paraconsistency and gamescanbaskent.net/logic/lectures/paragamesnotes.pdf · paraconsistency and...

Paraconsistency and Games

a quick course

Can BASKENT

Department of Computer Science, University of Bath

can@canbaskent.net canbaskent.net/logic @topologically

April 5th & 6th, 2016

CLE, UNICAMP - Campinas, Brazil

Slogan: Paraconsistency for Game Theory!

Paraconsistency helps us understand

game theory better,

game theory helps us understand

paraconsistency better.

Lecture One

Lecture OneParaconsistent Game Sematics

Lecture One

Outlook of Lecture One

É Logic of Paradox

É First-Degree Entailment

É Belnap’s 4-valued Logic

É Connexive Logic

Lecture One

Classical Game Semantics

During the semantic verification game, the givenformula is broken into subformulas by two players(Abelard and Heloise) step by step, and the gameterminates when it reaches the propositional atoms.

If we end up with a propositional atom which is true,then Heloise the verifier wins the game. Otherwise,Abelard the falsifier wins. We associate conjunctionwith Abelard, disjunction with Heloise.

A win for the verifier is when the game terminates witha true statement. The verifier is said to have a winningstrategy if she can force the game to her win,regardless of how her opponent plays.

Lecture One

Just because the game may end with a true/false atomdoes not necessarily suggest the truth/falsity of thegiven formula in general.

In classical logic, however, the major result of gametheoretical semantics states that the verifier has awinning strategy if and only if the given formula istrue in the model.

Lecture One

Classical Games

Classical semantic games are

É Two-player,

É Determined,

É Sequential,

É Zero-sum,

É Complete: winning strategies necessarily andsufficiently guarantee the truth value.

Question How do these attributes of semantical gamesdepend on the underlying logical structure? How canwe give game semantics for deviant logics?

Lecture One

Logic of Paradox

Lecture One

Logic of Paradox

Logic of Paradox and GTS

Consider Priest’s Logic of Paradox (LP) (Priest, 1979).

LP introduces an additional truth value P, calledparadoxical, that stands for both true and false.

¬T FP PF T

∧ T P FT T P FP P P FF F F F

∨ T P FT T T TP T P PF T P F

Lecture One

Logic of Paradox

Game Models

We define the verification game as a tupleΓ = (π, ρ, δ, σ) where

- π is the set of players,

- ρ is the set of well-defined game rules,

- δ is the set of designated truth values: the truthvalues preserved under validities: they determine thetheorems of the logic.

- σ is the set of positions: subformula and player pairs.

It is possible to extend it to concurrent games as well.

Lecture One

Logic of Paradox

Game Rules for LP

The introduction of the additional truth value P requiresan additional player in the game, let us call himAstrolabe (after Abelard and Heloise’s son).

Since we have three truth values in LP, we need threeplayers forcing the game to their win. If the game endsup in their truth set, then that player wins.

Then, how to associate moves with the connectives?

Lecture One

Logic of Paradox

Game Rules for LP

Denote this system with GTSLP.

p whoever has p in their extension, wins¬F Abelard and Heloise switch rolesF∧G Abelard and Astrolabe choose between

F and G simultaneouslyF∨G Heloise and Astrolabe choose between

F and G simultaneously

Lecture One

Logic of Paradox

An Example

Consider the conjunction. Take the formula p∧ q wherep,q are P,F respectively. Then, p∧ q is F.

p∧ q

Astrolabe

Abelard

Abelard makes a move and chooses q which is false.This gives him a win. Interesting enough, Astrolabechooses p giving him a win.

In this case both seem to have a winning strategy.Moreover, the win for Abelard does not entail a loss forAstrolabe.

Lecture One

Logic of Paradox

Correctness

Theorem

In GTSLP verification game for φ,

É Heloise has a winning strategy if φ is true,

É Abelard has a winning strategy if φ is false,

É Astrolabe has a winning strategy if φ is paradoxical.

The contra-positive of the above theorem is also useful.

Lecture One

Logic of Paradox

Correctness

Theorem

In a GTSLP game for a formula φ in a LP model M,

É If Heloise has a winning strategy, but Astrolabedoes not, then φ is true (and only true) in M,

É If Abelard has a winning strategy, but Astrolabedoes not, then φ is false (and only false) in M,

É If Astrolabe has a winning strategy, then φ isparadoxical in in M.

Lecture One

First-Degree Entailement

First-Degree Entailment

Lecture One

Semantic valuations are functions from formulas totruth values.

If we replace the valuation function with a valuationrelation, we obtain First-degree entailment (FDE) whichis due to Dunn (Dunn, 1976).

We use φr1 to denote the truth value of φ (which is 1 inthis case).

Since, r is a relation, we allow φr∅, and both φr0 andφr1.

