Possible World Semantics for Modal Logic

1. Possible World Semantics for Modal Logic Intermediate Logic April 14

2. 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.

3. 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!

4. Tautology, Consequence, etc. • A statement  is called a (modal) tautology iff for any W, R, and h: hw() = true for all wW • A statement  is a (modal) consequence of a statement  iff for any W, R, and h: if hw() = true then hw() = 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: In w1, □p is true, but p is false p w1

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 hw(□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: hw(□p  p) = true for any W, reflexive R, h and wW • Proof by contradiction: Suppose hw(□p  p) = false for some W, reflexive R, h and wW. 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.

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

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 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.

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: In w1, ◊p is true, but □◊p is false p p w2 w1

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 w1 w3 w2 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 w1 w3 w2 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 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.

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.