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**The current king of France, propositions, sentences, and missing subjects**

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Tue 29 Sep, 2009 09:09 am

Posted on my blog.

**The current king of France, propositions, sentences, and missing subjects**

Consider this set of propositions:

[INDENT] 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.

[/INDENT] 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.

**About (P3)**2 It is placed in the second sentence part, and it is *that*3

Why (P6) is true

Additionally we can now see why (P6) is true. Before we just thought (P6) evidenced by our intuitions. But since both (P9) and (P10) imply (by simplification) that:

[INDENT] P11. (∃x)(Kx)

P11'. There exists at least one current king of France.

[/INDENT] And, as we know there is no current king of France, therefore, (P11) is false. If (P11) is false, then so are (P9) and (P10). If they are both false, then (P6) is true.

We can also generalize a bit from this. What seems to confuse people about (S1) and (S2) is that the subject, that is, the current king of France, is missing, that is, he does not exist.

Appendix

Here is a predicate logic version of the original set.

Translation key

Ex ≡ x expresses a proposition.

C(AB) ≡ A and B are contradictories.

a ≡ (The sentence) "The current king of France is bald."

b ≡ (The sentence) "The current king of France is not bald."

A ≡ (The proposition) "The current king of France is bald."

B ≡ (The proposition) "The current king of France is not bald."

The set**Notes**

1I complicate matters a bit with bivalance because it is difficult to formulate things without bivalance.

2More on this in a forthcoming essay.

3The set may still be inconsistent for other reasons, though it doesn't seem inconsistent to me.

Consider this set of propositions:

[INDENT] 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.

[/INDENT] 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.

Why (P6) is true

Additionally we can now see why (P6) is true. Before we just thought (P6) evidenced by our intuitions. But since both (P9) and (P10) imply (by simplification) that:

[INDENT] P11. (∃x)(Kx)

P11'. There exists at least one current king of France.

[/INDENT] And, as we know there is no current king of France, therefore, (P11) is false. If (P11) is false, then so are (P9) and (P10). If they are both false, then (P6) is true.

We can also generalize a bit from this. What seems to confuse people about (S1) and (S2) is that the subject, that is, the current king of France, is missing, that is, he does not exist.

Appendix

Here is a predicate logic version of the original set.

Translation key

Ex ≡ x expresses a proposition.

C(AB) ≡ A and B are contradictories.

a ≡ (The sentence) "The current king of France is bald."

b ≡ (The sentence) "The current king of France is not bald."

A ≡ (The proposition) "The current king of France is bald."

B ≡ (The proposition) "The current king of France is not bald."

The set

1I complicate matters a bit with bivalance because it is difficult to formulate things without bivalance.

2More on this in a forthcoming essay.

3The set may still be inconsistent for other reasons, though it doesn't seem inconsistent to me.

mickalos

Reply
Sun 29 Nov, 2009 02:44 pm

@Emil,

Quote:

The correct interpretation of the propositions expressed by (S1) and (S2) is therefore:

P9. (∃x)(Kx∧Bx)

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."

Quote:

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.

No, Russell's theory of description has been pretty dominant for quite some time; indeed, mind published a special "On Denoting edition" on the 100th anniversary of "On Denoting" the article in which he first presented it. I'm more inclined to agree with Strawson. While Russell does provide a complete list of necessary conditions for the truth of "The current King of France is Bald":

1. There exists an x such that x is a present King of France (∃x(Kx))

2. For all x that is a present King of France and every y that is a present King of France, x equals y (i.e., there is at most one present King of France) (∀x(Kx → ∀y(Ky → y=x)))

3. For all x that is a present King of France, x is bald. (∀x(Kx → Bx))

This is not the same thing as saying Russell has provided an accurate analysis of the sentence "The current King of France is bald", which I'm going to refer to as S henceforth. When the sentence S was uttered under Louis XIV, it is conceivable that it expresses a statement with a different truth value to when the sentence was uttered under Louis XV (i.e. If one was bald and the other was not). Therefore, it seems that we cannot talk of S being true or false, but only of it being used to express a true or false proposition.