Thus, FDE is a paraconsistent (inconsistency-tolerant)and paracomplete (incompleteness-tolerant) logic.

Lecture One

For formulas φ,ψ, we define r as follows.

¬φr1 iff φr0¬φr0 iff φr1(φ∧ ψ)r1 iff φr1 and ψr1(φ∧ ψ)r0 iff φr0 or ψr0(φ∨ ψ)r1 iff φr1 or ψr1(φ∨ ψ)r0 iff φr0 and ψr0

Lecture One

Game Semantics for FDE

The truth values {0},{1} and {0,1} work exactly asthe truth values F,T,P respectively in LP. In fact, LP canbe obtained from FDE by introducing a restriction thatno formula gets the truth value ∅.

Recall that for GTSLP, we allowed parallel plays forselected players depending on the syntax of theformula: we associated conjunction with Abelard andAstrolabe, disjunction with Heloise and Astrolabe.

Lecture One

Game Semantics for FDE

For FDE, the idea is to allow each player play at eachnode.

Therefore, it is possible that both players (or none) mayhave a winning strategy.

Lecture One

An Example

Consider two formulas with the following relationalsemantics: φr0, φr1 and ψr1. In this case, we have(φ∧ ψ)r1 and (φ∧ ψ)r0.

We expect both Abelard and Heloise have winningstrategies, and allow each player make a move at eachnode.

φ∧ ψ

Abelard

Heloise

Lecture One

Game Rules for FDE

p whoever has p in their extension, wins¬F players switch rolesF∧G Abelard and Heloise choose between F and G

simultaneouslyF∨G Abelard and Heloise choose between F and G

simultaneously

Lecture One

Correctness

Theorem

In a GTSFDE verification game for a formula φ, we havethe following:

É Heloise has a winning strategy if φr1

É Abelard has a winning strategy if φr0

É No player has a winning strategy if φr∅

Lecture One

Belnap’s 4-Valued Logic

Lecture One

Belnap’s 4-Valued system, call it B4, introduces twonon-classical truth values. Traditionally, P stands forboth truth values and N stands for neither of the truthvalues.

¬T FP PN NF T

∧ T P N FT T P N FP P P F FN N F N FF F F F F

∨ T P N FT T T T TP T P T PN T T N NF T P N F

Notice that P and N are the fixed-points under negation.

Lecture One

Game Rules for B4

From a game-semantics perspective, the problems withB4 include

É Two fixed-points for negation

É Non-monotonicity: two truth values may produce athird truth value under binary connectives

In particular, we have P∧N = F and P∨N = T.

Lecture One

Game Rules for B4

Let us have 4 players for 4 truth values:

The truth value T is forced by Heloise, F by Abelard, Pby Astrolabe and N by Bernard1.

Two negation-fixed-points suggest that Astrolabe andBernard both will be the concurrent players.

1After Abelard’s rival Bernard of Clairvaux.

Lecture One

Game Rules for B4p whoever has p in their extension, wins¬F Heloise assumes Abelard’s role,

Abelard assumes Heloise’s role,Astrolabe and Bernard keep their previous roles,and the game continues with F,

F∧G if Bernard has a winning strategy for Fand Astrolabe has a winning strategy for G,then Abelard wins,

F∧G otherwise Abelard, Astrolabe and Bernard choosesimultaneously between F and G,

F∨G if Bernard has a winning strategy for Fand Astrolabe has a winning strategy for G,then Heloise wins,

F∨G otherwise Heloise, Astrolabe and Bernard choosesimultaneously between F and G.

Lecture One

Correctness

Theorem

For the evaluation games for a formula φ in Belnap’s4-valued logic, we have the following:

É Heloise the verifier has a winning strategy if φevaluates to T,

É Abelard the falsifier has a winning strategy if φevaluates to F,

É Astrolabe the paradoxifier has a winning strategy ifφ evaluates to P,

É Bernard the nullifier has a winning strategy if φevaluates to N.

Lecture One

Connexive Logic

Lecture One

Connexive Logic

McCall’s Connexive Logic

Connexive logic is a “comparatively little-known and tosome extent neglected branch of non-classical logic"(Wansing, 2015). Even if it is under-studied, itsphilosophical roots can be traced back to Aristotle andBoethius.

Connexive logic is defined as a system which satisfiesthe following two schemes of conditionals:

É Aristotle’s Theses: ¬(¬φ→ φ)

É Boethius’ Theses: (φ→ ¬ψ)→ ¬(φ→ ψ)

In this work, we discuss one of the earliest examples ofconnexive logics CC, which is due to McCall (McCall,1966).

Lecture One

Connexive Logic

CC is axiomatized by adding the scheme(φ→ φ)→ ¬(φ→ ¬φ) to the propositional logic. Therules of inference for CC is modus ponens andadjunction, which is given as ` φ,` ψ∴ ` φ∧ ψ.

