Friday, April 16, 2010

Frege's slingshot, a minor fix (in case you care about details)

In one of the classes, I'm talking about slingshot arguments, and I was thinking about a possibly clear way of laying out Frege's argument that (if sentences have denotations) propositions aren't in general denotations of sentences, showing why the conclusion technically speaking doesn't follow, and how once an extra assumption is added the argument flies. Here's the result. (If you know and understand Frege, this probably won't surprise you, but I'm only concerned with a way of presenting this to students here).

Frege in On Sense and Nominatum argues:
Thus far we have considered sense and nominatum only of such expressions, words, and signs which we called proper names. We are now going to inquire into the sense and the nominatum of a whole declarative sentence. [...] Is this thought to be regarded as the sense or the nominatum of the sentence? Let us for the moment assume that the sentence has a nominatum! If we then substitute a word in it by another word with the same nominatum but a different sense, then this substitution cannot affect the nominatum of the sentence. But we realize that in such cases the proposition is changed; e.g., the proposition of the sentence "the morning star is a body illuminated by the sun" is different from that of "the evening star is a body illuminated by the sun''. Someone who did not know that the evening star is the same as the morning star could consider the one proposition true and the other false. The proposition can therefore not be the nominatum of the sentence; it will instead have to be regarded as its sense.
To put this in a step-wise manner: consider two sentences:

[A] The morning star is a body illuminated by the sun.
[B] The evening star is a body illuminated by the sun.

The assumptions are:

[F1] If one can accept a sentence p but deny a sentence q simultaneously (in a given context), p and q do not express the same proposition.

[F2] Substitution of co-referential designators in larger expressions preserves the denotation of whole expressions.

[F3] In one and the same context, one can accept [A] and deny [B].

[F4] The morning star = The evening star

The first two steps in reasoning are quite straightforward. From [F1] and [F3] it follows:

[F5] [A] and [B] express different propositions.

and from [F2] and [F4] we obtain:

[F6] [A] and [B] have the same denotation.

Now for the slightly problematic step (supposedly from [F5] and [F6]):

[F7] The proposition expressed by [A] is not the denotation of [A] and the proposition expressed by [B] is not the denotation of [B]

To see why [F7] doesn't follow from [F5] and [F6], let's represent these more formally. "D(p)" stands for "the denotation of "p" " and "S(p)" stands for "the proposition expressed by (the sense of) "p" ".

[F5'] S(A) ≠S(B)
[F6'] D(A)=D(B)
[F7'] D(A)≠S(A) & D(B)≠S(B)

Now consider two models, in which:

[model 1] D(A)=S(A) & D(B)≠S(B)
[model 2] D(B)=S(B) & D(A)≠S(A)

In each of those models [F5'] and [F6'] are true, and yet [F7'] false (it's helps to draw a diagram here). Of course, there are at least two worries. One is that each of the models is not too uniform: the denotation of one of the sentences is identical with its sense, whereas the denotation of the other isn't identical with its sense. Another worry is that there are no clear reasons why any of these situations is to be preferred over the other. Hence, it seems sensible to use an extra assumption which captures the idea that the answer to the question whether denotations of sentences are their senses should have a general answer, that is, that either all sentences denote their senses, or no sentence does:
[F+] D(A)=S(A) ⇔D(B)=S(B)

In fact, [F7'] follows from [F5'], [F6'] and [F+]. For suppose that [F5'], [F6'] and [F+] are true and yet [F7'] is false:

[F7-] ~[D(A)≠S(A) & D(B)≠S(B)]

DeMorgan applied to [F7-] gives us:

[F7-a] D(A)=S(A) v D(B)=S(B)

Now, let's do a proof by cases. Suppose:

[A1] D(A)=S(A)

With [F+] this entails:

[A2] D(B)=S(B)

But transitivity of identity applied to [A1], [F6'] and [A2] gives us:

[A3] S(A)=S(B)

which contradicts [F5'].

An argument for the other case is analogous. Suppose:

[B1] D(B)=S(B)

An application of [F+] yields:

[B2] D(A)=S(A)

Transitivity of identity applied to [B1], [F6'] and [B2] gives us:

[B3] S(A)=S(B)

which contradicts [F5'].

If you find this formulation useful, feel free to use it in class (it would be nice if you let me know if you do, though).

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