2

# Is that which is not equal to itself, nothing, equal to itself or not?

Sun 3 Oct, 2010 03:36 am
Leibniz's identity of indiscernibles, (all F: Fx <-> Fy) -> x=y, fails if x or y is a non-referring described object.

For example, (all F: F(the x:~(x=x) <-> F(the x:~(x=x)) -> (the x:~(x=x)) = (the x:~(x=x)), is a contradiction.
The premise is tautologous and the conclusion is contradictory.

(all F: F(the x:~(x=x) <-> F(the x:~(x=x)), is tautologous for each and every F, ie. (all p: p <-> p).

If we define: nothing =df (the x:~(x=x)) then...

~(nothing = nothing), is a theorem.

Proof:

D1. G(the x:Fx) =df Ey(Ax(x=y <-> Fx) & Gy).
D2. nothing =df (the x:~(x=x)).

1. (nothing = nothing) <-> Ey(Ax(x=y <-> ~(x=x)) & y=y). By D1.
2. (nothing = nothing) <-> EyAx(x=y <-> ~(x=x)). By: 1, (all y: y=y).
3. (nothing = nothing) <-> EyAx(x=y <-> contradiction). By, (all x: x=x).
4. (nothing = nothing) <-> EyAx~(x=y). By, (p <-> contradiction) <-> ~p.
5. (nothing = nothing) <-> ~AxEy(x=y).
By: 4, Ax~Fx <-> ~ExFx, Ex~Fx <-> ~AxFx.
6. AxEy(x=y), is a theorem.
7. (nothing = nothing) <-> contradiction. By: 5, 6.
8. ~(nothing = nothing). By, (p <-> contradiction) <-> ~p.
QED.

Clearly, (the x:~(x=x)) is indiscernible from (the x:~(x=x)) and yet it is false that it is equal to itself.

D3. (the x:Fx) exists =df EyAx(x=y <-> Fx).

(nothing exists) <-> (the x:~(x=x)) exists.
(nothing exists) <-> EyAx(x=y <-> ~(x=x)).
(nothing exists) <-> EyAx(x=y <-> contradiction)
~((the x:~(x=x)) exists). QED.

~((the present king of France) = (the present king of France)), is proven in the same way.
The present king of France is indiscernible from itself and not equal to itself.

If the laws of identity are to include 'descriptions' (free logic?) then we need a new definition of identity.

D3a. E!(the x:Fx) =df EyAx(x=y <-> Fx).
D4. (the x:Fx) = (the x:Gx) =df E!(the x:Fx) & E!(the x:Gx) & (all H:H(the x:Fx) <-> H(the x:Gx)).

If (the x:Fx) exists and (the x:Gx) exists and they are indiscernible then they are equal.

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djjd62

1
Sun 3 Oct, 2010 05:37 am 0 Replies

north

1
Sun 3 Oct, 2010 10:42 pm
@Owen phil,
Owen phil wrote:

Leibniz's identity of indiscernibles, (all F: Fx <-> Fy) -> x=y, fails if x or y is a non-referring described object.

For example, (all F: F(the x:~(x=x) <-> F(the x:~(x=x)) -> (the x:~(x=x)) = (the x:~(x=x)), is a contradiction.
The premise is tautologous and the conclusion is contradictory.

(all F: F(the x:~(x=x) <-> F(the x:~(x=x)), is tautologous for each and every F, ie. (all p: p <-> p).

If we define: nothing =df (the x:~(x=x)) then...

~(nothing = nothing), is a theorem.

Proof:

D1. G(the x:Fx) =df Ey(Ax(x=y <-> Fx) & Gy).
D2. nothing =df (the x:~(x=x)).

1. (nothing = nothing) <-> Ey(Ax(x=y <-> ~(x=x)) & y=y). By D1.
2. (nothing = nothing) <-> EyAx(x=y <-> ~(x=x)). By: 1, (all y: y=y).
3. (nothing = nothing) <-> EyAx(x=y <-> contradiction). By, (all x: x=x).
4. (nothing = nothing) <-> EyAx~(x=y). By, (p <-> contradiction) <-> ~p.
5. (nothing = nothing) <-> ~AxEy(x=y).
By: 4, Ax~Fx <-> ~ExFx, Ex~Fx <-> ~AxFx.
6. AxEy(x=y), is a theorem.
7. (nothing = nothing) <-> contradiction. By: 5, 6.
8. ~(nothing = nothing). By, (p <-> contradiction) <-> ~p.
QED.