The semantics for CC is given with 4 truth values: T, t, fand F which can be viewed as “logical necessity",“contingent truth", “contingent falsehood", and “logicalimpossibility" respectively (Routley & Montgomery,1968).

In CC, the designated truth values are T and t.

Lecture One

Connexive Logic

¬T Ft ff tF T

∧ T t f FT T t f Ft t T F ff f F f FF F f F f

∨ T t f FT t T t Tt T t T tf t T F fF T t f F

First, we introduce 4 players for 4 truth values: T isforced by Heloise, F by Abelard, t by Aristotle and f byBoethius.

Lecture One

Connexive Logic

Game Rules for CC

As the trues and falses are closed under the binaryoperations respectively, we suggest the followingcoalitions.

Truth-maker Coalition:

Heloise (T) and Aristotle (t)

False-maker Coalition:

Abelard (F) and Boethius (f )

Lecture One

Connexive Logic

Game Rules for CC

p whoever has p in their extension, wins¬F switch the roles: Heloise assumes Abelard’s role,

Aristotle assumes Boethius’ role,Boethius assumes Aristotle’s role,Abelard assumes Heloise’s role, andthe game continues with F

F∧G false-makers coalition chooses betweenF and G

F∨G truth-makers coalition chooses betweenF and G

Lecture One

Connexive Logic

Correctness

Theorem

For the evaluation games for a formula φ in McCall’sConnexive logic, we have the following:

É truth-makers have a winning strategy if and only ifφ has the truth value t or T in M,

É false-makers have a winning strategy if and only ifφ has the truth value f or F in M.

Lecture One

Conclusion

What Have We Observed?

É Failure of the biconditional correctness

É Multiplayer semantic games in a nontrivial way

É Non-sequential / paralel / concurrent plays

É Variable sum games

É Coalitions

If winning strategies are proofs, game semantics forparaconsistent logics present a constructive way togive proofs for inconsistencies.

Lecture One

Conclusion

Difficult Logics

É Da Costa systems, Logics of Formal Inconsistency

É Preservationism

É First-order paraconsistent logics

É Infinitary, fixed-point non-classical logics

Lecture One

Conclusion

Reference

CB, Game Theoretical Semantics for Paracon-sistent Logics, in “Preceedings of the Fifth Interna-tional Conference on Logic, Rationality and Inter-action" (LORI-V), Edited by W. van der Hoek and W.Holliday and W. Wang, pp. 14-26, Springer, 2016.

Lecture Two

Lecture TwoA Self-Referential Paradox inGames

Lecture Two

Outlook of Lecture Two

É The Brandenburger - Keisler Paradox

É Paraconsistent Models

É A Countermodel

Lecture Two

Statement

The Brandenburger-KeislerParadox

Lecture Two

Statement

The Paradox

The Brandenburger-Keisler paradox (BK paradox) is atwo-person self-referential paradox in epistemic gametheory (Brandenburger & Keisler, 2006).

The following configuration of beliefs is impossible:

The Paradox

Ann believes that Bob assumes that Ann believes thatBob’s assumption is wrong.

The paradox appears if you ask whether “Ann believesthat Bob’s assumption is wrong".

Notice that this is essentially a 2-person Russell’sParadox.

Lecture Two

Statement

The Model

Brandenburger and Keisler use belief sets to representthe players’ beliefs.

The model (Ua,Ub,Ra,Rb) that they consider is called abelief structure where Ra ⊆ Ua × Ub and Rb ⊆ Ub × Ua.

The expression Ra(x,y) represents that in state x, Annbelieves that the state y is possible for Bob, andsimilarly for Rb(y,x). We will put Ra(x) = {y : Ra(x,y)},and similarly for Rb(y).

At a state x, we say Ann believes P ⊆ Ub if Ra(x) ⊆ P.

Lecture Two

Statement

The Semantics

A modal logical semantics for the interactive beliefstructures can be given.

We use two modalities � and ♡ for the belief andassumption operators respectively with the followingsemantics.

x |= �abφ iff ∀y ∈ Ub.Ra(x,y) implies y |= φx |= ♡abφ iff ∀y ∈ Ub.Ra(x,y) iff y |= φ

Note the bi-implication in the definition of theassumption modality!

Lecture Two

Statement

Completeness

A belief structure (Ua,Ub,Ra,Rb) is called assumptioncomplete with respect to a set of predicates Π on Ua

and Ub if for every predicate P ∈ Π on Ub, there is astate x ∈ Ua such that x assumes P, and for everypredicate Q ∈ Π on Ua, there is a state y ∈ Ub such thaty assumes Q.

