2

p = Smith is a brother

q = Smith has a sibling

i) □(p → q)

ii) (p → □q)

Which in English form would be:

i') Necessarily, if Smith is a brother then Smith has a sibling

ii') If Smith is a brother then necessarily Smith has a sibling

While i) is true, ii) is not.

It is said that i) doesn't imply ii), but why is this the case?

Could there be a counter-example to ii)? How is it that Smith is a brother but not necessarily have a sibling?

  • Just a complement to Mauro's answer below: (ii) would mean that if Smith is actually a brother as a matter of fact, then he would have had a sibling whatever the circumstances (i.e. even in counterfactual circumstances where in fact he wouldn't be a brother!). While (i) merely says that in whatever circumstances, if he was a brother, he would have a sibling. – Quentin Ruyant May 2 '15 at 14:00
  • @HunanRostomyan's argument pointed up the potential error in better detail. You are taking names that appear in propositions as part of the notation. If the identity of 'Smith' is part of the theory, then models are going to preserve statements like "Smith is a brother". We don't want them to. With the quantification allowed by the name Smith coupled in this way, i') does imply ii'). Try 'any given person' in place of Smith. – user9166 May 3 '15 at 20:03
2

Assume that (p → □q) is valid and "contrapose" it :

(~□q → ~p).

But ~□ is equivalent to ◊~, and thus we have :

(◊~q → ~p).

Consider now your example; under the above assumption, we have :

"if it is possible for Smith to have not a sibling, then Smith is not a brother"

that "sounds" wrong.

  • OK, but is Smith a free variable or is he a person? If Smith is a real person and it is possible for Smith to not have a sibling right here and now, he doesn't have one, because he is real, not potential. At worst if the reasons it is possible for him not to have one is because his sister is petting Schroedinger's cat, then it is not known whether or not he is a brother, right here and now. But that is notation pressing up against the boundaries of sophistry. You need more context than this problem gives us about the supposed range of unmodalized truths. – user9166 May 3 '15 at 18:05
1

Before giving a counterexample, let's unpack the meaning of the box, since it's not obvious as the truth-conditions of the conditional. Given a modal model M, and a world w ∈ M, we say:

(M, w) ⊧ □φ if and only if for all v ∈ |M| s.t. wRv, (M, v) ⊧ φ.

In other words, formula □φ is true at a world if and only if it's true at all the worlds accessible (this relation is represented by "R" in the formula) from that world. Let's apply this definition:

(1) (M, w) ⊧ □(p → q) iff for all v ∈ |M|, if wRv then (M, v) ⊧ (p → q),

(2) (M, w) ⊧ (p → □q) iff either (M, w) |/= p or for all v ∈ |M|, if wRv then (M, v) ⊧ q.

Intuitively, (1) says that with respect to world w, the formula is true just in case every world w sees has either p false or q true. The (2) says, again with respect to w, that the formula is true just in case either p is false at w or q is true in all the worlds w sees. The counterexample:

Counterexample. ( to (1) ⊢ (2) ). Let M = {w, v} and let R = {(w,v)}. Here we have a model M, with only two worlds in it: w and v, and the only accessibility relation is from w to v. Let p be true at w and false at v, and q false everywhere. In other words, V(p) = {w}, V(q) = ∅. The pointed model (M, w) constitutes a counterexample to the (1) to (2) inference, because (1) is true and (2) is false at (M, w):

  • (M, w) ⊧ □(p → q) just in case every world w sees has either p false or q true. Since w only sees v and v has p false, we know that (1) is true.

  • (M, w) ⊧ (p → □q) just in case either p is false or all the worlds that w sees make q true. Since w makes p true, we look at the second disjunct for the answer: is q true in all the worlds w sees? No, because w sees only v and v doesn't make q true. So (2) is false.

  • The question is whether Smith is a named point in the theory of your model, no? If Smith is like zero in group theory, an object that must be treated the same by all models, then you can't come up with a counterexample, right? Basically, we do not know whether to quantify Smith across all worlds or just this one, because modal statements with named elements are just never clear about this. – user9166 May 3 '15 at 18:11
  • @jobermark Thanks. I think I understand your worry. Consider what I proposed as a counterexample to the propositional form of the argument. I don't exclude that a sub-sentential analysis of p and q, along with an interesting theory of modality, can validate the argument. – Hunan Rostomyan May 3 '15 at 18:37
0

English is one of the worse languages to consider mood in, so this is a total misunderstanding.