Clearly, (the x:~(x=x)) is indiscernible from (the x:~(x=x)) and yet it is false that it is equal to itself.

D3. (the x:Fx) exists =df EyAx(x=y <-> Fx).

(nothing exists) <-> (the x:~(x=x)) exists.
(nothing exists) <-> EyAx(x=y <-> ~(x=x)).
(nothing exists) <-> EyAx(x=y <-> contradiction)
~((the x:~(x=x)) exists). QED.

~((the present king of France) = (the present king of France)), is proven in the same way.
The present king of France is indiscernible from itself and not equal to itself.

If the laws of identity are to include 'descriptions' (free logic?) then we need a new definition of identity.

D3a. E!(the x:Fx) =df EyAx(x=y <-> Fx).
D4. (the x:Fx) = (the x:Gx) =df E!(the x:Fx) & E!(the x:Gx) & (all H:H(the x:Fx) <-> H(the x:Gx)).

If (the x:Fx) exists and (the x:Gx) exists and they are indiscernible then they are equal.

hmmm... without the mathematics , because I don't understand , explain

simply

look I see " nothing " as a non-physical enity in the end

mathematics don't mean squat if the mathematics don't produce a physical thing or represent a physical thing , entity

mathematical imagination is one thing

give me something , physical, to grasp
Owen phil

1
Mon 4 Oct, 2010 05:23 am
@north,
north wrote:

hmmm... without the mathematics , because I don't understand , explain

simply

look I see " nothing " as a non-physical enity in the end

mathematics don't mean squat if the mathematics don't produce a physical thing or represent a physical thing , entity

mathematical imagination is one thing

give me something , physical, to grasp

North: ..look I see " nothing " as a non-physical enity in the end ..

Logical analysis requires symbolic logic, in order to clarify meanings.
The is no mathematics in my post at all...first order predicate logic with identity is all that is needed.

North: .. mathematics don't mean squat if the mathematics don't produce a physical thing or represent a physical thing , entity..

I don't agree. Although physics is a major application of mathematics, pure mathematics is independent of the physical world.

It is not the case that 'nothing' is something.
There is no entity, physical or non-physical, that nothing is.
(some x: x= nothing) is a contradiction.
There is no property that 'nothing' has.
'Nothing' does not exist!

It is a described entitiy at best, much like the present king of France.
"Nothing" is often used as though it named an object..e.g. that which is and is not.

(nothing is nothing) is a common expression that we assume has truth in virtue of the logical truth that: everything is equal to itself (A=A).
Note that 'nothing' is not a value of A in A=A.
But, everything is equal to itself requires that 'it' exists, that is to say: ~(nothing = nothing), because it (nothing) does not exist.

If there is a true statement that has x as its subject, then x exists,
There is no true statement which has 'nothing' as its subject.

An entity described by a contradictory predication (eg. the x: Fx & ~Fx) cannot exist.

Nothing is in my fridge, is false...even if the fridge is empty.
Nothing is in my fridge, means, It is not the case that something (some food thing) is in my fridge...which is often true of my fridge.

How do you define 'nothing'?
Fil Albuquerque

1
Mon 4 Oct, 2010 06:07 am
@Owen phil,
Well...I certainly agree that nothing does n´t exist to the point at it cannot even equal itself...Nothing can only be a figure of speech. No real value.
0 Replies

Etude

1
Mon 4 Oct, 2010 01:44 pm
@Owen phil,
hmm... i would say that zero best describes 'nothing'. because in A=A -> A-A=0.
so something deleted by the same 'something' will give you nothing. which is only true in mathematics land.
however in the physical realm, i don't think it's possible to create nothing from something. wait... how do you want us to describe 'nothing'? physically or abstractly?
Owen phil

1
Tue 5 Oct, 2010 04:23 am
@Etude,
Etude wrote:

hmm... i would say that zero best describes 'nothing'. because in A=A -> A-A=0.
so something deleted by the same 'something' will give you nothing. which is only true in mathematics land.
however in the physical realm, i don't think it's possible to create nothing from something. wait... how do you want us to describe 'nothing'? physically or abstractly?

