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

Question

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.

Answer 1

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).     ■

Answer 2

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.

Answer 3

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).

Answer 4

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.

Answer 5

Some English sentences are ambiguous. A single sentence may have more than one meaning. Also what seems like a single meaning may be expressed by different English sentences.

Harry Gensler suggests doing the following when one has to symbolize or translate an ambiguous sentence into modal logic: (page 148)

When you translate an ambiguous English sentence, say that it's ambiguous and give both translations. When you do an English argument with an ambiguous statement give both translations and work out both arguments.

One needs to see if the incongruity between the English interpretation can be understood as ambiguity that one can work with or if there is a more serious paradox or puzzle involved.

From the perspective of the symbolism, independent of the English sentence, the goal is to combine the □ operator and a proposition p along with the negation symbol ~. There are four ways to insert the negation symbol in front of □ and in front of p.

It is to these combinations of symbols that one tries to assign some English word recognizing there may be multiple words one could use for each. Here is one or more ways to assign an English sentence to each of these combinations:

1. □p (It is necessary that p.)
2. ~□p (It is not necessary that p.)
3. □~p (It is impossible that p. Or, perhaps, more literally, it is necessary that not p.)
4. ~□~p = ◊p (It is possible that p. Or, more literally, it is not necessary that not p.)

Paul McNamara illustrates the usefulness of having these four patterns by describing how they can be placed into a "modal square of opposition" which may aid in understanding them. In this square of opposition (1) and (2) are contradictories as are (3) and (4).

The question raised by the OP is how to approach an English language interpretation of the modal definition

◇P ↔ ~◻~P

where P is It will be sunny tomorrow. Using (4) above, this might be viewed as It will not necessarily not be sunny tomorrow as one way to paraphrase Possibly, it will be sunny tomorrow.

The OP continues:

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)?

A person may understand Possibly, it will be sunny tomorrow as It is not necessarily going to be sunny tomorrow or as It is not necessarily not going to be sunny tomorrow. An English speaker may tolerate these ambiguities, but the modal logic does not unless they can be shown to be equivalent through the rules of that logic.

Does it matter? At this point the ambiguity may be simply in saying the same thing in more than one way and so it may not matter. However, there are derivations that appear as puzzles or paradoxes in some modal logics such as Ross's paradox in deontic modal logic. Here the consequent of a conditional when translated back into English raises questions about the usefulness of deontic logic.

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Gensler, H. J. (2002). Introduction to logic. Routledge.

McNamara, Paul, "Deontic Logic", The Stanford Encyclopedia of Philosophy (Fall 2018 Edition), Edward N. Zalta (ed.), URL = https://plato.stanford.edu/archives/fall2018/entries/logic-deontic/.

Answer 6

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.)

There can be an interpretation where it is false, so it is not a logical entailment.

In short, we interpret ◇P as "there exists a possible tomorrow where it is sunny."

Assuming that is true, well, it still may be that all of them are.

[ It is similar to the fact that Ǝx Px does not entail ~Ɐx Px. ]