# In modal logic, why not 'possibly p' → 'not necessarily p'?

I'm told that if ◇ means 'possible' and ◻ means 'necessary' and ~ means 'not' and ↔ means 'if and only if', then

◇P ↔ ~◻~P

I get that if it is not necessarily not going to be sunny tomorrow, then it is possible that it will be sunny. But: what is wrong with saying that if ◇P then ~◻P? (Note no ~ between ◻ and P there.)

In other words, I understand the left-pointing arrow on the ↔, but not the right-pointing arrow. If it is possibly going to be sunny tomorrow, can I not just as well say that it is not necessarily going to be sunny (implying that it may or may not be sunny), as that I can say that is not necessarily not going to be sunny (also implying that it may or may not be sunny)?

Now I can see that if P is something that is always true (say 1+2=3), the fact that P is possible (◇P) cannot imply that it is not necessary (~◻P), because P must always necessarily be true. So that would perhaps answer my question. But what is going on with the sunshine example then? The always-true case seems a bit sneaky... is it even right to say "◇P" if you already know that P is not merely possible, but in fact necessary, such as with P meaning 1+2=3 ?

Can somebody clarify? Many thanks.

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Logic, though intuitive in many ways, exhibits precise terminology that sometimes seems to defy colloquial language. For example, "A or B" tends to have different meanings depending on whether you're speaking conversationally or are writing a formal logic paper. "Possibly P" is effectively the same difference. – commando Mar 28 '14 at 4:24
@commando I'm glad something about that has been said. Thank you. – Hunan Rostomyan Mar 28 '14 at 4:26
Thank you all! The main point seems indeed the difference between possibility and contingency, and that in natural language when we say 'possibly P' it tends to imply 'possibly not P', which is not 'possibility' as defined in Hunan's post, but 'contingency'. I think my confusion also stems from the Wikipedia article not actually defining possibility ♢φ as ¬▢¬φ, but positing a two-way implication (↔), implicitly assuming that we already know what we mean by 'possible'. I may have a look at the Kripke model proofs later. David Richerby's way of putting it is very helpful too! Thanks again. – Diploria Mar 28 '14 at 14:07

The box and the diamond are duals (in the usual systems), so if you have the box, you can define:

Definition 1. (Possibility)   ♢φ     =def    ¬▢¬ φ.

If you have the diamond as primitive, you can define the box in the same way. Now, suppose we take the box as a primitive. Then the above definition gives us a notion of possibility that differs from what has since Aristotle, at least, been called contingency and can then be defined as follows:

Definition 2. (Contingency)  ♢2 φ     =def     ♢ φ ∧ ♢ ¬ φ.

Lucas raised the good question as to whether the first is really a notion of possibility (is it?). But those are at least two versions we need to distinguish to answer your questions. Following David Richerby's suggestion, instead of focusing on the sun example, I'll simply state and prove some claims about the relationship between the two diamonds and the box that will help you make sense of the examples.

Fact 1. The following is invalid: ♢(φ) → ¬ ▢(φ).

Proof. To show that (1) is invalid, we need to find a Kripke model M = (W, R, V) and a world w ∈ M such that (M, w) satisfies ♢(φ) but not ¬ ▢(φ). Consider the following model M = (W, R, V) where W = {w,v}, R = {(w,v)}, V(φ) = {v}, and V(¬φ) = ∅. Since wRv and (M, v) |= φ, we have it that (M, w) |= ♢(φ). But since there is no world u ∈ W s.t. wRu and (M, u) |= ¬φ, we also have it that (M, w) |= ▢(φ). (M, w) is thus a counterexample to (1), because: (M, w) |= ♢(φ) and (M, w) |= ¬¬▢(φ).          ■

Fact 2. The following is valid: ♢2(φ) → ¬ ▢(φ).

Proof. This is rather trivial. Consider an arbitrary pointed model (M, w) s.t. (M, w) |= ♢2(φ). We want to show that (M, w) |= ¬ ▢(φ). From the above definition of ♢2, we know that: (M, w) |= ♢(φ) ∧ ♢¬φ, which implies that (M, w) |= ♢¬φ. That means that there is a world v ∈ |M| s.t. wRv and (M, v) |= ¬φ. Suppose, for contradiction, that (M, w) |= ▢(φ). If that's true, then for all u ∈ |M| s.t. wRu we have: (M, u) |= φ. But v is one such u and (M, v) |= ¬φ, so the supposition that (M, w) |= ▢(φ) must be false. That gives us: (M, w) |= ¬▢(φ). Since (M, w) was arbitrary, we conclude that: |= ♢2(φ) → ¬ ▢(φ).    ■

Fact 3. The following (converse of Fact 1) is invalid: ¬ ▢(φ) → ♢(φ).

Proof. To show that (3) is invalid, we need to find a Kripke model M = (W, R, V) and a world w ∈ M such that (M, w) satisfies ¬ ▢(φ) but not ♢(φ). Consider the following model M = (W, R, V) where W = {w,v}, R = {(w,v)}, V(¬φ) = {v}, and V(φ) = ∅. Since wRv and (M, v) |= ¬φ, we have it that (M, w) |= ♢(¬φ) ≡ ¬ ▢(φ). But since there is no world u ∈ W s.t. wRu and (M, u) |= φ, we also have it that (M, w) |= ▢(¬φ) ≡ ¬♢(φ). We have a counterexample, because: (M, w) |= ¬ ▢(φ) and (M, w) |= ¬♢(φ).  ■

Fact 4. The following (converse of Fact 2) is invalid: ¬ ▢(φ) → ♢2(φ).