We will use special propositions Ua and Ub with thefollowing meaning: w |=Ua if w ∈ Ua, and similarly forUb. Namely, Ua is true at each state for player Ann, andUb for player Bob.

Lecture Two

Statement

Incompleteness

Brandenburger and Keisler showed that no belief modelis complete for its (classical) first-order language.

Therefore, “not every description of belief can berepresented" with belief structures (Brandenburger &Keisler, 2006).

Lecture Two

Statement

Incompleteness

The incompleteness of the belief structures is due tothe holes in the model. A model, then, has a hole at φ ifeither Ub ∧ φ is satisfiable but ♡abφ is not, or Ua ∧ φ issatisfiable but ♡baφ is not.

Namely, φ is true for b, but cannot be assumed by a (orvice versa).

Lecture Two

Statement

Some Remarks

É BK paradox is a game theoretical example of aself-referential paradox

É It is a simple step towards the possibility ofparaconsistent games - a broader research programin progress

É It raises the possibility of discussingdiscursive/dialogical logics within game theoryproper

É Provides an interesting take on Hintikka’sinterrogative theory - how to inquire about aparadoxical sentence?

Lecture Two

Paraconsistent Approach

Paraconsistent TopologicalApproach

Lecture Two

What is a Topology?

Definition

The structure ⟨S, σ⟩ is called a topological space if itsatisfies the following conditions.

1. S ∈ σ and ∅ ∈ σ

2. σ is closed under finite unions and arbitraryintersections

Collection σ is called a topology, and its elements arecalled closed sets.

Lecture Two

Paraconsistent Topological Semantics

Use of topological semantics for paraconsistent logic isnot new. To our knowledge, the earliest work discussingthe connection between inconsistency and topologygoes back to Goodman (Goodman, 1981).

In classical modal logic, only modal formulas producetopological objects.

However, if we stipulate that:

extension of any propositional variable to be a closedset (Mortensen, 2000; Mortensen, 2010), we get aparaconsistent system.

Lecture Two

Problem of Negation

Negation can be difficult as the complement of a closedset is not generally a closed set, thus may not be theextension of a formula in the language.

For this reason, we will need to use a new negationsymbol ∼ that returns the closed complement (closureof the complement) of a given set.

Lecture Two

Topological Belief Models

The language for the logic of topological belief modelsis given as follows.

φ := p | ∼φ | φ∧ φ | �a | �b | �a | �b

where p is a propositional variable, ∼ is theparaconsistent topological negation symbol which wehave defined earlier, and �i and �i are the belief andassumption operators for player i, respectively.

Lecture Two

Topological Belief Models

For the agents a and b, we have a correspondingnon-empty type space A and B, and define closed settopologies τA and τB on A and B respectively.Furthermore, in order to establish connection betweenτA and τB to represent belief interaction among theplayers, we introduce additional constructionstA ⊆ A× B, and tB ⊆ B× A. We then call the structureF = (A,B, τA, τB, tA, tB) a paraconsistent topologicalbelief model.

A state x ∈ A believes φ ⊆ B if {y : tA(x,y)} ⊆ φ.Furthermore, a state x ∈ A assumes φ if{y : tA(x,y)} = φ. Notice that in this definition, weidentify logical formulas with their extensions.

Lecture Two

Semantics

For x ∈ A, y ∈ B, the semantics of the modalities aregiven as follows with a modal valuation attached to F.

x |= �aφ iff ∃Y ∈ τB with tA(x,Y)→ ∀y ∈ Y.y |= φx |= �aφ iff ∃Y ∈ τB with tA(x,Y)↔∀y ∈ Y.y |= φy |= �bφ iff ∃X ∈ τA with tB(y,X)→ ∀x ∈ X.x |= φy |= �bφ iff ∃X ∈ τA with tB(y,X)↔∀x ∈ X.x |= φ

Lecture Two

The Result

Theorem

The BK sentence is satisfiable in some paraconsistenttopological belief models.

Namely, we can construct a state which satisfies the BKsentence - push the holes that create theinconsistencies to the boundaries.

Lecture Two

Conclusion

This was a self-referential paradox in games.

What about non-self-referential paradoxes in games?

Lecture Two

Conclusion

Reference

CB, Some Non-Classical Approaches toBranderburger-Keisler Paradox, Logic Journal ofthe IGPL, vol. 23, no. 4, pp. 533-552, 2015.

Lecture Three

Lecture ThreeA Non-Self-Referential Paradox inGames

Lecture Three

Outlook of Lecture Two

É Yablo’s Paradox

É What is the big deal?

É A Countermodel

Lecture Three

Yablo’s Paradox

Lecture Three

Yablo’s Paradox

Yablo’s Paradox, according to its author, is a non-selfreferential paradox (Yablo, 1985; Yablo, 1993).

Yablo considers the following sequence of sentences.

