The following excerpt from wikipedia says:

In modal logic the distinction between de dicto and de re is one of scope. In de dicto claims, any existential quantifiers are within the scope of the modal operator, whereas in de re claims the modal operator falls within the scope of the existential quantifier. For example:

De dicto: ◻ ∃ x A x Necessarily, some x is such that it is A

De re: ∃ x ◻ A x Some x is such that it is necessarily A

Based on this logical distinction, Anthony Kenny in his A New History of Western Philosophy says the following:

However, ancient epistemology is bedevilled by two different but related fallacies. Both of them are generated by a misunderstanding of the truth that whatever is knowledge must be true. One of the fallacies haunts classical epistemology, up to the time of Aristotle; the other fallacy haunts Hellenistic and imperial epistemology. The first fallacy is this. "Whatever is knowledge must be true" may be interpreted in two ways.

(1) Necessarily, if p is known, p is true or (2) If p is known, p is necessarily true.

(1) is true and (2) is false. It is a necessary truth that if I know you are sitting down, then you are sitting down; but if I know you are sitting down it is not a necessary truth that you are sitting down; you may get up at any moment. Plato and Aristotle, over and over again, seem to regard (2) as indistinguishable from (1).

As I am only interested in this fallacy, I am not quoting the other one, as well.

So my question is actually simple: what is the difference between the two of them? As Plato and Aristotle, poor me, I am, too, quite unable to distinguish (1) from (2).


Although I don’t know the wider context in which Kenny discusses this, I don’t think the distinction between (1) and (2) has much to do with the distinction between de dicto and de re. In (1), the modal operator ‘necessarily’ applies to the entire conditional ‘If p is known, p is true’. By contrast, in (2), it applies only to the consequent of the conditional. Think about it in terms of ‘not’, and note that (1a) is different from (2a):

(1a) Not: (If p, then q).

(2a) If p, then not-q.

As far as I can see, (1) differs from (2) in much the same way in which (1a) differs from (2a) – and that difference has nothing to do with de dicto vs de re. It does, however, have to do with scope, as in (1a), but not (2a), the negation takes scope over the conditional. Hence, (1) says that the conditional as a whole is necessary: there is no situation/possible world in which it is false. By contrast, (2) says that if p is known, p is necessary: for all (2) says, there is a situation/possible world in which the conditional ‘If p is known, p is true’ is false.

In epistemology, the de dicto vs de re distinction is often invoked when so-called singular thoughts are discussed. Suppose Sally suffers from paranoia. Then we might say:

(3) Sally believes she is being persecuted by a crazy clown.

Now, what we are saying is presumably (3a), not (3b):

(3a) Sally believes the following claim to be true: ‘There is a crazy clown, and he is persecuting me’.

(3b) There is a crazy clown, and Sally believes that he is persecuting her.

In other words, when we assert (3), we aren’t saying that a person exists of whom Sally believes that he is persecuting her. We are merely saying that Sally holds a particular (false) belief. So, (3a) is de dicto while (3b) is de re – if we buy into that distinction and find it useful here.

  • +1 I think your answer is good, and provides a nice and simple way of explaining the difference between the two claims. My one qualm is that I do think there is a connection to the de re/de dicto distinction, which is primarily a distinction concerning scope -- whether the modal operator is inside our outside the scope of a quantifier. Since there's implicit quantification over all propositions, that seems relevant here. Also, the de re/de dicto distinction would apply if the negation in your example were intuitionistic negation (certain conditionals also exhibit similar behavior).
    – Dennis
    Jul 7 '17 at 20:13
  • @Dennis Agreed: If we read (1) and (2) as quantified statements – and not as schemata –, the quantifier in (1) can take scope over the modal operator, or vice versa. However, I’m not sure this means the difference between (1) and (2) is connected to the de re / de dicto distinction. Compare: (1*) For all p, necessarily, if p is known, then p is true. (2*) For all p, if p is known, then p is necessarily true. The quantifier takes scope over the modal operator in both (1*) and (2*). In that sense, both are de re statements. And yet, (1*) and (2*) are different!
    – MarkOxford
    Jul 8 '17 at 13:59
  • @Dennis. Also, the de re / de dicto distinction probably can’t capture all the subtleties of scope. Consider Kripke’s example (‘Speaker’s ref & semantic ref.’). K The number of planets might have been necessarily odd. That’s a three-way scope ambiguity: K1 It’s P that it’s N that there are exactly x planets and x is odd. K2 There are exactly x planets, and it is P that it’s N that x is odd. K3 It’s P that there are exactly x planets and it’s N that x is odd. While K is true only if read as K3, it’s unclear whether (K3) is de dicto, de re, or neither.
    – MarkOxford
    Jul 8 '17 at 14:07
  • Agreed on both counts. I don't think the difference between (1) and (2) is primarily due to a de re/de dicto ambiguity, just that it's in there and plausibly playing some role in the confusion. Thanks for the Kripke example, I need to take a look back at that paper.
    – Dennis
    Jul 8 '17 at 15:36

