∪ { ψ | K a ψ ∈ w } ∪ { χ | N χ ∈ X } . (24)Assume the opposite. Then, there are formulae K a ψ , . . . , K a ψ m ∈ w (25)and formulae N χ , . . . , N χ n ∈ X (26)such that ψ , . . . , ψ m , χ , . . . , χ n (cid:96) ϕ. Thus, by Lemma 10, K a ψ , . . . , K a ψ m , K a χ , . . . , K a χ n (cid:96) K a ϕ. Hence, by assumption (25), w, K a χ , . . . , K a χ n (cid:96) K a ϕ. (27)Consider any integer i ≤ n . Note that N χ i → K a χ i is an instance of theKnowledge of Necessity axiom. Then, (cid:96) N ( N χ i → K a χ i ) by the Necessitationinference rule. Thus, (cid:96) NN χ i → NK a χ i by the Distributivity axiom and theModus Ponens inference rule. Note that (cid:96) N ϕ → NN ϕ by Lemma 11. Hence, (cid:96) N χ i → NK a χ i by the laws of propositional reasoning. Then, X (cid:96) NK a χ by assumption (26). Thus, NK a χ ∈ X because set X is maximal. Hence, K a χ i ∈ w by Definition 9 for any integer i ≤ n . Then, statement (27) impliesthat w (cid:96) K a ϕ . Thus, K a ϕ ∈ w because set w is maximal, which contradictsassumption K a ϕ / ∈ w of the lemma. Therefore, set X is consistent.By Lemma 14, set X can be extended to a maximal consistent set u . Then, { χ | N χ ∈ X } ⊆ X ⊆ u by equation (24). Thus, u ∈ W by Definition 9. Also, { ψ | K a ψ ∈ w } ⊆ X ⊆ u by equation (24). Hence, w ∼ a u by Definition 10.Finally, ¬ ϕ ∈ X ⊆ u also by equation (24). Therefore, ϕ / ∈ u because set u isconsistent. (cid:2) H and S Recall that by E we denote one of the two emotional modalities: H and S . Lemma 34.
For any agent a ∈ A , any worlds w, u ∈ W , and any formula ϕ ∈ Φ , if E a ϕ ∈ w and w ∼ a u , then ϕ ∈ u .Proof. By Lemma 8 and the Modus Ponens inference rule, the assumption E a ϕ ∈ w implies w (cid:96) K a ϕ . Thus, K a ϕ ∈ w because set w is maximal. There-fore, ϕ ∈ u by Lemma 32 and the assumption w ∼ a u . (cid:2) emma 35. For any world w ∈ W , and any formula E a ϕ ∈ w , there is a world u ∈ W such that ϕ / ∈ u .Proof. By the Counterfactual axiom and the Modus Ponens inference rule, as-sumption E a ϕ ∈ w implies w (cid:96) ¬ N ϕ . Thus, N ϕ / ∈ w because set w is consistent.Therefore, by Lemma 31, there is a world u ∈ W such that ϕ / ∈ u . (cid:2) H Lemma 36.
For any agent a ∈ A , any worlds w, u, u (cid:48) ∈ W , and any formula ϕ ∈ Φ , if H a ϕ ∈ w , ϕ / ∈ u and ϕ ∈ u (cid:48) , then u ≺ a u (cid:48) .Proof. The assumption H a ϕ ∈ w implies ¬ H a ϕ / ∈ w because set w is consistent.Thus, N ¬ H a ϕ / ∈ X by Definition 9 because w ∈ W . Hence, ¬ N ¬ H a ϕ ∈ X because set X is maximal. Then, NH a ϕ ∈ X by the definition of modality N .Thus, ϕ ∈ ∆ a by Definition 11. Therefore, u ≺ a u (cid:48) by Definition 13 and theassumptions ϕ / ∈ u and ϕ ∈ u (cid:48) of the lemma. (cid:2) Lemma 37.
