A collection of infinitely many distinct elements

[ddt]

The "law of contradiction" would seem to live a busy life with you around, Doron. The law of non-contradiction, on the other hand, appears to have been quite excluded from consideration.

Salient point about the nomenclature, but they're two names for the same law. From Law of noncontradiction^WP:

law of noncontradiction (also called the law of contradiction) states, in the words of Aristotle, that "one cannot say of something that it is and that it is not in the same respect and at the same time". In the symbolism of propositional logic, this is expressed as:

[latex]$$\neg (P \wedge \neg P)$$[/latex]

 

[PixyMisa]

Yes, I was just wondering if Doron had taken the name a bit too literally...

 

[doronshadmi]

(bolding mine). Notice the difference in verb? "has" vs. "is"? You earlier said that "x is a property". Employ some rigour.

If you mean to http://www.internationalskeptics.com/forums/showpost.php?p=3783866&postcount=850, that was no more than a typo. But as usual you ignore the rest of my post and conclude only what fits to your context dependent reasoning, isn't it?

Yes it has, as you're stubbornly repeating over and over again a formula which is logically false as a consequence of the law of contradiction. Whenever "x belongs to A" is true, "x belongs not to A" must be false.

Finally we are at the root of the disagreement between us.

"x belongs to A" is true, "x belongs not to A" must be false only if x is local.

I have an idea, let us do it step by step:

Definition 5: x is local if (x € A) xor (x ~€ A).

Definition 6: x is non-local if (x € A) nxor (x ~€ A).

 

[ddt]

If you mean to http://www.internationalskeptics.com/forums/showpost.php?p=3783866&postcount=850, that was no more than a typo. But as usual you ignore the rest of my post and conclude only what fits to your context dependent reasoning, isn't it?

I did not ignore the rest, I spent 3 other sentences on the rest of your post. Your attitude when confronted with typos is very insulting too. First of all, you should proofread your posts. Secondly, when a typo slips through you should notice when someone reacts to it. Thirdly, you should then correct it in the first place and an apology is in place too.

Finally we are at the root of the disagreement between us.

"x belongs to A" is true, "x belongs not to A" must be false only if x is local.

I'm at a loss what you exactly mean by this.

I have an idea, let us do it step by step:

Definition 5: x is local if (x € A) xor (x ~€ A).

Definition 6: x is non-local if (x € A) nxor (x ~€ A).

And what brings this? Let's make a truth table:

(x € A) || (x ~€ A) || (x € A) nxor (x ~€ A)
-------------------------------------------------
   F    ||    T     ||     F
   T    ||    F     ||     F

So you still have that x cannot be non-local.

And you've still not incorporated the "w.r.t. A" part.

 

[doronshadmi]

Instead of Ture\False let us talk about the true values:

€ ~€
F  F
F  T
T  F
T  T

If the true values are the same, then x is non-local.

If the true values are the not the same, then x is local.

 

[PixyMisa]

Instead of Ture\False let us talk about the true values:

€ ~€
F  F
F  T
T  F
T  T

If the true values are the same, then x is non-local.

If the truth values are the same, you have a contradiction. That's exactly what contradiction means.

You've abandoned the very basis for rational thought.

 

[ddt]

Instead of Ture\False let us talk about the true values:

€ ~€
F  F
F  T
T  F
T  T

If the true values are the same, then x is non-local.

If the true values are the not the same, then x is local.

But they can't be the same. You write it yourself in TOUM, just a couple of lines above Definition 6:

€ is a member of an element, but not necessarily a sub-element of an element.
₡ is the negation of €.

So when one is true, the other is false, vice versa.

You've painted yourself in a corner you can't get out of.

BTW, three cheers for the poster who thought of pounding on Definition 6.

 

[Reality Check]

x is an element that has some proprty.

If x (an element that has some proprty) is both belong to A and does not belong to A, then x is non-local (it has non-local property).

A is not spesific because it can be urelemnt or a set, but it does not chage the fact that x has non-local property if x (an element that has some proprty) is both belong to A and does not belong to A.

According to TOUM.pdf x and A are the same types of placeholders - placeholders for elements which can be urelemnt or a set.