S1 : ∀k > 1,Sk is untrue,S2 : ∀k > 2,Sk is untrue,S3 : ∀k > 3,Sk is untrue,...

Lecture Three

Yablo’s Paradox

Why is it a Paradox?

By using reductio, Yablo argues that the above set ofsentences is contradictory. Here, the infinitary nature ofthe paradox is essential as the each finite set of Sn issatisfiable.

The scheme of this paradox is not new. To the best ofour knowledge, the first analysis of this paradox wassuggested in 1953 (Yuting, 1953).

Lecture Three

Yablo’s Paradox

Impact of Yablo’s Paradox

Ketland showed that the paradox is ω-inconsistent(Ketland, 2005).

Barrio showed that Yablo’s Paradox in first-orderarithmetic has a model and not inconsistent, but it isω-inconsistent (Barrio, 2010).

It is easy to see how. Since every finite set of Snsentences is satisfiable, then, by compactness thereexists a model for the Yablo sentences. Byω-inconsistency, it can be argued that the model weare looking for is a non-standard model of arithmetic.

Lecture Three

Yablo’s Paradox

Impact of Yablo’s Paradox

As Hardy puts it “Is Yablo’s paradox Liar-like? In someways yes, and in other ways no" (Hardy, 1995).

Priest offers another analysis regarding the infinitarylanguage that it requires, and suggests a reading of theparadox that does indeed involve circularity (Priest,1997).

Sorensen disagrees and point out the hierarchical viewof Tarskian truth theory arguing that Yablo’s paradox ineffect “exploit[s] an alternative pattern of semanticdependency" (Sorensen, 1998).

Lecture Three

Yablo’s Paradox

Applications of Yablo’s Paradox

Goldstein presents a set theoretical yabloesqueparadox for class membership (Goldstein, 1994).

Leitgeb suggests a yabloesque paradox fornon-well-founded definitions underlining the settheoretical limitations of the logical toolbox (Leitgeb,2005).

Picollo discusses the paradox in second-order logicgeneralizing the ω-inconsistency results (Picollo, 2013).

Non-well-founded Yablo chains form a topological spaceencouraging Bernardi’s topological approach to theparadox (Bernardi, 2009).

Cook and Beall consider Curry-like versions of theparadox (Cook, 2009; Beall, 1999).

Lecture Three

A Non-Self Referential Epistemic Game Theoretical Paradox

A Yabloesque Paradox inEpistemic Games

Lecture Three

A Yabloesque Paradox in Epistemic Games

Consider the following sequence of assumptions wherenumerals represent game theoretical players.

A1 : 1 believes that ∀k > 1,k’s assumption Al about ∀l > k is untrue,A2 : 2 believes that ∀k > 2,k’s assumption Al about ∀l > k is untrue,A3 : 3 believes that ∀k > 3,k’s assumption Al about ∀l > k is untrue,...

Lecture Three

An Interpretation

Imagine a queue of players, where players areconveniently named after numerals, holding beliefsabout each player behind them, but not aboutthemselves. In this case, each player i believes thateach player k > i behind them has an assumption abouteach other player l > k behind them and i believes thateach k’s assumption is false.

This statement is perfectly perceivable for games, andinvolves a specific configuration of players’ beliefs andassumptions, which can be expressible in the language.However, as we shall show, similar to Yablo’s paradoxand the BK paradox, this configuration of beliefs isimpossible.

Lecture Three

Modal Logically

Let us start with an informal argument.

Now, for a contradiction, assume An is true for some n.Therefore, player n believes that ∀k > n, k’s assumptionis untrue. In particular, player n+ 1’s assumption isuntrue. In other words, n+ 1’s assumption

An+1 : n+ 1 believes that ∀k > n+ 1,k’s assumption Alabout l > k is untrue

is untrue.

Lecture Three

Modal Logically

Therefore, n+ 1 believes that for some k′ > n+ 1, whatk′ assumes about some l′ > k′ is true. But, thiscombination of players k′ and l′, both of which arebigger than n+ 1, thus n, is accessible from n by meansof the belief-assumption modalities. We assumed An istrue, which entails that what k′ assumes about somel′ > k′ is untrue. Contradiction. The choice of n wasarbitrary, so each An in the sequence is untrue.

However, if each An is untrue, they can be assumeduntrue. But, if for all n, n’s assumption is untrue, thenA1 is indeed true. Yet, we just argued that each An isuntrue.

Contradiction.

Lecture Three

The Syntax

The Yabloesque Brandenburger - Keisler paradox (‘YBKParadox’, henceforth) requires ω-many players i ∈ I. Thesyntax of this language is given in the Backus-Naurform as follows for a set of propositional variables P:

φ := p | ¬φ | φ∧ φ | �ijφ | ♡ijφ

where p ∈ P and i 6= j for i, j ∈ I with |I| = ω.