For any agent a ∈ A , any world w ∈ W , and any formula ϕ ∈ Φ ,if H a ϕ / ∈ w , K a ϕ ∈ w , and N ϕ / ∈ w , then there are worlds u, u (cid:48) ∈ W such that ϕ / ∈ u , ϕ ∈ u (cid:48) , and u (cid:54)≺ a u (cid:48) .Proof. By Lemma 28, relation (cid:118) forms a total pre-order on set ∆ a . ByLemma 26, set ∆ a is countable. Thus, by the axiom of countable choice, thereis an ordering of all formulae in set ∆ a that agrees with pre-order (cid:118) . Generallyspeaking, such ordering is not unique. We fix any such ordering: δ (cid:118) δ (cid:118) δ (cid:118) δ (cid:118) . . . (28)If set ∆ a is finite, the above ordering has some finite number n of elements. Inthis case, the ordering is isomorphic to ordinal n . Otherwise, it is isomorphicto ordinal ω . Let α be the ordinal which is the type of ordering (28). Ordinal α is either finite or is equal to ω .For any ordinal k ≤ α , we consider set Y k = { ϕ } ∪ {¬ δ i | i < k } ∪ { ψ | N ψ ∈ X } . (29) Claim 1.
If there is no finite ordinal k < α such that Y k is consistent and Y k +1 is inconsistent, then Y α is consistent.Proof of Claim. To prove that Y α is consistent, it suffices to show that Y k isconsistent for each ordinal k ≤ α . We prove this by transfinite induction. Zero Case:
Suppose that Y is not consistent. Thus, there are formulae N ψ , . . . , N ψ n ∈ X such that ψ , . . . , ψ n (cid:96) ¬ ϕ . Hence, N ψ , . . . , N ψ n (cid:96) N ¬ ϕ by Lemma 10. Then, X (cid:96) N ¬ ϕ by the assumption N ψ , . . . , N ψ n ∈ X . Thus,because set X is maximal, N ¬ ϕ ∈ X . (30)39ence, ¬ ϕ ∈ w , by Definition 9. Then, w (cid:96) ¬ K a ϕ by the contraposition of theTruth axiom and propositional reasoning. Therefore, K a ϕ / ∈ w because set w isconsistent, which contradicts the assumption K a ϕ ∈ w of lemma. Successor Case:
Suppose that set Y k is consistent for some k < α . By theassumption of the claim, there is no finite ordinal k < α such that Y k is consistentand Y k +1 is inconsistent. Therefore, set Y k +1 is consistent. Limit Case:
Suppose that set Y ω is not consistent. Thus, there are formulae N ψ , . . . , N ψ n ∈ X and some finite k < ω such that ψ , . . . , ψ n , ¬ δ , . . . , ¬ δ k − (cid:96)¬ ϕ . Therefore, set Y k is not consistent. (cid:2) Let k (cid:48) be any finite ordinal k (cid:48) < α such that Y k (cid:48) is consistent and Y k (cid:48) +1 isinconsistent. If such finite ordinal does not exist, then let k (cid:48) be ordinal α . Notethat in either case, set Y k (cid:48) is consistent by Claim 1. Let u (cid:48) be any maximalconsistent extension of set Y k (cid:48) . Such extension exists by Lemma 14. Note that ϕ ∈ Y k (cid:48) ⊆ u (cid:48) by equation (29) and the choice of set u (cid:48) . Claim 2. u (cid:48) ∈ W .Proof of Claim. Consider any formula N ψ ∈ X . By Definition 9, it sufficesto show that ψ ∈ u (cid:48) . Indeed, ψ ∈ Y k (cid:48) by equation (29) and the assumption N ψ ∈ X . Thus, ψ ∈ u (cid:48) because Y k (cid:48) ⊆ u (cid:48) by the choice of set u (cid:48) . (cid:2) Consider the following set of formulae: Z = {¬ ϕ } ∪ { δ i | k (cid:48) ≤ i < α } ∪ { ψ | N ψ ∈ X } . (31) Claim 3.