• Where does TOUM.pdf mention that x has a proprty?
• Where does TOUM.pdf mention that A is urelemnt or a set and does not have a proprty?
• Where does TOUM.pdf define a proprty?

 

[doronshadmi]

But they can't be the same. You write it yourself in TOUM, just a couple of lines above Definition 6:

So when one is true, the other is false, vice versa.

You've painted yourself in a corner you can't get out of.

BTW, three cheers for the poster who thought of pounding on Definition 6.

F or T of

€ ~€
F  F
F  T
T  F
T  T

Do not describe membership status. Thay describe x as a placeholder of nothing (empty thing) or somthing (full thing).

See TOUM.pdf pages 2,3.

You still do not get the notion the non-locality and locality are more general than nothing or somthing.

 

[ddt]

F or T of

€ ~€
F  F
F  T
T  F
T  T

Do not describe membership status. Thay describe x as a placeholder of nothing (empty thing) or somthing (full thing).

It doesn't matter what exactly they describe. The point is - and I wrote that loud and clear - that ~€ is the negation of €. You wrote that yourself on page 4 of TOUM. Therefore, they cannot be both false or both true at the same time.

The rest of your post is irrelevant to that argument.

 

[doronshadmi]

It doesn't matter what exactly they describe. The point is - and I wrote that loud and clear - that ~€ is the negation of €. You wrote that yourself on page 4 of TOUM. Therefore, they cannot be both false or both true at the same time.

The rest of your post is irrelevant to that argument.

It does matter what exactly they describe.

They exactly not describe € or ~€ so you cannot conclude anything about ~€ or the negation of €.

Let us be more accurate.

If you look at pages 2,3 you will find that:

0 = nothing
1 = somthing

in = €
out = ~€

0 or 1 do not refer to membership (in or out) but to the element ( 0 or 1).

If the element it both 0 or both 1 ( no matter if it is in or out ) the element is non-local (in and out are the same from the point of view of the element).

If the element it not both 0 or not both 1 ( no matter if is in or out ) the element is local (in and out are the not the same from the point of view of the element).

Here it is:

€ ~€
0 0 --> [ ] (in,out are the same) (non-locality)
0 1 --> [ ]. (in,out are not the same (locality)
1 0 --> [.] (in,out are not the same (locality)
1 1 --> [_]_ (in,out are the same) (non-locality)

 

[jsfisher]

Definition 6: If x € A AND x ~€ A, then x is non-local.

Let us look at definition 6 from this point of view:

1) The property of x is shown according to its relation to A (where both €\~€ are examined).

You are belaboring the obvious. Is that because it isn't obvious to you?

2) It does not mean that this property depends on the particular logical connective (AND, in this case) that is used, if we can show that this property can be shown by other logical connectives.

Sure the membership property (and its negation) is dependent on the, dare I say it, the membership property. However, the AND operator is there because you made it a part of your definition for non-local. Change it to something else and you are no longer working with Definition #6.

By the way, we are not two full steps away from the starting point, the definition, and already you have forgotten it. This is just more evidence of a point I raised earlier.

3) "There exists x such that x and ~x" does not stand at the basis of definition 6, because of these reasons:

Actually, it was "there exists Y such that Y and ~Y", but you couldn't grasp the change of variable concept, so we reverted to the original terms, (x € A) and ~(x € A).

a) At definition 6 we examine x properties according to its relation to another thing, called A.

I see you still insist on animating things. No, we don't need to examine properties of x. Definition 6 is a theorem. It states the equivalence of (non-local(x)) and ((x € A) AND ~(x € A)). That said, if we accept your colloquial usage, then yes, the relations between x and A are at the heart of definition 6.

b) A is not x, because "x is a member of itself" is equivalent to {} € {} --> {{}}, and {} is not {{}}.

Why do you even bring this up? It is completely irrelevant to the discussion.

c) Since x is not A, it is not identical to any particular x\A relation, and we cannot assume that x is identical to (x € A).

More irrelevance.

d) In that case "There exits (x € A) such that (x € A) AND ~(x € A)" has nothing to do with definition 6.