Either or both.

(the x: Fx & ~Fx) = { }, is a contradiction.
(the x: Fx & ~Fx) = {{ }}, is a contradiction.
(some y:(the x: Fx &~Fx) =y), is a contradiction.

There is no y such that (y= nothing).
0 Replies

kennethamy

1
Tue 5 Oct, 2010 11:28 am
@Owen phil,
Owen phil wrote:

north wrote:

hmmm... without the mathematics , because I don't understand , explain

simply

look I see " nothing " as a non-physical enity in the end

mathematics don't mean squat if the mathematics don't produce a physical thing or represent a physical thing , entity

mathematical imagination is one thing

give me something , physical, to grasp

North: ..look I see " nothing " as a non-physical enity in the end ..

Logical analysis requires symbolic logic, in order to clarify meanings.
The is no mathematics in my post at all...first order predicate logic with identity is all that is needed.

North: .. mathematics don't mean squat if the mathematics don't produce a physical thing or represent a physical thing , entity..

I don't agree. Although physics is a major application of mathematics, pure mathematics is independent of the physical world.

It is not the case that 'nothing' is something.
There is no entity, physical or non-physical, that nothing is.
(some x: x= nothing) is a contradiction.
There is no property that 'nothing' has.
'Nothing' does not exist!

It is a described entitiy at best, much like the present king of France.
"Nothing" is often used as though it named an object..e.g. that which is and is not.

(nothing is nothing) is a common expression that we assume has truth in virtue of the logical truth that: everything is equal to itself (A=A).
Note that 'nothing' is not a value of A in A=A.
But, everything is equal to itself requires that 'it' exists, that is to say: ~(nothing = nothing), because it (nothing) does not exist.

If there is a true statement that has x as its subject, then x exists,
There is no true statement which has 'nothing' as its subject.

An entity described by a contradictory predication (eg. the x: Fx & ~Fx) cannot exist.

Nothing is in my fridge, is false...even if the fridge is empty.
Nothing is in my fridge, means, It is not the case that something (some food thing) is in my fridge...which is often true of my fridge.

How do you define 'nothing'?

In PM one way of expressing the proposition that X exists is "X=X", I believe, and so one way to expressing the proposition that X does not exist is ~(X-X).
Owen phil

1
Tue 5 Oct, 2010 12:16 pm
@kennethamy,
Ken:
"In PM one way of expressing the proposition that X exists is "X=X", I believe, and so one way to expressing the proposition that X does not exist is ~(X=X). "

Yes, Russell and Whitehead show that: (the x:Fx) = (the x:Fx) <-> E!(the x:Fx), and that (the x: x=a) = a. That is, a = a <-> E!a.

We can also demonstrate that: E!a <-> (some F: Fa).
kennethamy

1
Tue 5 Oct, 2010 01:28 pm
@Owen phil,
Owen phil wrote:

Ken:
"In PM one way of expressing the proposition that X exists is "X=X", I believe, and so one way to expressing the proposition that X does not exist is ~(X=X). "

Yes, Russell and Whitehead show that: (the x:Fx) = (the x:Fx) <-> E!(the x:Fx), and that (the x: x=a) = a. That is, a = a <-> E!a.

We can also demonstrate that: E!a <-> (some F: Fa).

Thank you. So the answer to the OP is, "nothing". Is that right?
Owen phil

1
Tue 5 Oct, 2010 02:02 pm
@kennethamy,
Ken: Thank you. So the answer to the OP is, "nothing". Is that right?

If we define 'nothing' as (the x such that ~(x=x)) then (nothing = nothing) is false.
That is, (the x:~(x=x)) = (the x:~(x=x)) is contradictory.

I think Carnap called (the x:~(x=x)) the null object, and (the x's:~(x=x)) the null set.

kennethamy

1
Tue 5 Oct, 2010 06:14 pm
@Owen phil,
Owen phil wrote:

Ken: Thank you. So the answer to the OP is, "nothing". Is that right?