Set Z is consistent.Proof of Claim. We consider the following two cases separately:
Case 1: k (cid:48) < α . Suppose that set Z is not consistent. Thus, there are finiteordinals m < α and n < ω and formulae N ψ , . . . N ψ n ∈ X (32)such that k (cid:48) ≤ m and δ k (cid:48) , δ k (cid:48) +1 , . . . , δ m , ψ , . . . , ψ n (cid:96) ϕ. Hence, by the Modus Ponens inference rule applied m − k (cid:48) times, δ k (cid:48) , δ k (cid:48) → δ k (cid:48) +1 , δ k (cid:48) +1 → δ k (cid:48) +2 , δ k (cid:48) +2 → δ k (cid:48) +3 , . . . , δ m − → δ m , ψ , . . . , ψ n (cid:96) ϕ. Then, by Lemma 9, δ k (cid:48) → δ k (cid:48) +1 , δ k (cid:48) +1 → δ k (cid:48) +2 , . . . , δ m − → δ m , ψ , . . . , ψ n (cid:96) δ k (cid:48) → ϕ. Thus, by Lemma 10, N ( δ k (cid:48) → δ k (cid:48) +1 ) , N ( δ k (cid:48) +1 → δ k (cid:48) +2 ) , . . . , N ( δ m − → δ m ) , N ψ , . . . , N ψ n (cid:96) N ( δ k (cid:48) → ϕ ) . δ k (cid:48) (cid:118) δ k (cid:48) +1 (cid:118) · · · (cid:118) δ m by assumption (28). Hence, it follows that N ( δ k (cid:48) → δ k (cid:48) +1 ) , N ( δ k (cid:48) +1 → δ k (cid:48) +2 ) , . . . , N ( δ m − → δ m ) ∈ X by Definition 12.Then, X , N ψ , . . . , N ψ n (cid:96) N ( δ k (cid:48) → ϕ ) . Thus, by assumption (32), X (cid:96) N ( δ k (cid:48) → ϕ ) . (33)At the same time, k (cid:48) < α by the assumption of the case. Hence, set Y k (cid:48) +1 is notconsistent by the choice of the finite ordinal k (cid:48) . Then, by equation (29), theremust exist formulae N ψ (cid:48) , . . . , N ψ (cid:48) p ∈ X (34)such that ¬ δ , ¬ δ , ¬ δ , . . . , ¬ δ k (cid:48) , ψ (cid:48) , . . . , ψ (cid:48) p (cid:96) ¬ ϕ. In other words, ¬ δ k (cid:48) , ¬ δ k (cid:48) − , ¬ δ k (cid:48) − , . . . , ¬ δ , ψ (cid:48) , . . . , ψ (cid:48) p (cid:96) ¬ ϕ. Thus, by applying the Modus Ponens inference rule k (cid:48) times, ¬ δ k (cid:48) , ¬ δ k (cid:48) → ¬ δ k (cid:48) − , ¬ δ k (cid:48) − → ¬ δ k (cid:48) − , . . . , ¬ δ → ¬ δ , ψ (cid:48) , . . . , ψ (cid:48) p (cid:96) ¬ ϕ. Hence, by Lemma 9, ¬ δ k (cid:48) → ¬ δ k (cid:48) − , ¬ δ k (cid:48) − → ¬ δ k (cid:48) − , . . . , ¬ δ → ¬ δ , ψ (cid:48) , . . . , ψ (cid:48) n (cid:96) ¬ δ k (cid:48) → ¬ ϕ. Then, by Lemma 10, N ( ¬ δ k (cid:48) → ¬ δ k (cid:48) − ) , N ( ¬ δ k (cid:48) − → ¬ δ k (cid:48) − ) , . . . , N ( ¬ δ → ¬ δ ) , N ψ (cid:48) , . . . , N ψ (cid:48) n (cid:96) N ( ¬ δ k (cid:48) → ¬ ϕ ) . Recall that δ (cid:118) δ (cid:118) · · · (cid:118) δ k (cid:48) by assumption (28). Thus, it follows that N ( ¬ δ → ¬ δ ) , N ( ¬ δ → ¬ δ ) , . . . , N ( ¬ δ k (cid:48) → ¬ δ k (cid:48) − ) ∈ X by Lemma 27.Hence, X , N ψ (cid:48) , . . . , N ψ (cid:48) n (cid:96) N ( ¬ δ k (cid:48) → ¬ ϕ ) . Then, by assumption (34), X (cid:96) N ( ¬ δ k (cid:48) → ¬ ϕ ) . (35)Thus, by item 1 of Lemma 12 and statement (33), X (cid:96) N ( δ k (cid:48) ↔ ϕ ) . (36)Note that δ k (cid:48) ∈ ∆ a because (28) is an ordering of set ∆ a . Hence, by Def-inition 11, either NH a δ k (cid:48) ∈ X or NS a ¬ δ k (cid:48) ∈ X . Then, by items 1 and 2of Lemma 13 and statement (44), either X (cid:96) NH a ϕ or X (cid:96) NS a ¬ ϕ . Thus,either X (cid:96) NNH a ϕ or X (cid:96) NNS a ¬ ϕ by the definition of modality N , the41egative Introspection axiom, and the Modus Ponens inference rule. Hence,either NNH a ϕ ∈ X or NNS a ¬ ϕ ∈ X because set X is maximal. Then, either NH a ϕ ∈ w or NS a ¬ ϕ ∈ w by Definition 9 and assumption w ∈ W of the lemma.Thus, w (cid:96) K a ϕ → H a ϕ by the first Emotional Predictability axiom and propo-sitional reasoning. Hence, w (cid:96) H a ϕ by assumption K a ϕ of the lemma and theModus Ponens inference rule. Therefore, H a ϕ ∈ w because set w is maximal,which contradicts assumption H a ϕ / ∈ w of the lemma. Case 2: α ≤ k (cid:48) . Recall that k (cid:48) ≤ α by the choice of ordinal k (cid:48) made after theend of the proof of Claim 1. Thus, k (cid:48) = α . Hence, Z = {¬ ϕ } ∪ { ψ | N ψ ∈ X } by equation (31). Then, inconsistency of set Z implies that there are formulae N ψ , . . . N ψ n ∈ X (37)such that ψ , . . . , ψ n (cid:96) ϕ . Thus, N ψ , . . . , N ψ n (cid:96) N ϕ by Lemma 10. Hence, X (cid:96) N ϕ (38)by the assumption (37). Then, X (cid:96) NN ϕ by Lemma 11 and the Modus Ponensinference rule. Hence, NN ϕ ∈ X because set X is maximal. Therefore, N ϕ ∈ w by Definition 9, which contradicts the assumption of the lemma. (cid:2) Let u be any maximal consistent extension of set Z . Note that ¬ ϕ ∈ Z ⊆ u byequation (31) and the choice of set u . Claim 4. u ∈ W .Proof of Claim. Consider any formula N ψ ∈ X . By Definition 9, it suffices toshow that ψ ∈ u . Indeed, ψ ∈ Z by equation (31) and the assumption N ψ ∈ X .Thus, ψ ∈ u because Z ⊆ u by the choice of set u . (cid:2) Claim 5. u (cid:54)≺ a u (cid:48) .Proof of Claim. Suppose that u ≺ a u (cid:48) . Thus, by Definition 13, there is a formula δ ∈ ∆ a such that δ / ∈ u and δ ∈ u (cid:48) . Recall that (28) is an ordering of the set∆ a . Hence, there must exist an integer i < α such that δ i / ∈ u and δ i ∈ u (cid:48) . (39)We consider the following two cases separately: Case 1: i < k (cid:48) . Then, ¬ δ i ∈ Y k (cid:48) ⊆ u (cid:48) by equation (29) and the choice of set u (cid:48) .Thus, δ i / ∈ u (cid:48) because set u (cid:48) is consistent, which contradicts to statement (39). Case 2: k (cid:48) ≤ i . Then, δ i ∈ Z ⊆ u by equation (31) and the choice of set u ,which contradicts to statement (39). (cid:2) This concludes the proof of the lemma. (cid:2) S Lemma 38.