Your conclusion, in addition to being completely wrong, is in no way connected to any of your previous points. It follows no scheme of inference.

Contrary to your statement, it has everything to do with definition 6. Definition 6 is the theorem ((x € A) AND ~(x € A)) <==> Nonlocal(x). You have assured us that there are non-local things, therefore \exists x : Nonlocal(x) is also a theorem. The rest follows by simple inference.

4) I explicitly showed that non-locality is also shown by NOR,NXOR connectives, so according to (2) AND,NXOR,NOR depends on non-locality (it is their common value) and they are no more than particular representations of it.

If by "explicitly showed" you mean stated without basis, then yes, we agree.

By the way, in addition to definition 6 being self-contradictory, it also is involved in circular reasoning. You have stated that non-local is defined by logical connectives and the logical connectives are defined by non-local.

5) The universality of the law of contradiction is kept (again, x is not identical to some particular relation with A).

So, you have reversed your position regarding the Law of Non-contradiction?

Be that as it may, why did you include that unrelated nonsense between the parentheses?

 

[doronshadmi]

You are belaboring the obvious. Is that because it isn't obvious to you?



Sure the membership property (and its negation) is dependent on the, dare I say it, the membership property. However, the AND operator is there because you made it a part of your definition for non-local. Change it to something else and you are no longer working with Definition #6.

By the way, we are not two full steps away from the starting point, the definition, and already you have forgotten it. This is just more evidence of a point I raised earlier.



Actually, it was "there exists Y such that Y and ~Y", but you couldn't grasp the change of variable concept, so we reverted to the original terms, (x € A) and ~(x € A).



I see you still insist on animating things. No, we don't need to examine properties of x. Definition 6 is a theorem. It states the equivalence of (non-local(x)) and ((x € A) AND ~(x € A)). That said, if accept your colloquial usage, then yes, the relations between x and A are at the heart of definition 6.



Why do you even bring this up? It is completely irrelevant to the discussion.



More irrelevance.



Your conclusion, in addition to being completely wrong, is in no way connected to any of your previous points. It follows no scheme of inference.

Contrary to your statement, it has everything to do with definition 6. Definition 6 is the theorem ((x € A) AND ~(x € A)) <==> Nonlocal(x). You have assured us that there are non-local things, therefore \exists x : Nonlocal(x) is also a theorem. The rest follows by simple inference.



If by "explicitly showed[sic]" you mean stated without basis, then yes, we agree.

By the way, in addition to definition 6 being self-contradictory, it also is involved in circular reasoning. You have stated that non-local is defined by logical connectives and the logical connectivies are defined by non-local.



So, you have reversed your position regarding the Law of Non-contradiction?

Be that as it may, why did you include that unrelated nonsense between the parentheses?

Your post is non-relevant because you did not understand pages 2,3 of TOUM.pdf.

In order to get it please read post http://www.internationalskeptics.com/forums/showpost.php?p=3784168&postcount=871 .

I think it will help you to understand why the law of contradiction is not relevant in this case.

 

[ddt]

It doesn't matter what exactly they describe. The point is - and I wrote that loud and clear - that ~€ is the negation of €. You wrote that yourself on page 4 of TOUM. Therefore, they cannot be both false or both true at the same time.

The rest of your post is irrelevant to that argument.

It does matter what exactly they describe.

They exactly not describe € or ~€ so you cannot conclude anything about ~€ or the negation of €.

Let us be more accurate.

If you look at pages 2,3 you will find that:

What's written on pages 2 and 3 is far from accurate - it's incomprehensible. What's written on page 4, however, is (pretty) accurate and comprehensible. And just above Definition 6, it says:

₡ is the negation of €.

In other words, it says that for all x and A:

x ₡ A = not (x € A)

and from that, my point follows. Until you renounce that sentence, I'm gonna keep you to it.

0 = nothing
1 = somthing

in = €
out = ~€

0 or 1 do not refer to membership (in or out) but to the element ( 0 or 1).

So you have a 2-element universe? Who do you think you're kidding?

If the element it both 0 or both 1 ( no matter if it is in or out ) the element is non-local (in and out are the same from the point of view of the element).

