Witt

Scrambles:
x exists is only meaningful when applied to a description.
P.S. I guess Exists(Vulcan) is valid, but Exists is then a second order predicate.

You and Russell are wrong on this point.

Russell: Principia Mathematica, page 175.

"It would seem that the word "existence" cannot be significantly applied to subjects immediately given; i.e. not only does our definition give no meaning to "E!x," but ther is no reason, in philosophy, to suppose that a meaning of existence could be found which would be applicable to immediately given objects."

But, Russell includes *14.2 (the x: x=y)=y, as a theorem of Principia Mathematica.

Surely then, E!(the x:x=y) <-> E!y, by Leibnitz's law.

How can it be that E!(the x x=y) has sense and E!y does not ??


Undoubtably, Vulcan is a described object.

But, all objects can be considered described objects.

On this basis, Quine suggests that objects can be eliminated ??

Clearly, existence is a predicate of all objects. in spite of Kant.

Kant's refutation of 'Anslem's ontological proof' fails.


Existence can be expressed in first and in second order logic.

Witt

Scrambles

Because (the x : x = y) is an incomplete symbol and y is not (when y is given and not of the form (the z : phi(z)). And E!z is not a predicatable function.

(the x : x=y) is of the form (the x: phi(x)). Every time this occurs in a proposition like
psi(the x : phi(x)) it is to be replaced with

((E x) : phi(x)) & (z)(phi(z) --> z = x) & psi(x)

The domain of (x) and (E x) is all real individuals. Now, since (the x : phi(x)) is an incomplete symbol and only has meaning in the context of a statement, E! (the x : phi(x)) was created so as to just extract the first bit, without the & psi(x) on the end. E!(the x : phi(x)) is a variation on E!(alpha), where alpha is a class. E!(alpha) expresses that alpha has at least one member.



Scrambles

Primal

I am not quite sure what you are trying to say.

Witt

Witt: How can it be that E!(the x x=y) has sense and E!y does not ??

Scrambles: Because (the x : x = y) is an incomplete symbol and y is not

Are you denying that: (the x:x=y)=y?

(the x:x=y)=y ->. F(the x:x=y) <-> Fy, see PM *14.15, page 179.
therefore,
(the x:x=y)=y ->. E!(the x:x=y) <-> E!y.

Scrambles: (when y is given and not of the form (the z : phi(z)). And E!z is not a predicatable function.

Why do you claim that E!x is not predicatable function?

(the x : x=y) is of the form (the x: phi(x)). Every time this occurs in a proposition like
psi(the x : phi(x)) it is to be replaced with

((E x) : phi(x)) & (z)(phi(z) --> z = x) & psi(x)

The domain of (x) and (E x) is all real individuals. Now, since (the x : phi(x)) is an incomplete symbol and only has meaning in the context of a statement, E! (the x : phi(x)) was created so as to just extract the first bit, without the & psi(x) on the end. E!(the x : phi(x)) is a variation on E!(alpha), where alpha is a class. E!(alpha) expresses that alpha has at least one member.


E!(the x:Fx) is defined EyAx(x=y <-> Fx), see:14.01 PM.
E!(the x:x=y) <-> EzAx(x=z <-> x=y).
Ax(x=z <-> x=y) <-> z=y, see *13.183 PM, page 170.

E!(the x:x=y) <-> Ez(z=y)

But, E!x is defined in first order logic as Ey(x=y).

Therefore: E!(the x:x=y) <-> E!y, is a theorem.

Witt

Witt

Witt

quote:
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Your right on! That "common" language brought a end to the language in logic systems, as advocated by G.E. Moore and in late life Wittgenstein in my opinion.
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I am not quite sure what you are trying to say.
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I di not write this, I believe it was John Page.

Witt

Scrambles

Because it isn't.

Your definition right? If it is defined as such then whenever we have E!(the x: x=y) we are stating

(Ex)(z)(z=y <-> x=z) & (Ez)(z = y)

But,

This definition immediately above looks distinctly different from,

(Ex)(z)(z=y <-> x=z) & (Ez)(z = y).

If you define it as you define it then it is a predicable function. But your E! is distinctly different from Russel's E!. If we replace your E! with E?, since it is completely different to Russel's E!, we see that your last theorem (which is true) is,

E!(the x:x=y) <-> E?y.

But who cares?

John Page

'twas cobrashock

Witt

Witt: E!x is defined in first order logic as Ey(x=y)

Scrambles: Your definition right? If it is defined as such then whenever we have E!(the x: x=y) we are stating

(Ex)(z)(z=y <-> x=z) & (Ez)(z = y)

Not quite.
E!(the x:x=y) <-> (Ex)(z)(z=y <-> x=z .& (Ez)(z=x))

From the definition of descriptions:
G(ix:Fx) <-> (Ey)(x)(x=y <-> Fx .& Gy), PM *14.1

E!(ix:Fx) <-> (Ey)(x)(x=y <-> Fx .& (Ez)(z=y))
But, (Ez)(z=x), is a theorem, PM *13.15.
E!(ix:Fx) <-> (Ey)(x)(x=y <-> Fx), which is Russell's definition *14.02.

Therefore,

E!(the x:x=y) <-> (Ex)(z)(z=y <-> x=z)
E!(the x:x=y) <-> (Ex)(x=y)
E!(the x:x=y) <-> E!y.

QED

(x)(E!x) is an implied axiom of FOPL.
It follows from the axiom (x)(x=x).

Witt

Scrambles

Yeah..I probably needed more brackets and got it slightly wrong. I think it's more correctly.,
(Ex) [(z)(z=y <-> x=z)] & (Ez)(z=x)

Definition is not two way implication. If it were, then that would be a primitive proposition rather than a definition. And

(Ey)(x)(x=y <-> F(x)) & (Ez)(z=y)

is not the same as (Ey)(x)(x=y <-> F(x)), even though they have the same truth value. Also, Russel would not need a separate definition of E!(ix:Fx) if he had E!y defined as (Ez)(z=y), because the form of E!(ix:Fx) would then automatically follow. Certainly the following statements hold:

E!(ix : x=y) <-> E?(ix : x=y)
E!(ix : x=y) <-> E?y
E?(ix: x=y) <-> E?y

But E! as defined by russel does not make sense when applied to y. You are using E! as two different symbols. Now, you can argue that your E? has meaning and it means that y exists. But since E? is a predicate, it can only apply to real individuals, so it doesn't tell us anything useful.

You may say it's perfectly legitimate to say E?(ix : phi(x)) and this may or may not be a real individual. But (ix: phi(x)) is an incomplete symbol. It never occurs in the statement, because

E?(ix : phi(x)) is defined as
(Ex)[(y)(phi(y) <-> y=x)] & E?(x).
So E?(x) is only operating on real individuals.


Scrambles

Scrambles

Witt,

With regards to your claim that A=A is invalid. You are rejecting A = A because

(ix : phi(x)) = (ix : phi(x))

can be false. But this disingenuous as it treats (ix : phi(x)) as a complete symbol. With regards E!y...whatever, you win. It is secondary to the thread.

That's all I'm going to say. This seems a fruitless argument.



Scrambles