For any agent a ∈ A , any worlds w, u, u (cid:48) ∈ W , and any formula ϕ ∈ Φ , if S a ϕ ∈ w , ϕ ∈ u and ϕ / ∈ u (cid:48) , then u ≺ a u (cid:48) .Proof. Note that ϕ ↔ ¬¬ ϕ is a propositional tautology. Thus, (cid:96) N ( ϕ ↔ ¬¬ ϕ )by the Necessitation inference rule. Hence, (cid:96) S a ϕ → S a ¬¬ ϕ by the Substi-tution axiom and the Modus Ponens inference rule. Then, w (cid:96) S a ¬¬ ϕ bythe Modus Ponens inference rule and the assumption S a ϕ ∈ w of the lemma.Thus, ¬ S a ¬¬ ϕ / ∈ w because set w is consistent. Hence, N ¬ S a ¬¬ ϕ / ∈ X byDefinition 9. Then, ¬ N ¬ S a ¬¬ ϕ ∈ X because set X is maximal. Thus, NS a ¬¬ ϕ ∈ X by the definition of modality N . Hence, ¬ ϕ ∈ ∆ a by Defini-tion 11. Therefore, u ≺ a u (cid:48) by Definition 13 and the assumptions ¬ ϕ / ∈ u and ¬ ϕ ∈ u (cid:48) of the lemma. (cid:2) Lemma 39.
For any agent a ∈ A , any world w ∈ W , and any formula ϕ ∈ Φ ,if S a ϕ / ∈ w , K a ϕ ∈ w , and N ϕ / ∈ w , then there are worlds u, u (cid:48) ∈ W such that ϕ ∈ u , ϕ / ∈ u (cid:48) , and u (cid:54)≺ a u (cid:48) .Proof. The proof of this lemma is similar to the proof of Lemma 37. Here weoutline the differences. The choice of ordering (28) and of ordinal α remains thesame. Sets Y k for any ordinal k ≤ α is now defined as Y k = {¬ ϕ } ∪ {¬ δ i | i < k } ∪ { ψ | N ψ ∈ X } . (40)This is different from equation (29) because set Y k now includes formula ¬ ϕ instead of formula ϕ .The statement of Claim 1 remains the same. The proof of this claim is alsothe same except for the Zero Case. In the Zero Case, the proof is similar tothe original till equation (30). Because set Y k now contains formula ¬ ϕ insteadof formula ϕ , equation (30) will now have the form N ϕ ∈ X . Thus, in ourcase, X (cid:96) NN ϕ by Lemma 11 and the Modus Ponens inference rule. Hence, NN ϕ ∈ X because set X is maximal. Then, N ϕ ∈ w by Definition 9, whichcontradicts the assumption N ϕ / ∈ w of the lemma.The statement and the proof of Claim 2 remain the same. Set Z will nowbe defined as Z = { ϕ } ∪ { δ i | k (cid:48) ≤ i < α } ∪ { ψ | N ψ ∈ X } . (41)This is different from equation (31) because set Z now includes formula ϕ insteadof formula ¬ ϕ .The statement of Claim 3 remains the same. The Case 1 of the proof ofthis case is similar to the original proof of Claim 3 till formula (35), except forformula ϕ will be used instead of formula ¬ ϕ and formula ¬ ϕ instead of formula ϕ everywhere in that part of the proof. In particular formula (33) will now havethe form X (cid:96) N ( δ k (cid:48) → ¬ ϕ ) . (42)43nd formula (35) will now have the form X (cid:96) N ( ¬ δ k (cid:48) → ϕ ) . (43)The argument after formula (35) will change as follows. By item 2 of Lemma 12and statements (42) and (43), X (cid:96) N ( δ k (cid:48) ↔ ¬ ϕ ) . (44)Note that δ k (cid:48) ∈ ∆ a because (28) is an ordering of set ∆ a . Hence, by Defi-nition 11, either NH a δ k (cid:48) ∈ X or NS a ¬ δ k (cid:48) ∈ X . Then, by items 1 or 3 ofLemma 13 and statement (44), either X (cid:96) NH a ¬ ϕ or X (cid:96) NS a ϕ . Thus,either X (cid:96) NNH a ¬ ϕ or X (cid:96) NNS a ϕ by the definition of modality N , theNegative Introspection axiom, and the Modus Ponens inference rule. Hence,either NNH a ¬ ϕ ∈ X or NNS a ϕ ∈ X because set X is maximal. Then, ei-ther NH a ¬ ϕ ∈ w or NS a ϕ ∈ w by Definition 9 and assumption w ∈ W of thelemma. Thus, w (cid:96) K a ϕ → S a ϕ by the second Emotional Predictability axiomand propositional reasoning. Hence, w (cid:96) S a ϕ by assumption K a ϕ of the lemmaand the Modus Ponens inference rule. Therefore, S a ϕ ∈ w because set w ismaximal, which contradicts assumption S a ϕ / ∈ w of the lemma.The Case 2 of the proof of Claim 3 will be similar to the original proof ofClaim 3 till formula (38) except that formula except for formula ϕ will be usedinstead of formula ¬ ϕ and formula ¬ ϕ instead of formula ϕ everywhere in thatpart of the proof. Statement (38) will now have the form X (cid:96) N ¬ ϕ . Fromthis point, the proof will continue as follows. Statement X (cid:96) N ¬ ϕ implies that N ¬ ϕ ∈ X because set X is maximal. Then, ¬ ϕ ∈ w by Definition 9. Hence, w (cid:96) ¬ K a ϕ by the contraposition of the Truth axiom. Therefore, K a ϕ / ∈ w because set w is consistent, which contradicts the assumption K a ϕ ∈ w of thelemma.The statements and the proofs of Claim 4 and Claim 5 remain the same asin the original proof. (cid:2) We are now ready to state and to prove the “induction” or “truth” lemma.
Lemma 40. w (cid:13) ϕ iff ϕ ∈ w .Proof. We prove the lemma by structural induction on formula ϕ . If ϕ is apropositional variable, then the required follows from Definition 14 and item 1of Definition 2. If formula ϕ is a negation or an implication, then the statementof the lemma follows from the induction hypothesis using items 2 and 3 ofDefinition 2 and the maximality and the consistency of the set w in the standardway.Suppose that formula ϕ has the form K a ψ .( ⇐ ) : By Lemma 32, assumption K a ψ ∈ w implies that ψ ∈ u for any world u ∈ W such that w ∼ a u . Thus, by the induction hypothesis, u (cid:13) ψ for anyworld u ∈ W such that w ∼ a u . Therefore, w (cid:13) K a ψ by item 5 of Definition 2.44 ⇒ ) : Assume that K a ψ / ∈ w . Thus, by Lemma 33, there is a world u ∈ W suchthat w ∼ a u and ψ / ∈ u . Hence, u (cid:49) ψ by the induction hypothesis. Therefore, w (cid:49) K a ψ by item 5 of Definition 2.If formula ϕ has the form N ψ , then the proof is similar to the case K a ψ exceptthat Lemma 30 and Lemma 31 are used instead of Lemma 32 and Lemma 33respectively. Also, item 4 of Definition 2 is used instead of item 5.Assume that formula ϕ has the form H a ψ .( ⇐ ) : Assume that H a ψ ∈ w . To prove that w (cid:13) H a ψ , we verify conditions (a),(b), and (c) from item 6 of Definition 2.(a) The assumption H a ψ ∈ w implies w (cid:96) K a ψ by Lemma 8 and the ModusPonens inference rule. Thus, K a ψ ∈ w because set w is maximal. Hence,by Lemma 32, for any world u ∈ W , if ψ ∈ u , then w ∼ a u . Therefore, bythe induction hypothesis, u (cid:13) ψ for any world u ∈ W such that w ∼ a u .