The disjunction and implication are taken asabbreviations in the standard way.

Lecture Three

The Model

The extended belief model is a tupleM = ({Ui}i∈I,{Rij}i 6=j∈I,V) where Rij ⊆ Ui × Uj and V is avaluation function.

As before, the expression Rij(x,y) represents that instate x, the player i believes that the state y is possiblefor player j.

We prevent (a trivial form of) self-reference bydisallowing players having beliefs about themselves.

Lecture Three

The Semantics

The semantics for the modal operators is given asfollows in a similar way.

x |= �ijφ iff ∀y ∈ Uj.Rij(x,y) implies y |= φx |= ♡ijφ iff ∀y ∈ Uj.Rij(x,y) iff y |= φ

Lecture Three

An Illustrative Example I

Let w |= A3. Therefore, w |=∧

k>3�3k{∧

l>k♡kl¬Al}.

Let us spell this out.

w |=∧

�3k{∧

♡kl¬Al}

w |= �34(♡45¬A5 ∧♡46¬A6 ∧♡47¬A7 ∧ . . . ) ∧

�35(♡56¬A6 ∧♡57¬A7 ∧♡58¬A8 ∧ . . . ) ∧

Lecture Three

An Illustrative Example II

In particular, for example, w |= �34♡47¬A7 andw |= �35♡57¬A7. Therefore, it can be seen that for all3 < a < 7, w |= �3a♡a7¬A7. Simply put, from agent 3,through each other agent between 3 and 7, it ispossible to reach ¬A7 via belief-assumption chain. Wesimply focused on player 7 and his assumption A7, butthe argument works for any player n > 4 in ourexample.

The contradiction simply occurs when A7 is hit by twodifferent players in two different ways. In order to seeit, consider A5 (which can also reach A7 by abelief-assumption chain).

Lecture Three

An Illustrative Example III

So, we havew |= �34♡45¬A5

where ¬A5 is given as follows:∨

◊5k{∨

¬♡kl¬Al}.

Therefore, for all v, if R34wv then v |= ♡45¬A5. Then, forall u, R45vu if and only if u |= ¬A5. The use of theassumption modality here is crucial. It associates theset of states that falsifies A5 with what is accessiblefrom v with R45.

Lecture Three

An Illustrative Example IV

Spelling this out, we have the following.

u |=◊56[¬♡67¬A7 ∨¬♡68¬A8 ∨ . . . ]∨

◊57[¬♡78¬A8 ∨¬♡79¬A9 ∨ . . . ]∨ . . .

The first disjunct above suggests that there is a t suchthat R56ut and t |= ¬♡67¬A7 ∨¬♡68¬A8 ∨ . . . .

However, this is impossible. The first disjunct(¬♡67¬A7) cannot be the case at t.

Lecture Three

An Illustrative Example VBecause it reduces to the following.

t |= ¬♡67¬A7 iff ∃y ∈ U7.[ (R67(x,y)∧ y |= A7)∨

(¬R67(x,y)∧ y |= ¬A7) ]

But, this is impossible by our earlier observation: thereis a state accessible via R67 that satisfies ¬A7, and allthe states accessible from u satisfies A7 due to thedefinition of the ♡ modality.

The argument can easily be extended to other disjunctsand their disjuncts. Thus, contradiction. Therefore,each An is false. As we observed earlier, then Ans arealso true by definition.

This is the paradox.

Lecture Three

Further Results

Some further results:

1. ♡ijφ→ �ijφ

2. ♡ij(φ∧ ψ) ≡ ♡ijφ∧♡ijψ

3. �ij(♡jkφ∧♡jlψ) ≡ �ij♡jkφ∧�ij♡jlψ.

4. ◊ij(♡jkφ∨♡jlψ) ≡ ◊ij♡jkφ∨ ◊ij♡jlψ

5. ♡ij(�jkφ∧�jlψ) ≡ ♡ij�jkφ∧♡ij

�jlψ

Lecture Three

Further Results

If w |= An, then for all p,q with n < p < q;w |= �np♡pq¬Aq.

Theorem

If w |= An, then for all p,p′,q with n < p < q andn < p′ < q; we have Rpq(v) = Rp

′q(v′) for all v ∈ Up andall v′ ∈ Up′ .

Lecture Three

Further Results

Corollary

If w |= An, then �np♡pqφ↔�np′♡p′qφ for n < p < q andn < p′ < q.

Lecture Three

Discussion

Discussion and FurtherRemarks

Lecture Three

Discussion

Assumption Modality and the Diagonal Formula

Assumption modality ♡ij is an essential part of theconstruction of the paradox. Without it, it is notpossible to generate the YBK paradox.

For example, the following set of sentences aboutplayers’ beliefs is not inconsistent.

