A collection of infinitely many distinct elements

[doronshadmi]

It's not clear to me what could come before logic. Logic is the meta-level of mathematics;

In other words you do not get Symmetry\Asymmetry as the basis of logical connectives.

Also you are unable to say how X and its negation exist in a one framework without immediately contradict (eliminating) each other (in that case they actually do not exist, and no logical value can be found between them).

You did not provide any answer which is not "there simply are", which is nothing but and arbitrary answer with no reasoning behind it.

 

[jsfisher]

I am sure that you don't get non-locality.

Show me that I am wrong about that.

No more dodges and weaves, please. Whether I "get" non-locality to your satisfaction is not relevant. The topic is foundational logic¹, specifically your foundational logic.

You have explicitly rejected all standard forms of Mathematical Logic. Now is the time to stop telling us what you don't believe in, and start telling us what you do believe in.

In your logic framework: How do the operations work? How can well-formed formulae be constructed? How are inferences made?

_________________________
¹Foundational logic is now a candidate for phrase scavenging.

 

[doronshadmi]

Your definition is abundantly clear: it cannot occur. We already established that.


That means you're a louse stick thrower.



You present this as if there is some paradox, but there really isn't.

The relational approach
Define
O(t) := "part of the stick is out of the circle at time t"
and
I(t) := "part of the stick is in the circle at time t"

Then we can express "all of the stick is in the circle at time t" as
I(t) AND ~ O(t)
This even accounts for a stick that moves with time.

The stochastic approach
Define the stochasts
X_t := "the fraction of the stick that is in the circle at time t"
and kick the stick at every clock tick. Now you have a Markov chain^WP of stochasts, provided you don't memorize how you kicked last time.

The calculus approach
Define the indicator function
i(x,y) := "at point (x, y) lies a piece of stick"
and integrate this function over the area of the circle (the function is Riemann integrable).

Do you want more approaches?

1) There is no paradox here, but simply non-local atom.

2) No case that you used deals with non-local atom, which is both belongs and does not belong to some domain:

[latex]$$$ \exists$$$[/latex] __ : (_)_

There are no parts, fractions, points or time here, but only non-local atom.

Without the non-local atom, logical connectives like NXOR, NOR, AND do not hold.

 

[doronshadmi]

You have explicitly rejected all standard forms of Mathematical Logic.

Wrong.

Please be aware of how your reasoning is closed under excluded-middle way of thinking (You say: "You have explicitly rejected all ...").

Actually I said is that the law of contradiction is not universal (it does not hold without any exception).

I gave the example of non-locality :


There exist x such that it is both belongs and does not belong to some domain.

Now please show that you are able to get non-locality, notated as:

[latex]$$$ \exists$$$[/latex] __ : (_)_

 

[ArmillarySphere]

Except that drawn circles are neither domains nor sets, nor do they consists of sticks or line segments. Your analogy is flawed from the start.

 

[ddt]

It's not clear to me what could come before logic. Logic is the meta-level of mathematics;

In other words you do not get Symmetry\Asymmetry as the basis of logical connectives.

Thus far, your "symmetry\asymmetry" is only hot air. In the truth tables of the connectives you've shown nothing unusual.

Also you are unable to say how X and its negation exist in a one framework without immediately contradict (eliminating) each other (in that case they actually do not exist, and no logical value can be found between them).

What is X? A proposition? Both X and ~X exist; they can't be both true at the same time though. They don't eliminate each other - it's not an electron and a positron.

Apropos "no logical value can be found between them": I don't go searching for logical values like one searches for mushrooms in the forest.

You did not provide any answer which is not "there simply are", which is nothing but and arbitrary answer with no reasoning behind it.

Huh? I see words but barely correct grammar and no meaning.

 

[ddt]

1) There is no paradox here, but simply non-local atom.

Define "non-local atom".

2) No case that you used deals with non-local atom, which is both belongs and does not belong to some domain:

Sentence does not parse.

 

[jsfisher]

Wrong.

Please be aware of how your reasoning is closed under excluded-middle way of thinking (You say: "You have explicitly rejected all ...").

Actually I said is that the law of contradiction is not universal (it does not hold without any exception).

And since all standard frameworks for Mathematical Logic include the Law of Non-contradiction as a basic, universal fact, you have explicitly rejected all standard frameworks for Mathematical Logic.

You don't get to pick and choose which axioms and which laws you like and which you don't like. It doesn't work that way. You can built your own logic framework that's similar to one that already exists, but that's it.

Without the Law of Non-contradiction, the standard models cease to work. Since you claim to have a logic that works without the Law of Non-contradiction being a universal fact, tell us what it is.

How do the logic operators behave in your logic framework? How are well-formed formulae constructed? How are inferences made?

 

[ddt]

You have explicitly rejected all standard forms of Mathematical Logic.

Wrong.

Please be aware of how your reasoning is closed under excluded-middle way of thinking ("You have explicitly rejected all ...").

You confuse the "law of excluded middle" with the "law of contradiction". The law of excluded middle says that A OR ~A for every formula A. The law of contradiction says that A AND ~A is false for every formula A. There is a subtle difference which you would have known had you actually studied some logic (intuitionistic logic, in this case).

All I said is that the law of contradiction is not universal (it does not hold without any exception).

You may say all you like. You haven't convinced anyone though.

I gave the example of non-locality :

The silence following the colon? You haven't given an example thus far. You haven't even given a definition.

There exist x such that it is both belongs and does not belong to some domain.

Where?

