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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 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: –h w ( ) = true iff h w ( ) = true and h w ( ) = true –Etc. Modal semantics is as follows: –h w (□ ) = true iff for all w’ such that wRw’: h w’ ( ) = true –h w (◊ ) = true iff there exists a w’ such that wRw’ and h w’ ( ) = 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: h w ( ) = true for all w W A statement is a (modal) consequence of a statement iff for any W, R, and h: if h w ( ) = true then h w ( ) = true for all w W 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: pp In w1, □p is true, but p is false w1w1

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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 h w (□p p) = true for any W, R, and h and w W: 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: h w (□p p) = true for any W, reflexive R, h and w W Proof by contradiction: Suppose h w (□p p) = false for some W, reflexive R, h and w W. Then h w (□p) = true and h w (p) = false. But, since R is reflexive, wRw. So, since h w (□p) = true, then by the semantics of □, it must also be the case that h w (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 In w1, □p is true, but □□p is false w1w1 w2w2 pp w3w3

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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 w 1 such that h w1 (□p □□p) = false. Then h w1 (□p) = true and h w1 (□□p) = false. Since h w1 (□□p) = false, there must be a world w 2 such that w 1 Rw 2 and h w2 (□p) = false. So there must also be a world w 3 such that w 2 Rw 3 and h w3 (p) = false. But since R is transitive, we must have w 1 Rw 3. And since h w1 (□p) = true, we know that h w3 (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: pp p In w1, ◊p is true, but □◊p is false w1w1 w2w2

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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: pp p w2w2 w1w1 pp w3w3 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: pp p w2w2 w1w1 pp w3w3 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 h w1 (◊p) = true and h w1 (□◊p) = false. Since h w1 (□◊p) = false, there must be a world w 2 such that w 1 Rw 2 and h w2 (◊p) = false. And since h w1 (◊p) = true, there must be a world w3 such that w 1 Rw 3 and h w3 (p) = true. Now, since R is symmetrical, we must have w 2 Rw 1, and by transitivity, we must then have w 2 Rw 3. But since h w2 (◊p) = false, we must have h w3 (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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