A′1 :=∧

�1k{∧

�kl¬A′

A′2 :=∧

�2k{∧

�kl¬A′

A′3 :=∧

�3k{∧

�kl¬A′

Lecture Three

Discussion

Non-Well-Foundedness

As Yablo also argued, what we would have is a“downward facing tree with ω branches descendingfrom each node" (Yablo, 2004).

The set of sentences above, in other words, generatetrees that are infinitely-branching which satisfy it.

Lecture Three

Discussion

Categoricity

As argued by Ketland, the set of Yablo sentences is notsatisfiable on the standard model of arithmetic, thusthey are “ω-inconsistent" (Ketland, 2005).

This observation suggests that the YBK paradox can besatisfied in a game with ω+ 1 players. As every finiteset of Ans in the YBK Sentence are satisfiable, bycompactness, there must exist a model for the Yablosentences.

Lecture Three

Countermodels

Lecture Three

Countermodels

Topological Countermodels

Now we construct paraconsistent models for the YBKparadox. For agent i, we take the correspondingnon-empty type space Si and define topologies withclosed sets σi. For example, for player 3, the type spacewill be denoted by S3 with a topology σ3. In order tomake this approach interactive, we define a functionsij ⊆ Si × Sj which associates states for player i with thestates of j. For example, for player i, at states from Si,sij returns a closed set K ∈ σj. We write sij(w,K) meansthat at state w ∈ Si, player i believes that states k ∈ Kare possible for j.

Lecture Three

Countermodels

The Model

The model is a tuple ({Si}i∈I,{σi}i∈I,{sij}i,j∈I,V) where Vis a valuation defined in the standard way. The syntaxfor this system is similar to what we have given earlierwith the paraconsistent negation symbol ∼. Here p ∈ Pwhere i 6= j ∈ I:

φ := p | ∼φ | φ∧ φ | �ijφ | ♡ijφ

The dual modalities are defined as usual with theparaconsistent negation.

Lecture Three

Countermodels

The Semantics

The paraconsistent topological semantics for thislanguage is given as follows for negation and the modaloperators as the Booleans are standard. For a set X, thecomplement of X will be denoted by Xc.

|∼φ| =Clo(Kc)

|�ijφ| ={w ∈ Si : ∃K ∈ σj with sij(w,K) and K ⊆ |φ|}

|♡ijφ| ={w ∈ Si : ∃K ∈ σj with sij(w,K) and K = |φ|}

Lecture Three

Countermodels

The Semantics

For easy read, we give the semantics for the modalitiesin the traditional sense as follows.

w |= �ijφ iff ∃K ∈ σj with sij(w,K)→ ∀v ∈ K.v |= φw |= ◊ijφ iff ∀K ∈ σj with sij(w,K),∃v ∈ K such that v |= φw |= ♡ijφ iff ∃K ∈ σj with sij(w,K)↔∀v ∈ K.v |= φ

Lecture Three

Countermodels

The Topological Countermodel

Now we construct a counter-model using topologicalparaconsistent models for the YBK paradox. Let usreconsider the set of modal formulas again.

A1 :=∧

�1k{∧

♡kl¬Al}

A2 :=∧

�2k{∧

♡kl¬Al}

A3 :=∧

�3k{∧

♡kl¬Al}

Lecture Three

Countermodels

We will construct a model and a state w in it that satisfythe above set of formulas step by step starting withplayer 1 and A1. Now, for player 1, take w1 ∈ S1 andconsider A1. For each k > 1, construct K1k ∈ σk suchthat s1k(w1,K1k) and each vk ∈ K1k satisfies

l>k♡kl∼Al.

Therefore, for each l > k, there exists Ukl that skl(vk,Ukl)such that every ul ∈ Ukl if and only if ul ∈ |∼Al|. Let usunravel ∼Al as follows: ∼Al =

p>l◊lp{∨

q>p∼♡pq∼Aq}.This is a disjunctive statement. As our goal is toconstruct a counter-model, we will try to satisfy onlyone of the disjuncts and nested-disjuncts.

Lecture Three

Countermodels

Now at ul, include (ul,K1p)in slp for all p > l. Thus, wehave slp(ul,K1p). By construction of A1 (hence of eachAi), each vp ∈ K1p satisfies

q>p♡pq∼Aq, hence ♡pq∼Aq

for each q > p. Similarly, include (vp,xq) for eachxq ∈ ∂(|Aq|) into spq in a way that spq(vp, ∂(|∼Aq|)). Thus,vp |= ∼♡pq∼Aq ∧♡pq∼Aq. We constructed a model inwhich we have w1 |= A1.

Lecture Three

Countermodels

This methodology can be extended inductively for eachassumption Ai which in turn builds the counter-modelthat satisfy each and every formula in the set of gametheoretical Yablo sentences.