Proof. Consider the same model from the proof of Fact 3. Since wRv and (M, v) |= ¬φ, we have it that (M, w) |= ♢(¬φ). We want to disprove that (M, v) |= ♢2(φ), which by Definition 2 means that we want to show that (M, v) |= ¬[♢(φ) ∧ ♢(¬φ)] ≡ [¬♢(φ) ∨ ¬♢(¬φ)]. To prove that it will suffice to prove that (M, v) |= ¬♢(φ), since then we can ∨-introduce ¬♢(¬φ) and obtain the required negation. To prove that we need to show that there is no world u ∈ |M| s.t. wRu and (M, u) |= φ. Observing that our countermodel is such that V(φ) = ∅, we know that there cannot be any such u, so we can conclude that: (M, w) |= ¬♢(φ). Lastly, as promised, we introduce the second disjunct: (M, w) |= [¬♢(φ) ∨ ¬♢(¬φ)], and de Morgan it: (M, v) |= ¬[♢(φ) ∧ ♢(¬φ)], which gives us: (M, v) |= ¬♢2(φ).     ■

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So, in general, ♢ φ does not follow from ♢ ¬ φ (or vice-versa)? If that's right, reading ♢ as "possibly" is quite misleading. Although I can see that x being possible does not mean ¬x is possible - 'possibly' makes me think "maybe", which would imply the opposite is possible. – Lucas Mar 27 '14 at 21:16
That's right. That's because <M, w> |- ♢(φ) says that there is a world v accessible from w such that <M, v> |- φ. If you're in world w and possibly P, then that just means that you see a P-world, and says nothing about you seeing a ~P-world; and vice versa. – Hunan Rostomyan Mar 27 '14 at 22:26
another complementary way of looking at it. Consider the rule "if Box p, then diamond P." That seems obviously right. If it is necessarily the case that 2+2=4, then it should certainly be possible that 2+2=4. Adopting the axiom "diamond p, therefore not Box p" would be inconsistent with that very plausible assumption about how the modal operators should work. – shane Mar 28 '14 at 1:25
@Lucas: That's right, I too find it not a very intuitive use of the term 'possible' when ♢φ does not follow from ♢¬φ or vice-versa. I imagine logicians must have encountered this issue time and again, and wonder what the arguments were to stick to this terminology... are there no better alternatives? – Diploria Mar 28 '14 at 14:14
I took the time to get to grips with the possible worlds thing - everything's easier to understand when you can write it as a directed graph. – Lucas Mar 28 '14 at 14:38

A very simple explanation: "Possibly P" means "P might be true"; "Not necessarily P" means "P might be false". These are not equivalent. For example, if P is always true, then "Possibly P" is true but "Not necessarily P" is false; if P is always false, then "Possibly P" is false but "Not necessarily P" is true. On the other hand, "Not necessarily not P" means "not P might be false", i.e., "P might be true", which is equivalent to "Possibly P".

This intuition can be made formal using Kripke frames, which is how semantics of modal logics are defined.

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Due to the low number of votes, I'm wondering what is wrong with this explanation, because it is so easy to understand and makes sense!? – GreenAsJade Mar 28 '14 at 22:12
@GreenAsJade I don't think anything's wrong with my answer, but Hunan's is way more detailed. I've only given the intuition: in logic, intuition is a useful guide but far from the whole story. – David Richerby Mar 28 '14 at 23:34
For the record, I found this answer very clean and helpful (as mentioned/hidden? in a comment under the question). Voted for it. I 'accepted' neither this nor Hunan's answer as I feel they both bring useful ideas to the table. Would 'accept' both if possible. cheers – Diploria Mar 29 '14 at 17:53

While I think Hunan's answer is roughly exhaustive, I'm going to give an answer that proceeds more simply. Consider ◻P. From necessarily P, it follows that ◇P [P is possible]. Or at least it does in every system of modal logic I'm aware of. Given this, it would be strange for ◇P → ~◻P. Since that would mean ◻P ↔ ~◻~P.

The issue and the solution is to distinguish contingency from possibility. The former is non-necessity; the latter merely identifies a less strict condition. (All Scottish terriers are dogs but not vice-versa).

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I'm not quite following this. You write "it would be strange for ◇P → ~◻~P". But isn't ◇P → ~◻~P true by definition? – Diploria Mar 28 '14 at 14:10
You're right ... I've corrected the error. – virmaior Mar 28 '14 at 14:34

Most simply, take a look at the contrapositive of `◇P → ~◻P`, which is logically equivalent:

``````◻P → ~◇P
``````

Clearly it's false that if something is necessary then it is impossible. The contrapositive is false, and so `◇P → ~◻P` is false.

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Interesting. I wasn't aware of proof by contrapositive. Satisfyingly, when the same thing is done for contingency as in Hunan's Def. 2, one gets that the contrapositive of ◇₂P → ~◻P, namely ◻P → ~◇₂P, reads "if P is necessary, then it is not contingent". Entirely sensible. – Diploria Mar 29 '14 at 18:50