If the element it not both 0 or not both 1 ( no matter if is in or out ) the element is local (in and out are the not the same from the point of view of the element).

Here it is:

€ ~€
0 0 --> [ ] (in,out are the same) (non-locality)
0 1 --> [ ]. (in,out are not the same (locality)
1 0 --> [.] (in,out are not the same (locality)
1 1 --> [_]_ (in,out are the same) (non-locality)

Gibberish. Write out in full the formulae you're meaning - in the above table, is the 0 resp. 1 the x or the A in "x € A"? And for IPU's sake, please lose the utterly confusing underscores-notation.

 

[doronshadmi]

What's written on pages 2 and 3 is far from accurate - it's incomprehensible. What's written on page 4, however, is (pretty) accurate and comprehensible. And just above Definition 6, it says:

In other words, it says that for all x and A:

x ₡ A = not (x € A)

and from that, my point follows. Until you renounce that sentence, I'm gonna keep you to it.


So you have a 2-element universe? Who do you think you're kidding?


Gibberish. Write out in full the formulae you're meaning - in the above table, is the 0 resp. 1 the x or the A in "x € A"? And for IPU's sake, please lose the utterly confusing underscores-notation.

No ddt,

€ ~€
0 0 --> [ ] (in,out are the same) (non-locality)
0 1 --> [ ]. (in,out are not the same (locality)
1 0 --> [.] (in,out are not the same (locality)
1 1 --> [_]_ (in,out are the same) (non-locality)

is formal.

You simply do not get it because you force your reasoning on it.

€ ~€
0 0 --> [ ] (in,out are the same) (non-locality)
1 1 --> [_]_ (in,out are the same) (non-locality)

and we can use only one of them ( 0 0 or 1 1) in order to represent non-locality.

€ ~€
0 1 --> [ ]. (in,out are not the same (locality)
1 0 --> [.] (in,out are not the same (locality)

and we can use only one of them ( 0 1 or 1 0) in order to represent locality.

You still look at € ~€ instead of the x element.

 

[jsfisher]

I think it will help you to understand why the law of contradiction is not relevant in this case.

Before, you said the Law of [Non-]contradiction was not universal.

Is it universal or not?

 

[ddt]

Your post is non-relevant because you did not understand pages 2,3 of TOUM.pdf.

But those pages are really not relevant. That's chapter 1, "Introduction" - that can be skipped. The real meat starts at page 4 with chapter 2, "Locality and non-locality, basic terms and definitions". That is where you define the things.

 

[jsfisher]

No ddt,

€ ~€
0 0 --> [ ] (in,out are the same) (non-locality)
0 1 --> [ ]. (in,out are not the same (locality)
1 0 --> [.] (in,out are not the same (locality)
1 1 --> [_]_ (in,out are the same) (non-locality)

is formal.

Nonsense. It is not at all formal.

Just for starters, € and its negation are operators. They do not have values (0 or 1, in or out, false or true, whatever). They have meaning when expressing relationships between operands.

Try as you may, € cannot be offered up by itself as something meaningful as you have tried to do. It has to be in the form (A € B) -- it is a binary operator after all.

 

[ddt]

No ddt,

€ ~€
0 0 --> [ ] (in,out are the same) (non-locality)
0 1 --> [ ]. (in,out are not the same (locality)
1 0 --> [.] (in,out are not the same (locality)
1 1 --> [_]_ (in,out are the same) (non-locality)

is formal.

You simply do not get it because you force your reasoning on it.

Can't you read?

I specifically asked:

Write out in full the formulae you're meaning - in the above table, is the 0 resp. 1 the x or the A in "x € A"?

Your € is a binary relation. You just said that the 0's and 1's in this table are the elements of your domain of discourse:

0 or 1 do not refer to membership (in or out) but to the element ( 0 or 1).

But if 0 is an element, does that mean in the above table that it is the x of "x € A" or the A of "x € A"?

I'm not even touching now on the idiocy of

If the element it both 0 or both 1

which can only lead to the conclusion that 0 and 1 are the same element and your domain of discourse collapses to a one-element domain.

 

[Don At Work]

[Gumby] MY BRAIN HURTS[/Gumby]