The correct interpretation of the propositions expressed by (S1) and (S2) is therefore:
P9. (∃x)(Kx∧Bx)

Consider this set of propositions:
P1. The sentence "The current king of France is bald." expresses a proposition.
P2. The sentence "The current king of France is not bald." expresses a proposition.
P3. The propositions expressed by the sentences "The current king of France is bald."and "The current king of France is not bald." are contradictory.
P4. For all p and for all q, if p is a contradictory of q, then (p is true and q is false, or q is true and p is false).1
P5. Iff the sentence "The current king of France is bald." expresses a proposition, then the sentence "The current king of France is not bald." expresses a proposition.
P6. If the sentences "The current king of France is bald." and "The current king of France is not bald." express propositions, then the propositions are both false.
These propositions form an inconsistent set. When a person is confronted with that he believes is inconsistent, then that person will almost always stop believing at least one of the things he believes. However in this case people usually stop believing both (P1) and (P2), and that is because of (P5). To stop believing one of them only would be futile as the set would still be inconsistent. I suppose that the same people find the other propositions in the set much more justified than (P1) and (P2), and that is the reason why they are rejected.

Definite descriptions

D1. (G)(the x:Fx) =df Ey(Ax(x=y <-> Fx) & Gy).

1. (G)(the x:Fx) <-> Ey(Ax(x=y <-> Fx) & Gy).
2. ~(G)(the x:Fx) <-> ~Ey(Ax(x=y <-> Fx) & Gy).
3. (~G)(the x:Fx) <-> Ey(Ax(x=y <-> Fx) & ~Gy).

4. Kx means x is a current king of France,
5. By means y is bald,
then ..

The current king of France is bald, means,
Ey(Ax(x=y <-> Kx) & By).

~(The current king of France is bald), means,
~Ey(Ax(x=y <-> Kx) & By).

The current king of France is non-bald, means,
Ey(Ax(x=y <-> Kx) & ~By).

(1 & 2) is contradictory. (1 & 3) is not contradictory.


D2. E!(the x:Fx) =df EyAx(x=y <-> Fx). [(the x:Fx) exists]

[~(G)(the x:Fx) <-> (~G)(the x:Fx)] <-> E!(the x:Fx).

The not/non ambiguity wrt definite descriptions is resolved.

The current king of France is bald, is false.
The current king of France is non-bald, is false.
~(The current king of France is bald), is true.
~(The current king of France is non-bald), is true.

There are no properties that 'the current king of France' has.

The current king of France does not exist.

Alright. I forgot about the uniqueness part. Luckily it is a pretty easy fix. I don't need to do the complicated fix that you use. I can simply write (∃!x) as a placeholder.

I agree that sense and reference are different. (Frege there.)

This might seem a bit pedantic, but this claim could be misleading. Strictly speaking the correct formalisation of "The current King of France is bald" using your dictionary, at least if you're going to take Russell's approach to analysing sentences containing denoting phrases, is:

(∃x)(Kx∧Bx∧∀y(x=y))
"There exists an x such that x is a current King of France, x is bald, and there is only one x."

The last clause is necessary to capture the uniqueness implied by "The" in "The current King of France."

Funnily enough I've just noticed that my correction of the formailisation is wrong; at least the first part of it:


∃x(Kx∧Bx∧∀y(x=y)) says there exists an x such that x is the King of France, x is Bald, and x is the only thing. The formalisation that I promise is definitely correct is ∃x(Kx∧Bx∧∀y(Ky → y=x)). The irony. Too late to edit it now.

If you want to be more concise ∃x∀y((Ky ↔ y=x)∧Bx), as used in Owen's post, is logically equivelent, but I prefer the first as I prefer to think about meaning in idiomatic English and the second seems much more removed from that (at least it is harder for me to translate it).

But why don't you simply use (∃!x)(Kx∧Bx)? (There exists exactly one x such that x is the current king of France and x is bald) The "!" means a hidden (∀y)(Ky⇒y=x) clause? (For all y, that y is a current king of France logically implies that y is identical to x)

Also note that I use "→" for material implication, it should be logical implication, so "⇒" is used instead.

Yes,

B(the x:Kx) <->
Ex(Ay(x=y <-> Kx) & Bx) <-> Ex(Kx & Bx & Ay(Ky → y=x)).

Note: Kx <-> Ay(y=x -> Ky), that is...

Ex(Kx & Bx & ∀y(Ky → y=x)) <-> Ex(Ay(y=x -> Ky) & Ay(Ky -> y=x) & Bx)
ie.
Ex(Kx & Bx & ∀y(Ky → y=x)) <-> Ex(Ay(y=x <-> Ky) & Bx).


An easier translation is given by Russell in 'Introduction to Mathematical Philosophy' page 177.

B(the x:Kx) <-> (ExKx & AyAx((Ky & Kx) -> x=y) & Ax(Kx -> Bx)).

That is to say,

The current king of France is bald, means,
1. At least one x is a current king of France &
2. At most one x is a current king of France &
3. Whatever is a current king of France is bald.

---------- Post added 12-19-2009 at 06:43 AM ----------



(∃!x)(Kx) means ExAy(x=y <-> Kx), by definition.
(∃!x)(Kx∧Bx) means ExAy(x=y <-> (Kx & Bx))

(The x:(Kx & Bx)) is unique, does not mean B(the x:Kx).

(∀y)(Ky⇒y=x) means Ay[](Ky -> y=x), which is equivalent to
[]Ay(Ky -> y=x).

Necessarily(there is at most one x:Kx) does not imply B(the x:Kx).

This clause, []Ay(Ky -> y=x), is not implied by B(the x:Kx).

Why do you think that (∀y)(Ky⇒y=x) is a hidden clause?

But why don't you simply use (∃!x)(Kx∧Bx)? (There exists exactly one x such that x is the current king of France and x is bald) The "!" means a hidden (∀y)(Ky⇒y=x) clause? (For all y, that y is a current king of France logically implies that y is identical to x)

Also note that I use "→" for material implication, it should be logical implication, so "⇒" is used instead.

I had no idea what the (∃!x) symbol meant until now, and in fact I've never seen it used before. It's clearly not a conventional symbol, so if you use it you probably won't be understood. I'm not really sure how hidden clauses are supposed to work in a formal language. Usually when people learn logic they might be taught different bracketing conventions, semantics and proof methods, but generally we all use the same syntax.

I would advise not using (∃!x) any more, because hardly anybody uses it, and this is supposed to be a language. If you want to check out a more conventional language, the lecture slides at the bottom of this page give a pretty detailed outline The Logic Manual Lectures four and five give the syntax and semantics.

I found it here, here and some other places, and I've seen it before too some other places I can't remember.

That hardly anyone uses it is not a good reason not to use it, but it is a good reason not to use it without defining it first.

I had no idea what the (∃!x) symbol meant until now, and in fact I've never seen it used before. It's clearly not a conventional symbol, so if you use it you probably won't be understood. I'm not really sure how hidden clauses are supposed to work in a formal language. Usually when people learn logic they might be taught different bracketing conventions, semantics and proof methods, but generally we all use the same syntax.

I would advise not using (∃!x) any more, because hardly anybody uses it, and this is supposed to be a language. If you want to check out a more conventional language, the lecture slides at the bottom of this page give a pretty detailed outline The Logic Manual Lectures four and five give the syntax and semantics.