Strawson's Gap and How Grice Filled It

Speranza

Jones wrote:

"I am very bad in attributing "Principia Mathematica" to Russell alone", rather than to Whitehead and Russell.

No problem. It's just that I love Whitehead, especially the Whitehead of "Principia"!

Jones goes on:

"Usually I am talking about Russell's Theory of Types, or related doctrines of Russell's such as his theory of descriptions, which is my excuse for thinking just of Russell. But nevertheless, if I refer to "Principia Mathematica" rather than specifically to Russell's work I should credit both authors."

Well, perhaps Russell was mainly responsible for that bit that was wrong!

---- (Just joking).

It was worse with Grice. In "Prejudices and Predilections" he implicates that most of the material in Strawson's book "Individuals" derived from his joint seminars on "Categories", and there's not one single mention of Grice in it!

----

Jones goes on:

"I'm not convinced by the rejection of the implication from simple statements about definite descriptions to the existence of something satisfying the description, since this does seem to hold in [Principia Mathematica] and I would think in [Grice's] System Q [in Vacuous Names]. Can you tell us how Grice comes to the conclusion that the implication does not hold. The reason for including the definite description in the square brackets in indicating the scope is that there may be more than one definite description in that scope, and you need some way of telling which definite description is being assigned that particular scope. There may by another which occurs in that scope but whose scope is broader. You could allow that detail to be omitted and write only "[].s" if only one definite description appears in "s", but I don't think Russell does."

I see.

The use of the square brackets in Grice differs from the use of the square brackets in Whitehead's and Russell's "Principia Mathematica".

There is a THIRD use of square brackets that is more or less irrelevant here -- as when square brackets are used instead of simple brackets.

∃x[∀y(Ky ≡ y=x) & ~Bx]

~∃x[∀y(Ky ≡ y=x) & Bx] 

Note that in the above square brackets are used, but they can be replaced by simple brackets:

∃x(∀y(Ky ≡ y=x) & ~Bx))

~∃x(∀y(PKFy ≡ y=x) & Bx)

The way Grice uses the square brackets in WoW ("Presupposition and Conversational Implicature") seems to be the following.

He seems to be saying that if we start with

i. The King of France is bald.

and then apply "~", the only possible logical form is, indeed

~∃x(∀y(Ky ≡ y=x) & Bx)

-- and the considers variants which are not really isomorphically co-extensive, but apply, and not just

ii. The king of France is not bald.

But:

ii. It is false that the king of France is bald.

iii. It is not the case that the king of France is bald.

---- His views on negation may have been influenced by Aristotle, but they should have been influenced MORE than they _were_ by Aristotle.

So we have the canonical negation of the expression containing the definite description (already analysed alla Whitehead/Russell):

~∃x(∀y(Ky ≡ y=x) & Bx)

So, why is Strawson saying that (ii) The king of France is not bald, like (i) The king of France is bald, is NEITHER TRUE NOR FALSE?

Strawson's confusion (but don't tell Burton-Roberts Strawson is confused. Burton-Roberts defines himself as a neo-Strawsonian, and I think Paul Elbourne too) is double.

Re: the affirmative, Strawson seems unable to accommodate with the implicature, where the implicature is CANCELLED:

iv. The king of France is not bald; indeed, there is no king of France.

Grice provides another example. The Loyalty Examiner. If A knows that B knows that such examiner does not exist (he -- the Loyalty Examiner only summons traitors), and A knows that B is hardly a traitor, A can say:

"The Loyalty Examiner won't be summoning you, at any rate -- you mark my words".

Here, the implicature is cancelled in AN IMPLICIT way, as opposed to (iv) where the cancellation is explicit.

It is perhaps more difficult how to argue against Strawson's case for the AFFIRMATIVE "The king of France is bald" being NEITHER TRUE OR FALSE. But I suppose that Strawson's mistake here is never minding what Whitehead and Russell analyse in "Principia Mathematica".

Back to the square brackets as used by Grice.

If the only negation is:

~∃x(∀y(Ky ≡ y=x) & Bx)

how is that we 'feel' the utterance to read as per below?

∃x(∀y(Ky ≡ y=x) & ~Bx)

The answer Grice gives is that we tend to assign common ground status to uncontroversial material. He gives one example:

A: I love Kaufmann.
B: My aunt went to his concert. She loves him too.

When B says "My aunt" he is not expecting A to go on with:

A: I did not know you had an aunt.

He takes B's words for granted. They are part of the 'common ground', non-controversial.

∃x(∀y(Ky ≡ y=x) & Bx)

the topic is 'the king of France', and the COMMENT, which adds information is that he is bald.

In the negation, by default:

~∃x(∀y(Ky ≡ y=x) & Bx)

we tend to IMPORT the "~", so that it reads

∃x(∀y(Ky ≡ y=x) & ~Bx)

The use of square-brackets here then would be something like:

~[∃x(∀y(Ky ≡ y=x) &] Bx)

-- which has some odd syntax to it, and perhaps the below should be preferred:

[∃x(∀y(Ky ≡ y=x) &] ~Bx)

When it comes to the subscript device in "Vacuous Names", we seem to have two choices.

We may apply the subscript to the negation (this is where Elbourne thinks we go wrong) or to the iota operator.

The possible formulae would be something like:

I. APPLYING SUBSCRIPTS TO THE NEGATION SIGN:

A) ~_mP_n

-- where "~" has broader scope than the 'phrastic' P.

B) ~_nP_m

-- where "~" has narrower scope.


II. but then we can there's the option of APPLYING SUBSCRIPTS to the iota operator

Taking:

[(x)(Kx)] . ~B(x)(Kx)

~[(x)(Kx)] . B(x)(Kx)

we may find a way of NOT HAVING to repeat the use of the iota symbol, and merely indicate by means of a subscript whether this or that occurrence has either broad or narrow scope.

  (x)_m(Kx and  ~Bx)_n

 

  (x)_m(Kx and  ~Bx)_m

but I would have to revise in the literature how this could be done -- and I don't mean Shakespeare!
















∃x(∀y(Ky ≡ y=x) & ~Bx))

~∃x(∀y(PKFy ≡ y=x) & Bx)