(b) By Lemma 36, assumption H a ψ ∈ w implies that for any worlds u, u (cid:48) ∈ W ,if ψ / ∈ u and ψ ∈ u (cid:48) , then u ≺ a u (cid:48) . Thus, by the induction hypothesis, forany worlds u, u (cid:48) ∈ W , if u (cid:49) ψ and u (cid:48) (cid:13) ψ , then u ≺ a u (cid:48) .(c) By the Counterfactual axiom and the Modus Ponens inference rule, theassumption H a ψ ∈ w implies that w (cid:96) ¬ N ψ . Thus, N ψ / ∈ w because set w is consistent. Hence, by Lemma 31, there is a world u ∈ W such that ψ / ∈ u . Therefore, u (cid:49) ψ by the induction hypothesis.( ⇒ ) : Assume that H a ψ / ∈ w . We consider the following three cases separately: Case I: K a ψ / ∈ w . Thus, by Lemma 33, there is a world u ∈ W such that w ∼ a u and ψ / ∈ u . Hence, u (cid:49) ψ by the induction hypothesis. Therefore, w (cid:49) H a ψ by item 6(a) of Definition 2. Case II: N ψ ∈ w . Then, ψ ∈ u for any world u ∈ W by Lemma 30. Thus, bythe induction hypothesis, u (cid:13) ψ for any world u ∈ W . Therefore, w (cid:49) H a ψ byitem 6(c) of Definition 2. Case III: K a ψ ∈ w and N ψ / ∈ w . Thus, by the assumption H a ψ / ∈ w andLemma 37, there are worlds u, u (cid:48) ∈ W such that ψ / ∈ u , ψ ∈ u (cid:48) , and u (cid:54)≺ a u (cid:48) .Hence, u (cid:49) ψ and u (cid:48) (cid:13) ψ by the induction hypothesis. Therefore, w (cid:49) H a ψ byitem 6(b) of Definition 2.If formula ϕ has the form S a ψ , then the argument is similar to the oneabove, except that Lemma 38 and Lemma 39 are used instead of Lemma 36 andLemma 37 respectively. (cid:2) Theorem 5 (strong completeness). If X (cid:48) ϕ , then there is world w of anepistemic model with preferences such that w (cid:13) χ for each formula χ ∈ X and w (cid:49) ϕ .Proof. The assumption X (cid:48) ϕ implies that set X ∪ {¬ ϕ } is consistent. Thus,by Lemma 14, there is a maximal consistent set w such that X ∪ {¬ ϕ } ⊆ w .Consider canonical epistemic model with preferences M ( w ). By Lemma 24, set w is one of the worlds of this model. Then, w (cid:13) χ for any formula χ ∈ X and45 (cid:13) ¬ ϕ by Lemma 40. Therefore, w (cid:49) ϕ by item 2 of Definition 2. (cid:2)
15. Conclusion
In this article we proposed a formal semantics for happiness and sadness,proved that these two notions are not definable through each other and gave acomplete logical system capturing the properties of these notions. The approachto happiness that we advocated could be captured by famous saying “Success isgetting what you want, happiness is wanting what you get”. Although popular,this view is not the only possible. As we mentioned in the introduction, someview happiness as “getting what you want”.As defined in this article, happiness and sadness are grounded in agent’sknowledge. We think that an interesting next step could be exploring belief-based happiness and sadness. A framework for beliefs, similar to our epistemicmodels with preferences, has been proposed by Liu [17].