The crucial observation is that the extension of a♡-formula uniquely identifies with the extension of theformula in question. However, some of the points inthat extension may also satisfy the negation of theformula in question in paraconsistent models. Thismakes it quite easy and straight-forward to constructthe counter-model.

Lecture Three

Conclusion

Lecture Three

Conclusion

Yablo’s Paradox is

É An interactive, ω-player paradox,

É A modal paradox,

É A paradox of well-foundedness in some ways

What to do: Develope a Curryesque epistemic gametheoretical paradox in which negation and falsitypredicates are not used.

Lecture Three

Conclusion

Reference

CB, A Yabloesque Paradox in Epistemic GameTheory, under submission, 2016.

Closing Remarks

Conclusion

Game Theory relates non-classical logics tohomo-economicus.

It helps us understand how we make decisions, reachequilibria, reveal preferences and put utilities in goods.

Closing Remarks

Thank you!

Slides are available at:

www.CanBaskent.net/Logic

Closing Remarks

References I

Barrio, Eduardo Alejandro. 2010.

Theories of Truth without Standard Models and Yablo’s Sequences.

Studia Logica, 96(3), 375–391.

Beall, Jc. 1999.

Completing Sorensen’s Menu: A Non-Modal Yabloesque Curry.

Mind, 108(431), 737–739.

Bernardi, Claudio. 2009.

A Topological Approach to Yablo’s Paradox.

Notre Dame Journal of Formal Logic, 50(3), 331–338.

Brandenburger, Adam, & Keisler, H. Jerome. 2006.

An Impossibility Theorem on Beliefs in Games.

Studia Logica, 84, 211–240.

Closing Remarks

References II

Cook, Roy T. 2009.

Curry, Yablo and Duality.

Analysis, 69(4), 612–620.

Dunn, J. Michael. 1976.

Intuitive Semantics for First-Degree Entailments and ’Coupled Trees’.

Philosophical Studies, 29(3), 149–168.

Goldstein, Laurence. 1994.

A Yabloesque Paradox in Set Theory.

Analysis, 54(4), 223–7.

Goodman, Nicolas D. 1981.

The Logic of Contradiction.

Zeitschrift für Mathematische Logik und Grundlagen der Mathematik, 27(8-10), 119–126.

Closing Remarks

References III

Hardy, James. 1995.

Is Yablo’s Paradox Liar-Like?

Analysis, 55(3), 197–198.

Ketland, Jeffrey. 2005.

Yablo’s Paradox and ω-Inconsistency.

Synthese, 145(3), 295–302.

Leitgeb, Hannes. 2005.

Paradox by (non-well-founded) Definition.

Analysis, 65(4), 275–8.

McCall, Storrs. 1966.

Connexive Implication.

Journal of Symbolic Logic, 31(3), 415–433.

Closing Remarks

References IV

Mortensen, Chris. 2000.

Topological Seperation Principles and Logical Theories.

Synthese, 125(1-2), 169–178.

Mortensen, Chris. 2010.

Inconsistent Geometry.

College Publications.

Picollo, Lavinia María. 2013.

Yablo’s Paradox in Second-Order Languages: Consistency and Unsatisfiability.

Studia Logica, 101(3), 601–617.

Priest, Graham. 1979.

The Logic of Paradox.

Journal of Philosophical Logic, 8, 219–241.

Closing Remarks

References VPriest, Graham. 1997.

Yablo’s Paradox.

Analysis, 57(4), 236–242.

Routley, Richard, & Montgomery, H. 1968.

On Systems Containing Aristotle’s Thesis.

The Journal of Symbolic Logic, 33(1), 82–96.

Sorensen, Roy A. 1998.

Yablo’s Paradox and Kindred Infinite Liars.

Mind, 107(425), 137–155.

Wansing, Heinrich. 2015.

Connexive Logic.

In: Zalta, Edward N. (ed), The Stanford Encyclopedia of Philosophy, fall 2015 edn.

http://plato.stanford.edu/entries/logic-connexive:http://plato.stanford.edu/archives/spr2016/entries/logic-connexive.

Closing Remarks

References VI

Yablo, Stephen. 1985.

Truth and Reflection.

Journal of Philosophical Logic, 14(3), 297–349.

Paradox without Self-Reference.

Analysis, 53(4), 251–2.

Circularity and Paradox.

Pages 165–183 of: Bolander, Thomas, Hendricks, Vincent F., & Pedersen, Stig Andur (eds),Self-Reference.

CSLI Publications.

Yuting, Shen. 1953.

Paradox of the Class of All Grounded Classes.

The Journal of Symbolic Logic, 18(2), 114.

paraconsistency and gamescanbaskent.net/logic/lectures/paragamesnotes.pdf · paraconsistency and...

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