Conocimiento docente

1.10

Entailment

With the definition of truth at a world, we can define an entail- ment relation between formulas. A formula B entails A iff, when- ever B is true, A is true as well. Here, “whenever” means both “whichever model we consider” as well as “whichever world in that model we consider.”

Definition 1.23. If Γ is a set of formulas and A a formula, then Γ entails A, in symbols: Γ ⊨ A, if and only if for every model

M = ⟨W,R,V ⟩ and world w ∈ W , if M,w ⊩ B for every B ∈ Γ,

then M, w ⊩ A. If Γ contains a single formula B, then we write B ⊨ A.

Example 1.24. To show that a formula entails another, we have to reason about all models, using the definition of M, w ⊩. For instance, to show p → ♢p ⊨ □¬p → ¬p, we might argue as fol- lows: Consider a model M= ⟨W,R,V ⟩ and w ∈ W , and suppose

M, w ⊩ p → ♢p. We have to show that M,w ⊩ □¬p → ¬p. Sup-

pose not. Then M, w ⊩ □¬p and M,w ⊮ ¬p. Since M,w ⊮ ¬p,

M, w ⊩ p. By assumption, M,w ⊩ p → ♢p, hence M,w ⊩ ♢p. By

definition of M, w ⊩ ♢p, there is some w′ with Rww′ such that

M, w′⊩ p. Since also M,w ⊩ □¬p, M,w′⊩ ¬p, a contradiction. To show that a formula B does not entail another A, we have to give a counterexample, i.e., a model M = ⟨W,R,V ⟩ where we show that at some world w ∈ W , M,w ⊩ B but M,w ⊮ A. Let’s show that p → ♢p ⊭ □p → p. Consider the model inFigure 1.2. We have M, w1 ⊩ ♢p and hence M,w1⊩ p → ♢p. However, since

M, w1⊩ □p but M,w1 ⊮ p, we have M, w1 ⊮ □p → p.

Often very simple counterexamples suffice. The model M′ ₌

{W′, R′,V′} _{with W}′ = {w}, R′ = ∅, and V′_{(p) = ∅ is also a} counterexample: Since M′,w ⊮ p, M′,w ⊩ p →♢p. As no worlds are accessible from w, we have M′,w ⊩ □p, and so M′,w ⊮ □p → p.

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w1 ¬p

w2 p w3 p

Figure 1.2:Counterexample to p → ♢p ⊨ □p → p.

Problems

Problem 1.1. Consider the model of Figure 1.1. Which of the following hold? 1. M, w1⊩ q ; 2. M, w3⊩ ¬q ; 3. M, w1⊩ p ∨ q ; 4. M, w1⊩ □(p ∨ q ); 5. M, w3⊩ □q ; 6. M, w3⊩ □⊥; 7. M, w1⊩ ♢q ; 8. M, w1⊩ □q ; 9. M, w1⊩ ¬□□¬q .

Problem 1.2. Complete the proof ofProposition 1.8.

Problem 1.3. Let M = ⟨W,R,V ⟩ be a model, and suppose w1,w2 ∈ W are such that:

1. w1 ∈ V (p) if and only if w2 ∈ V (p); and

19 1.10. ENTAILMENT

Using induction on formulas, show that for all formulas A:

M, w1⊩ A if and only if M,w2⊩ A.

Problem 1.4. Let M= ⟨M,R,V ⟩. Show that M,w ⊩ ¬♢A if and only if M, w ⊩ □¬A.

Problem 1.5. Consider the following model M for the language comprising p1, p2, p3as the only propositional variables:

w1 p1 ¬p2 ¬p3 w2 p1 p2 ¬p3 w3 p1 p2 p3

Are the following formulas and schemas true in the model M, i.e., true at every world in M? Explain.

1. p → ♢p (for p atomic); 2. A → ♢A (for A arbitrary); 3. □p → p (for p atomic); 4. ¬p → ♢□p (for p atomic); 5. ♢□A (for A arbitrary); 6. □♢p (for p atomic).

Problem 1.6. Show that the following are valid:

1. ⊨ □p → □(q → p); 2. ⊨ □¬⊥;

20 CHAPTER 1. SYNTAX AND SEMANTICS

Problem 1.7. Show that A→□A is valid in the class C of models

M= ⟨W,R,V ⟩ where W = {w}. Similarly, show that B →□A and

♢A → B are valid in the class of models M = ⟨W, R,V ⟩ where R = ∅.

Problem 1.8. ProveProposition 1.20.

Problem 1.9. Prove the claim in the “only if” part of the proof ofProposition 1.22. (Hint: use induction on A.)

Problem 1.10. Show that none of the following formulas are valid: D: □p → ♢p; T: □p → p; B: p → □♢p; 4: □p → □□p; 5: ♢p → □♢p.

Problem 1.11. Prove that the schemas in the first column ofta- ble 1.1are valid and those in the second column are not valid.

Problem 1.12. Decide whether the following schemas are valid or invalid:

1. (♢A → □B) → (□A → □B); 2. ♢(A → B) ∨ □(B → A).

Problem 1.13. For each of the following schemas find a model

M such that every instance of the formula is true in M:

1. p → ♢♢p; 2. ♢p → □p.

21 1.10. ENTAILMENT

Problem 1.14. Show that □(A ∧ B) ⊨ □A.

Problem 1.15. Show that □(p → q ) ⊭ p → □q and p → □q ⊭ □(p → q ).

CHAPTER 2

Frame

Definability

2.1

Introduction

One question that interests modal logicians is the relationship be- tween the accessibility relation and the truth of certain formulas in models with that accessibility relation. For instance, suppose the accessibility relation is reflexive, i.e., for every w ∈ W , Rww. In other words, every world is accessible from itself. That means that when □A is true at a world w, w itself is among the accessible worlds at which A must therefore be true. So, if the accessibility relation R of M is reflexive, then whatever world w and formula A we take, □A → A will be true there (in other words, the schema □p → p and all its substitution instances are true in M).

The converse, however, is false. It’s not the case, e.g., that if □p → p is true in M, then R is reflexive. For we can easily find a non-reflexive model M where □p → p is true at all worlds: take the model with a single world w, not accessible from itself, but with w ∈ V (p). By picking the truth value of p suitably, we can make □A → A true in a model that is not reflexive.

The solution is to remove the variable assignment V from the equation. If we require that □p → p is true at all worlds in M, regardless of which worlds are in V (p), then it is necessary that