If we define 'nothing' as (the x such that ~(x=x)) then (nothing = nothing) is false.
That is, (the x:~(x=x)) = (the x:~(x=x)) is contradictory.

I think Carnap called (the x:~(x=x)) the null object, and (the x's:~(x=x)) the null set.

Which is just to say that it is false that anything is not self-identical.
0 Replies

wayne

1
Thu 7 Oct, 2010 12:54 am
@Owen phil,
Owen phil wrote:

Leibniz's identity of indiscernibles, (all F: Fx <-> Fy) -> x=y, fails if x or y is a non-referring described object.

For example, (all F: F(the x:~(x=x) <-> F(the x:~(x=x)) -> (the x:~(x=x)) = (the x:~(x=x)), is a contradiction.
The premise is tautologous and the conclusion is contradictory.

(all F: F(the x:~(x=x) <-> F(the x:~(x=x)), is tautologous for each and every F, ie. (all p: p <-> p).

If we define: nothing =df (the x:~(x=x)) then...

~(nothing = nothing), is a theorem.

Proof:

D1. G(the x:Fx) =df Ey(Ax(x=y <-> Fx) & Gy).
D2. nothing =df (the x:~(x=x)).

1. (nothing = nothing) <-> Ey(Ax(x=y <-> ~(x=x)) & y=y). By D1.
2. (nothing = nothing) <-> EyAx(x=y <-> ~(x=x)). By: 1, (all y: y=y).
3. (nothing = nothing) <-> EyAx(x=y <-> contradiction). By, (all x: x=x).
4. (nothing = nothing) <-> EyAx~(x=y). By, (p <-> contradiction) <-> ~p.
5. (nothing = nothing) <-> ~AxEy(x=y).
By: 4, Ax~Fx <-> ~ExFx, Ex~Fx <-> ~AxFx.
6. AxEy(x=y), is a theorem.
7. (nothing = nothing) <-> contradiction. By: 5, 6.
8. ~(nothing = nothing). By, (p <-> contradiction) <-> ~p.
QED.

Clearly, (the x:~(x=x)) is indiscernible from (the x:~(x=x)) and yet it is false that it is equal to itself.

D3. (the x:Fx) exists =df EyAx(x=y <-> Fx).

(nothing exists) <-> (the x:~(x=x)) exists.
(nothing exists) <-> EyAx(x=y <-> ~(x=x)).
(nothing exists) <-> EyAx(x=y <-> contradiction)
~((the x:~(x=x)) exists). QED.

~((the present king of France) = (the present king of France)), is proven in the same way.
The present king of France is indiscernible from itself and not equal to itself.

If the laws of identity are to include 'descriptions' (free logic?) then we need a new definition of identity.

D3a. E!(the x:Fx) =df EyAx(x=y <-> Fx).
D4. (the x:Fx) = (the x:Gx) =df E!(the x:Fx) & E!(the x:Gx) & (all H:H(the x:Fx) <-> H(the x:Gx)).

If (the x:Fx) exists and (the x:Gx) exists and they are indiscernible then they are equal.

The whole problem with nothing seems to be lack of definition.
Nothing is intended to refer to no thing. Therefore the term nothing, defines no thing. It appears to me as impossible to equate that which lacks definition as we would that which is defined.
It would be possible to equate void x with void x, void being a measurement of nothing. Or would it? Is nothing measurable?
A measurable absence of things is called void. Is there anything in a void? No, there is nothing in a void.
Is the human mind capable of grasping that which lacks definition?
I think not. Infinity stumps us all.
laughoutlood

1
Thu 7 Oct, 2010 04:48 am
@Owen phil,
i canardly wait until the next instalment on much ado about dux
0 Replies

kennethamy

1
Thu 7 Oct, 2010 06:19 am
@wayne,
wayne wrote:

Owen phil wrote:

Leibniz's identity of indiscernibles, (all F: Fx <-> Fy) -> x=y, fails if x or y is a non-referring described object.

For example, (all F: F(the x:~(x=x) <-> F(the x:~(x=x)) -> (the x:~(x=x)) = (the x:~(x=x)), is a contradiction.
The premise is tautologous and the conclusion is contradictory.

