Presentation is loading. Please wait.

Presentation is loading. Please wait.

Possible World Semantics for Modal Logic Intermediate Logic April 14.

Similar presentations


Presentation on theme: "Possible World Semantics for Modal Logic Intermediate Logic April 14."— Presentation transcript:

1 Possible World Semantics for Modal Logic Intermediate Logic April 14

2 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.

3 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!

4 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.

5 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.

6 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: pp In w1, □p is true, but p is false w1w1

7 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.

8 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 pp w3w3

9 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.

10 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: pp p In w1, ◊p is true, but □◊p is false w1w1 w2w2

11 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 w2w2 w1w1 pp w3w3 In w1, ◊p is true, but □◊p is false

12 S5 (cont’d) In fact, even symmetry plus reflexivity is not enough: pp p w2w2 w1w1 pp w3w3 In w1, ◊p is true, but □◊p is false

13 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.

14 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.


Download ppt "Possible World Semantics for Modal Logic Intermediate Logic April 14."

Similar presentations


Ads by Google