Thus we see that the thing that "the present King of France" refers to changes dependent on context. Russell wants to say that the meaning of a name is the actual thing that it denotes (i.e. "London" means London, the actual city) , so if a name fails to pick any object out of the world it is meaningless. Therefore as S is not meaningless (which it would be without a meaningful subject), but "the present King of France" fails to refer to anything in particular, "The present King of France" is fundamentally different from a name, it is a definite-description. However, if you were to ask me what does the name "John Smith's house" mean, I wouldn't point to a particular house; this name could be used to refer to any house belonging to any person named John Smith. Russell confuses meaning with reference.

To illustrate what I mean, consider the expression "this". If you ask me what "this" means, I'm not going to hand you the object I just used it to refer to, saying that the meaning of "this" changes each time it is used. Nor should I hand you every object it ever has and ever will be used to refer to. Instead, I should explain and illustrate the conventions governing the use of the

expression. The meaning of a denoting expression is the set of rules, habits and conventions for its use in referring. Thus, given the sentence "The library is contains philosophy books", it would be absurd to ask what object that sentence is about; this would depend on when, how and where the sentence was used. Similarly it would be absurd to ask whether or not the sentence was true for the same reasons. If, in fact, the sentence does fail to refer to anything, then it can only be a spurious pseudo-use (if for example, I uttered S now in an attempt to describe the world); the language habits, conventions and rules required that the sentence could be logically used to talk about something do not allow us to use "the present King of France" as a denoting phrase today do not exist (except in exceptional circumstances, perhaps if we wanted to mockingly refer to a particularly arrogant Frenchman), and thus when it is used so we say something neither true or false.

Emil

Reply
Thu 3 Dec, 2009 07:20 am

@Emil,

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.)

Owen phil

Reply
Fri 18 Dec, 2009 02:35 pm

@Emil,

Definite descriptionsD1. (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.

kennethamy

Reply
Fri 18 Dec, 2009 05:22 pm

@Owen phil,

Owen;112452 wrote:

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.

Yes. The scope of the negation sign should not be confused. Russell made that very clear.

mickalos

Reply
Sat 19 Dec, 2009 01:27 am

@kennethamy,

Emil;107800 wrote:

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.)

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

mickalos;106910 wrote:

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."

∃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).

Emil

Reply
Sat 19 Dec, 2009 02:46 am

@Emil,

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.

Owen phil

Reply
Sat 19 Dec, 2009 04:58 am

@mickalos,

mickalos;112574 wrote:

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).

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 ----------

Emil;112590 wrote:

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.

(∃!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?

Emil

Reply
Sat 19 Dec, 2009 11:03 am

@Owen phil,

Owen;112620 wrote:

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?

Sorry. Your symbols have me confused. Use these.

mickalos

Reply
Sat 19 Dec, 2009 02:55 pm

@Emil,

Emil;112590 wrote:

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.

Emil

Reply
Sat 19 Dec, 2009 10:24 pm

@mickalos,

mickalos;112716 wrote:

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.

Owen phil

Reply
Mon 21 Dec, 2009 04:37 am

@Emil,

Emil;112870 wrote:

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.

The unit quantifier is defined in Principia Mathematica (1910) by Russell and Whitehead...

*14.02 E!(the x:Fx) =df EyAx(x=y <-> Fx).

There are an endless number of uniquiness quantifiers such as:

There is exactly 0 x such that Fx...~ExFx.

There is exactly 1 x such that Fx... EyAx(x=y <-> Fx).

There is exactly 2 x's such that Fx ...

There is exactly 3 x's such that Fx ...

etc.

See: An Introduction to Symbolic Logic and Its Applications, R. Carnap, (1962). page 140.

(a is a) implies there is exactly one x such that it equals a.

a=a -> E!(the x=a).

Proof,

E!(the x=a) <-> EyAx(x=y <-> x=a)

E!(the x=a) <-> Ey(y=a)

E!(the x=a) <-> a=a.

E!(the x=a).

Therefore, AyE!(the x=y), ie. every value of the individual variable is unique.

Note, (the x=y)=y, is true for all y's.

[I now see a mistake in that proof, E!(the x=a) should replace E!x(x=a)]

kennethamy

Reply
Mon 21 Dec, 2009 08:02 am

@mickalos,

mickalos;112716 wrote:

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.

It is introduced in

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