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**Possible World Semantics for Modal Logic**

Intermediate Logic April 14

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**Kripke Models for Propositional Modal Logic**

A model <W,R,h> consists of: W: a set of worlds R W W: an accessibility relationship W and R together is called a frame h: W P {true, false}: a propositional truth-assignment function for each world Thus, in a Kripke model, statements are true relative to whichever world they are being evaluated in.

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**Semantics Propositional semantics is exactly as one would think, i.e:**

hw( ) = true iff hw() = true and hw() = true Etc. Modal semantics is as follows: hw(□) = true iff for all w’ such that wRw’: hw’() = true hw(◊) = true iff there exists a w’ such that wRw’ and hw’() = true That’s it!

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**Tautology, Consequence, etc.**

A statement is called a (modal) tautology iff for any W, R, and h: hw() = true for all wW A statement is a (modal) consequence of a statement iff for any W, R, and h: if hw() = true then hw() = true for all wW Etc.

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**Characteristic Axioms**

Systems T, S4, and S5 can be defined using ‘characteristic axioms’: T: □p p S4: □p □□p S5: ◊p □◊p On the next slides we will see that these axioms correspond to certain properties of the accessibility relationship R.

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T T is defined using the axiom □p p. In other words, in system T the statement □p p is considered a (modal) tautology. But, without any restrictions on R, we can easily build a model that shows that □p p is not a tautology: In w1, □p is true, but p is false p w1

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T (cont’d) On the previous slide we were able to construct a countermodel for the claim □p p using a non-reflexive accessibility relationship R. So, it is not true that hw(□p p) = true for any W, R, and h and wW: if we don’t know that R is reflexive, we can’t be certain of the principle. Reflexivity of R is therefore a necessary condition for □p p to always hold true. In fact, reflexivity is sufficient: hw(□p p) = true for any W, reflexive R, h and wW Proof by contradiction: Suppose hw(□p p) = false for some W, reflexive R, h and wW. Then hw(□p) = true and hw(p) = false. But, since R is reflexive, wRw. So, since hw(□p) = true, then by the semantics of □, it must also be the case that hw(p)=true. Contradiction.

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S4 The characteristic axiom for S4 is □p □□p. Again, however, this is not a tautology: p p p w1 w2 w3 In w1, □p is true, but □□p is false

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S4 (cont’d) However, as long as R is transitive, then □p □□p will always hold. Proof by contradiction: Suppose there is a world w1 such that hw1(□p □□p) = false. Then hw1(□p) = true and hw1(□□p) = false. Since hw1(□□p) = false, there must be a world w2 such that w1Rw2 and hw2(□p) = false. So there must also be a world w3 such that w2Rw3 and hw3(p) = false. But since R is transitive, we must have w1Rw3. And since hw1(□p) = true, we know that hw3(p) = true. Contradiction. Please note that S4 adds the axiom □p □□p to the axiom □p p. Hence, S4 assumes the accessibility relationship to be transitive and reflexive.

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S5 The characteristic formula of S5 is ◊p □◊p. Again, this gets added to the axioms of T and S4. What does this do to R? First, let’s see how the characteristic axiom of S5 is not a tautology in S4: In w1, ◊p is true, but □◊p is false p p w1 w2

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S5 (cont’d) Notice that R on the previous slide is reflexive and transitive, but not symmetric. Hence, if R is not symmetric, the characteristic formula does not hold. What if R is symmetric? Well, symmetry alone is not enough either: p p p w2 w1 w3 In w1, ◊p is true, but □◊p is false

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**S5 (cont’d) In fact, even symmetry plus reflexivity is not enough: p**

w2 w1 w3 In w1, ◊p is true, but □◊p is false

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**S5 (cont’d) However, symmetry plus transitivity will do the job.**

Proof by contradiction: Suppose hw1(◊p □◊p) = false. Then hw1(◊p) = true and hw1(□◊p) = false. Since hw1(□◊p) = false, there must be a world w2 such that w1Rw2 and hw2(◊p) = false. And since hw1(◊p) = true, there must be a world w3 such that w1Rw3 and hw3(p) = true. Now, since R is symmetrical, we must have w2Rw1, and by transitivity, we must then have w2Rw3. But since hw2(◊p) = false, we must have hw3(p) = false. Contradiction. Of course, S5 adds the axiom ◊p □◊p to the axioms of T and S4, requiring R to be reflexive, symmetrical, and transitive.

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HW 18 Use possible world semantics to see whether ◊(p ◊q) (□◊p ◊□q) is a tautology in T. In other words, try and find a model and a world in that model such that this statement is false in that world, or prove that no such model can be constructed.

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