There are actually two modalities involved: the obvious (logical or metaphysical) modality invoked by "necessarily" and the epistemic modality invoked by "known".

In possible worlds semantics for modal logic, "necessarily p" is interpreted the claim that p is true in every possible world. In epistemic logic "it is known (by some agent) that p" is usually interpreted as "in every world compatible with what the agent knows, p is true". So, the tricky bit here is how these two modalities interact when combined.

Using N to abbreviate "necessarily" and K for "it is known (by some agent)" the two sentences have the following forms:

  1. N(If K(p) then p), and
  2. If K(p), then N(p).

However, p is being used schematically here, so there is really a hidden universal quantifier over propositions. Incorporating that insight into those two forms yields the following:

  • 1'. N(For all p(If K(p) then p)), and

  • 2'. For all p(If K(p), then N(p)).

Recognizing the implicit quantification over propositions is what yields the connection to the de dicto/de re distinction. While (1') predicates necessity to the whole quantified sentence, (2') predicates necessity to a subformula of the quantified sentence -- the consequent of the conditional. So (1') is de dicto while (2') is de re since it moves the necessity operator into the scope of a quantifier.

In possible worlds terms, (1') says that every world is such that every proposition p, known by an agent in that world, is true in those worlds compatible with what the agent knows. (2'), however, says that every proposition p is such that if it is true in all worlds compatible with the agent's knowledge, it is true is every world whatsoever (not merely the one's compatible with what the agent knows).

This point can be made more clearly, I think, if we consider generalized quantifiers, popular in linguistics and closer to the ancients' syllogistic understanding of quantification, rather than the first-order objectual quantifiers familiar from logic. With generalized quantifiers, a quantified noun phrase like "every boy" is taken to denote the set of sets with all boys as members. So, "every proposition" is the set of sets with all propositions as members and "every known proposition" (again relativized to a particular agent) is the set of sets with all known (by that agent) propositions as members. This yields the following interpretation of (1) and (2):

  • 1''. Necessarily, every known proposition is a true proposition.
  • 2''. Every known proposition is a necessarily true proposition.

Since the worlds compatible with what an agent knows are a proper subset of all worlds (except when the agent only knows necessary truths like truths of logic), we can see that (1'') and (2'') say different things about the set of sets of all known propositions. (1'') says that in every world the set of sets of all known propositions is a subset of the set of sets of all propositions true in worlds compatible with what the agent knows -- clearly true. (2''), however, says that the set of sets of all known propositions is a subset of the set of sets of all necessarily true propositions.

Put in simplest terms, (1'') is true so long as you can only have knowledge of truths -- usually taken as definitional of knowledge. (2'') requires that you can only know necessary truths, which is a much stronger claim since it rules out the possibility of knowledge of contingent truths. The only way that (2'') looks plausible is if you take knowledge to require something like infallibile justification, a position called infallibilism and (arguably) held by Descartes. This rules out knowledge of pretty much anything you can't know by means of logical deduction alone. The other way (2'') could be true is if necessitarianism -- the view that there are no mere possibilities and the world is necessarily the way it is -- is true. Neither of these views is very popular, since they have many counterintuitive consequences.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.