[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).