(all F: F(the x:~(x=x) <-> F(the x:~(x=x)), is tautologous for each and every F, ie. (all p: p <-> p).

If we define: nothing =df (the x:~(x=x)) then...

~(nothing = nothing), is a theorem.

Proof:

D1. G(the x:Fx) =df Ey(Ax(x=y <-> Fx) & Gy).
D2. nothing =df (the x:~(x=x)).

1. (nothing = nothing) <-> Ey(Ax(x=y <-> ~(x=x)) & y=y). By D1.
2. (nothing = nothing) <-> EyAx(x=y <-> ~(x=x)). By: 1, (all y: y=y).
3. (nothing = nothing) <-> EyAx(x=y <-> contradiction). By, (all x: x=x).
4. (nothing = nothing) <-> EyAx~(x=y). By, (p <-> contradiction) <-> ~p.
5. (nothing = nothing) <-> ~AxEy(x=y).
By: 4, Ax~Fx <-> ~ExFx, Ex~Fx <-> ~AxFx.
6. AxEy(x=y), is a theorem.
7. (nothing = nothing) <-> contradiction. By: 5, 6.
8. ~(nothing = nothing). By, (p <-> contradiction) <-> ~p.
QED.

Clearly, (the x:~(x=x)) is indiscernible from (the x:~(x=x)) and yet it is false that it is equal to itself.

D3. (the x:Fx) exists =df EyAx(x=y <-> Fx).

(nothing exists) <-> (the x:~(x=x)) exists.
(nothing exists) <-> EyAx(x=y <-> ~(x=x)).
(nothing exists) <-> EyAx(x=y <-> contradiction)
~((the x:~(x=x)) exists). QED.

~((the present king of France) = (the present king of France)), is proven in the same way.
The present king of France is indiscernible from itself and not equal to itself.

If the laws of identity are to include 'descriptions' (free logic?) then we need a new definition of identity.

D3a. E!(the x:Fx) =df EyAx(x=y <-> Fx).
D4. (the x:Fx) = (the x:Gx) =df E!(the x:Fx) & E!(the x:Gx) & (all H:H(the x:Fx) <-> H(the x:Gx)).

If (the x:Fx) exists and (the x:Gx) exists and they are indiscernible then they are equal.

The whole problem with nothing seems to be lack of definition.
Nothing is intended to refer to no thing. Therefore the term nothing, defines no thing. It appears to me as impossible to equate that which lacks definition as we would that which is defined.
It would be possible to equate void x with void x, void being a measurement of nothing. Or would it? Is nothing measurable?
A measurable absence of things is called void. Is there anything in a void? No, there is nothing in a void.
Is the human mind capable of grasping that which lacks definition?
I think not. Infinity stumps us all.

But the definitions are crystal clear. The question asks whether anything can be not self-identical. The answer is just as clearly, no. The reason is that it is a logical truth that everything has the property of being self-identical (being identical with itself). The problem is solved.
0 Replies

Owen phil

1
Thu 7 Oct, 2010 07:05 am
@wayne,
wayne wrote:

The whole problem with nothing seems to be lack of definition.
Nothing is intended to refer to no thing. Therefore the term nothing, defines no thing. It appears to me as impossible to equate that which lacks definition as we would that which is defined.
It would be possible to equate void x with void x, void being a measurement of nothing. Or would it? Is nothing measurable?
A measurable absence of things is called void. Is there anything in a void? No, there is nothing in a void.
Is the human mind capable of grasping that which lacks definition?
I think not. Infinity stumps us all.

I am not convinced that 'nothing' lacks a definition.
Even if 'nothing' is not defined we can still talk about it.

For example:
Nothing exists, means, It is not the case that something exists.
But, something exists is tautologous or axiomatic.
therefore,
It is not the case that something exist, ie. nothing exists..is still a contradiction.

Something exists <-> ExEy(x=y).
Nothing exists <-> ~ExEy(x=y).

a=a -> EyEx(x=y).
a=a -> something exists.

(x exists) <-> x=x <-> Ey(x=y) <-> EF(Fx).

