Download presentation

Presentation is loading. Please wait.

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 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.

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 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.

8
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

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 w1 w2

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

12
**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

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.

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google