In a more inflected language like Latin or older forms of Greek the 'is' or the 'has' would have a subjunctive or optative inflection that indicate which of them is more highly hypothetical and therefore to which the necessity applies.

Although Germanic languages sort of try to have a subjunctive, they fail. Instead, mood is not expressed in a verb form, but with auxiliaries or adverbs, and it is assumed to be an attribute of the most primary verb from which it is not fully isolated. English has very weak isolation, so the mood almost always applies to the primary verb and selects a range of possible worlds for the whole sentence to be true or false in.

"I might be going to the store later" can mean that I am going somewhere later and that place might or might not be the store; or that I might or might not be going anywhere at all later, but if I am going, where I am going is to the store. It remains ambiguous because the modality is expressed with its own verb, just tossed in there independently, without being bound to anything.

It is unlikely to mean that I might or might not be going and where I am going, if I go, might or might not be the store, because there is still only one modal verb. To be that ambiguous, we have to say more. How silly is that? Luckily, the exact intention seldom matters to the hearer, who just gets the vague sense of uncertainty, if they care at all.

Beyond that, unlike languages with more hierarchical grammars, English just automatically moves attributes wherever they make the most sense. If I say "tomorrow he says the car will run", we totally forgive the illogic and put the 'tomorrow' on 'run', whether it is grammatically possible or not, because "tomorow he says..." is illogical and "tomorrow, he says, ..." is not.

So there is absolutely no natural way to say the equivalent of "p implies that necessarily q" and get heard that way consistently, because even when it is written just that way, the 'necessarily' automatically jumps back to the 'implies' and the sentence means "p necessarily implies that q", which is "box (p arrow q)". However you try to keep it bound, our speaking habits let it loose and it goes where it is most likely to belong.

You have to descend to indirect reference: "p arrow box q" means "p being true (in this reality) implies q is true in all possible (related) realities".

To some degree that seems just silly to even think about for many English speakers. "How can this reality affect all related realities, sentences hold in realities or they don't..." because moods generally are not that finely articulated in our language.

-1

In Bangladesh, and generally in India one calls people of your own peer-group 'brothers'.

This falsifies i') and ii')

In England, being a brother means that one has a sibling; this establishes the truth of i) and ii); and in fact in natural language their equivalence.

Here the truth is established by semantics; one has to establish what exactly brother means; in Kantian terms, it is not an analytic statement; a statement whose truth can be discovered by grammar; but whose truth is synthetic; ie established by experience.

Another answer is provided by interpreting via Model Theory; and in which a different interpretation of what neccessarily means; and this is by the introduction of possible worlds; under this interpretation a distinction can be brought out as by Rostomyan.

Another possibility is shown by Mauros answer which hints at the possibility of introducing time:

After all, I may not be a brother now; but might be in the future...

  • I'm a native speaker of English - and I don't see any real difference between those two sentences, when natively thought; I'm not a prescriptive grammarian. My point is that it matters how we wish to interpret the term necessarily - and this need not be in terms of Model Theory ie Possible Worlds; though I accept that this is a possibility- if you look carefully at what I wrote. – Mozibur Ullah May 2 '15 at 22:02
  • Right, you can't hear this because you are a native speaker of English, and the second concept is just weird to say in English, ever, in any way. – user9166 May 2 '15 at 22:14
  • Sure, different languages have different resources. – Mozibur Ullah May 2 '15 at 22:21
  • You are a Classisist, can you try it in Greek? If I say "Sam is a brother" in the present deduce "Sam has a sibling" in the aorist absolute, it is bizarre. I have either stepped out of reality and into a story, or I have moved the wrong direction in time. If I say "Sam is a brother" as a subjunctive hypothetical and "Sam has a sibling" as a consequent in the same tense, it is always true. – user9166 May 2 '15 at 22:26
  • Let's take this discussion to chat - I've just tried it but I can't get in. – Mozibur Ullah May 2 '15 at 22:32

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