# Does necessary possibility entail actuality?

In Modal logic,

Does being necessarily possible entail actuality?

## 3 Answers

The entailment does not hold in the modal logic S5 and so it also doesn't hold in all weaker normal modal logics (normal in the sense that they validate the K-Axiom □(A→B) → (□A→□B)).

Let a (propositional) S5 model be a pair M = (W, V), where W is some non-empty set of possible worlds and V a function from atoms to functions from W to {0,1}. V can be extended to the whole modal propositional language as usual. Note that, as usual, V(□A)(w) = 1 iff V(A)(w') = 1 for all w'∈ W. Let ⋄A := ¬□¬A and V(A, w) := V(A)(w). We take consequence to be local: A formula A is a S5 consequence of some set of formulas S (S ⊨ A), if for every S5 model M = (W, V) and every w ∈ W: If V(B,w) = 1, for all B ∈ S, then V(A, w) = 1.

The following S5 model shows that □⋄p ⊭ p (if you're taking actuality as an operator we have to move to two-dimensional modal logic, but the reasoning is almost the same): M = (W, V) such that W = {w, w'}, V(p, w) = 0, V(p, w') = 1 and V(q, w) = V(q, w') = 0, if q ≠ p. Clearly, there is some u ∈ W such that V(p, u) = 1. So, V(⋄p, w) = V(⋄p, w') = 1. So, V(□⋄p, w) = 1. But we also have that V(p, w) = 0.

This is a 2D variation on Sequitur's already complete answer. The question is whether:

1) ▢♢φ |= L

holds for modal logic L. The answer depends on which logic this L is. Confining our attention to normal modal logics, we choose the strongest (S5) among them and see whether (1) holds with respect to S5. If (1) does not hold with respect to S5, that is, if we can invalidate it, then we can invalidate it in all normal modal logics weaker than S5 (i.e. logics B, S4, T, D, and even K)! Before we get to the counterexample, let's understand the meaning of (1). We start, as usual, with the underlying language:

Definition 1. (Language) Given a propositional letter 'p', the language of modal propositional logic is generated by the following grammar:

φ   :=   p   |   φ′   |   ¬φ   |   (φ ∧ φ)   |   □φ   |   Aφ.

The addition to our usual setup here is the actuality operator A, which intuitively tells us whether a formula is true in the world of utterance; to see how it works pay attention to the semantics below.

Definition 2. (Models) A modal model is a triple M = (W, R, V), where W is a set of 'possible worlds', R is a binary accessibility relation on W, V is a function from propositional letters and worlds to {0,1}.

We've got the language and the models, so we just need to interpret the language in those models:

Definition 3. (2D Semantics) The truth of a formula φ of the modal propositional language in a modal model M = (W, R, V) at a pair of worlds (w,v) ∈ W2 (hence the name '2D') is defined as follows:

1. M, w, v |= p               iff         V(p, w) = 1;
2. M, w, v |= ¬φ            iff         ¬(M, w, v |= φ);
3. M, w, v |= φ ∧ ψ        iff         (M, w, v |= φ) and (M, w, v |= ψ);
4. M, w, v |= □φ            iff         ∀w′ ∈ W : wRw′ → M, w′, v |= φ;
5. M, w, v |= Aφ            iff         M, v, v |= φ.

The novel thing here is clause (5), which says that φ is true in model M at a world of evaluation w with respect to a world of utterance v just in case φ is true in M if we evaluate it at world v. For the usual applications in philosophy of language we'd define operators to shift the world of utterance to make the most out of v, but for our purposes here the above will be sufficient. Recall that S5 are those modal models that have an R that is an equivalence relation, which means that all worlds can access each other, effectively rendering the notion of accessibility useless. This allows us to simplify clause (4):

Clause (4) simplification: M, w, v |= S5 □φ iff ∀w′ ∈ W : M, w′, v |= S5 φ.

The box and the diamond are duals in this logic, so we introduce the abbreviation: ♢φ ≡ ¬▢¬φ. Now we're ready to invalidate (1) in (S5) in this 2D framework:

Fact 4. For all L ≤ S5, ▢♢φ |= L Aφ is false.

Proof. [Sequitur] Let consequence be S5-consequence. We want to consider an S5 model M and a pair of worlds (w,v) s.t. M, w, v |= S5 ▢♢φ but ¬(M, w, v |= S5 Aφ). Unpacking the meanings of the fancy symbols according to (Definition 3) we understand that (i) to make the antecedent true it will suffice to make M such that all of its worlds see a φ-world, and (ii) to make the consequent false we have to make sure the context of utterance (or actual) world is not a φ-world. Let φ be p and consider the model M = (W, V), where W = {a, b} and V = {(a,p)}. Since p is true at a according to V, and since every world can see every world in W, the antecedent is satisfied, i.e., M, a, b |= S5 ▢♢p. But since V doesn't assign (b,p) the value 1, M, b, b |= S5 p ≡ V(p, b) = 1 is falsified. Therefore, (1) is not valid with respect to S5 and by extension with respect to all logics between K and S5. ■

References (2D Semantics)

Davies, M. & L. Humberstone (1980) "Two Notions of Necessity", Philosophical Studies 38:1–30.
Kamp, H. (1971) "Formal Properties of 'Now'", Theoria 37: 237–273.
Kaplan, D. (1989) "Demonstratives", in Themes from Kaplan.

• Great answer! Just to add some more relevant references: Frank Vlach: 'Now' and 'Then':A Formal Study in the Logic of Tense and Anaphora. Ph.D. Dissertation, Philosophy Department, UCLA, 1973. Krister Segerberg: "Two-Dimensional Modal Logic". Journal of Philosophical Logic 2 (1):77-96 (1973). Max Cresswell: Entities and Indices. Dordrecht: Kluwer, 1990. – sequitur Jun 5 '14 at 13:35
• @sequitur Thank you very much. Wonderful suggestions. – Hunan Rostomyan Jun 5 '14 at 14:10

If P is actually true, it is true in this world. If P is possibly true, it is true in a possible world. If P is necessary true, it is true in all worlds. Therefore, if something is necessary true, then it is true in this world (and in all others).

Edit: As the comments point out, I missed the core of your question. However, let me just add the relevant line:

If P is necessarily possible, that means that it is true for all worlds that there exists a possible world in which P. This does not entail that P is true in the actual world.

It is necessarily possible that I'm drinking a beer tonight. There is at least one possible world in which I am drinking a beer tonight. And it is true for all worlds that there is this possible world in which I am drinking a beer tonight. But: I won't drink a beer tonight, to demonstrate to you that necessary possibility does not entail actuality.

• so there is no "necessarily possible"? – user6917 Jun 4 '14 at 11:48
• I'm pretty sure you can have stacked operators in modal logic, and he's asking if □ \Diamond P entails P. The answer is no but he's not confused about whether you can have more than one operator – virmaior Jun 4 '14 at 11:54
• The OP's question doesn't concern the implications of necessity but those of necessary possibility, which is a completely different notion. – sequitur Jun 4 '14 at 11:56
• misread the question, should be on topic now – Lukas Jun 4 '14 at 12:17
• Now you're on topic, but two problems remain (can easily be fixed though): (1) 'another' is unwarranted, and (2) you forgot a concluding sentence to the effect that: the answer to the question is 'no': just because every world can see a P-world, it doesn't mean that the actual world makes P true! So how to fix those: get rid of 'another' in your last sentence, and add a concluding sentence answering the question. – Hunan Rostomyan Jun 4 '14 at 17:12