~(x exists) <-> ~(x=x) <-> ~Ey(x=y) <-> ~EF(Fx).
Fil Albuquerque

1
Thu 7 Oct, 2010 07:08 am
@Owen phil,
In other words, this is all about SETS...
..."Does the list of all lists that do not belong to themselves, belongs to itself or not" ?
kennethamy

1
Thu 7 Oct, 2010 07:16 am
@Owen phil,
Owen phil wrote:

wayne wrote:

The whole problem with nothing seems to be lack of definition.
Nothing is intended to refer to no thing. Therefore the term nothing, defines no thing. It appears to me as impossible to equate that which lacks definition as we would that which is defined.
It would be possible to equate void x with void x, void being a measurement of nothing. Or would it? Is nothing measurable?
A measurable absence of things is called void. Is there anything in a void? No, there is nothing in a void.
Is the human mind capable of grasping that which lacks definition?
I think not. Infinity stumps us all.

The term "nothing" is defined in any dictionary of the English language. How can it not be? It is a word in the English language.

The term, "nothing" is grammatically a noun, but, in fact it acts as an adverb like the term, "not", since a phrase like "Nothing is in my drawer but a pair of old socks" means, "It is not the case that there is anything in my drawer but a pair of old socks". The grammatical form of "nothing" as a noun misleads some people into believing that it is the name of some mysterious object. But this is just a case of what Wittgenstein called, "language on holiday". That is, when we theorize about what a word means rather than examining how it is actually used.
I am not convinced that 'nothing' lacks a definition.
Even if 'nothing' is not defined we can still talk about it.

For example:
Nothing exists, means, It is not the case that something exists.
But, something exists is tautologous or axiomatic.
therefore,
It is not the case that something exist, ie. nothing exists..is still a contradiction.

Something exists <-> ExEy(x=y).
Nothing exists <-> ~ExEy(x=y).

a=a -> EyEx(x=y).
a=a -> something exists.

(x exists) <-> x=x <-> Ey(x=y) <-> EF(Fx).

~(x exists) <-> ~(x=x) <-> ~Ey(x=y) <-> ~EF(Fx).
kennethamy

1
Thu 7 Oct, 2010 07:17 am
@kennethamy,
kennethamy wrote:

Owen phil wrote:

wayne wrote:

The whole problem with nothing seems to be lack of definition.
Nothing is intended to refer to no thing. Therefore the term nothing, defines no thing. It appears to me as impossible to equate that which lacks definition as we would that which is defined.
It would be possible to equate void x with void x, void being a measurement of nothing. Or would it? Is nothing measurable?
A measurable absence of things is called void. Is there anything in a void? No, there is nothing in a void.
Is the human mind capable of grasping that which lacks definition?
I think not. Infinity stumps us all.

I am not convinced that 'nothing' lacks a definition.
Even if 'nothing' is not defined we can still talk about it.

For example:
Nothing exists, means, It is not the case that something exists.
But, something exists is tautologous or axiomatic.
therefore,
It is not the case that something exist, ie. nothing exists..is still a contradiction.

Something exists <-> ExEy(x=y).
Nothing exists <-> ~ExEy(x=y).

a=a -> EyEx(x=y).
a=a -> something exists.

(x exists) <-> x=x <-> Ey(x=y) <-> EF(Fx).

~(x exists) <-> ~(x=x) <-> ~Ey(x=y) <-> ~EF(Fx).

The term "nothing" is defined in any dictionary of the English language. How can it not be? It is a word in the English language.

The term, "nothing" is grammatically a noun, but, in fact it acts as an adverb like the term, "not", since a phrase like "Nothing is in my drawer but a pair of old socks" means, "It is not the case that there is anything in my drawer but a pair of old socks". The grammatical form of "nothing" as a noun misleads some people into believing that it is the name of some mysterious object. But this is just a case of what Wittgenstein called, "language on holiday". That is, when we theorize about what a word means rather than examining how it is actually used.
0 Replies

Fil Albuquerque

1
Thu 7 Oct, 2010 07:31 am
It is not the same to say that nothing refers to nothing, then to say, that nothing refers to the absence of something...although it just looks quite the same...

...and this is also the reason why the word showed up in our vocabulary...

0 Replies

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