Now please show that you are able to get non-locality, notated as:

[latex]$$$ \exists$$$[/latex] __ : (_)_

Nonsense notation.

You haven't reacted to the last part of jsfisher's post:

Now is the time to stop telling us what you don't believe in, and start telling us what you do believe in.

In your logic framework: How do the operations work? How can well-formed formulae be constructed? How are inferences made?

I've asked the same several times. You failed thus far to give an answer.

 

[ddt]

1) There is no paradox here, but simply non-local atom.

2) No case that you used deals with non-local atom, which is both belongs and does not belong to some domain:

[latex]$$$ \exists$$$[/latex] __ : (_)_

There are no parts, fractions, points or time here, but only non-local atom.

Without the non-local atom, logical connectives like NXOR, NOR, AND do not hold.

Again writing most of your post after you posted it?

And logical connectives do not "hold". They form formulae. The formulae may hold or not.

 

[ddt]

Except that drawn circles are neither domains nor sets, nor do they consists of sticks or line segments.

The sticks is just a mental preparation for his soon-to-be hunter/gatherer existence.

 

[ddt]

doronshadmi, you should really read the Nominations for the TLA thread. After all, you did intend to nominate my "agricultural revolution" post, didn't you?

 

[doronshadmi]

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).

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.

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

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

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

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).

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

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.

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

 

[PixyMisa]

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.

What, exactly, are you going to replace AND with?

 

[Reality Check]

Translating your TOUM.pdf document a bit (yet again):

Define and as a logical AND connective
Define an urelement as not a set, but can be a member of a set.
Define an element as either a set or an urelement.
Define a sub-element as an element that defines another element.
Define € be there exists a member of an element, but not necessarily a sub-element of an element.
Define ~€ be the negation of € (there does not exist a member of an element

Let x be an element.
Let A be an element.

(x € A) denotes that the element x exists as a member in the element A.
(x ~€ A) denotes that the element x does not exist as a member in the element A.

if (x € A) is true then (x ~€ A) is false.
Therefore any evaluation of the expression (x € A) AND (x ~€ A) will return false

Your "definition" (which is not a definition but a test to see if x is non-local with respect to A) 6 is:
"Definition" 6: if (x € A) and (x ~€ A) then (x ~€ A) is non-local.
which is the same as (see the first definition above)
"Definition" 6: if (x € A) AND (x ~€ A) then (x ~€ A) is non-local.

Therefore no x is non-local.

Note that the above and your TOUM.pdf document do not mention an XOR/NXOR complementation or the other stuff you have keep on repeating.
I ignore this because that section in TOUM.pdf ignores it.

If XOR/NXOR complementation etc. changes the meaning of test 6 then you need to add it to TOUM.pdf. Otherwise people will keep looking at this section and immediately deduce that you are a crackpot.

 

[doronshadmi]

What, exactly, are you going to replace AND with?

NOR,NXOR (see (4) at the previous post).

 

[doronshadmi]

Translating your TOUM.pdf document a bit (yet again):

Note that the above and your TOUM.pdf document do not mention an XOR/NXOR complementation or the other stuff you have keep on repeating.
I ignore this because that section in TOUM.pdf ignores it.

If XOR/NXOR complementation etc. changes the meaning of test 6 then you need to add it to TOUM.pdf. Otherwise people will keep looking at this section and immediately deduce that you are a crackpot.

Please read very carefully http://www.internationalskeptics.com/forums/showpost.php?p=3783689&postcount=833 .

if (x € A) is true then (x ~€ A) is false.

Only if you force the law of excluded middle on x, but then you get exactly what you forced, which is, locality.

Indeed locality is not non-locality, so you cannot conclude anything about non-locality by forcing locality, and you cannot conclude anything about locality by forcing non-locality.

 

[ddt]

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).

Good to see you finally notice. It has been mentioned at least half a dozen times by various posters that the definition of non-local you give here depends on which A you look at too. Now please change the wording to "... x is non-local w.r.t. A" in order to reflect that.

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.

It's a definition, FSM almighty! That's a fixed statement of what you call "non-local". It really means that x is non-local if and only if that formula holds. You can't change the formula on-the-go (did I already mention my brother and the games he designed as a kid?). If you really only meant this to be an "if" implication, it should be a theorem (or lemma, or proposition, or corollary) and not a definition - but then you have to give another statement as the definition of non-local.

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

Nobody said so. Nobody said anything remotely like that. How does that come into your skull? It's the definition when a particular x is non-local. There is no quantifier at work here, that is very clear. You might have noticed that most here actually know math and know how to write it up.

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

Yes. Nobody denied that.

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

Nobody claimed that A is equal to x. We actually know Russell's paradox and why it should be avoided, and it's not the reason you give after "because", since that is nonsense.

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).

Huh?

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

No, there is no "exists" quantifier and everybody knows that. Assume that any poster can better read than you can write.

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.

You haven't shown that. You've given a definition - above - of non-locality with the AND connective. That definition is set in stone and cannot be changed. Unless you change your mind, throw away the definition and come up with a new one.

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

Yesterday you had another story. You've been to Damascus last night and had an epiphany? Anyway, so much the better as it makes life so much easier.

Now how about the rest of your logic? Do you decide after all to stick with classical logic?

 

[PixyMisa]

NOR,NXOR (see (4) at the previous post).

"NOR,NXOR" isn't what I'd call "exact".

What exactly are you going to replace AND with?

 

[doronshadmi]

No, there is no "exists" quantifier and everybody knows that. Assume that any poster can better read than you can write.

Really?

Please explain your support of jsfisher's conclusion abuot x, expressed by:

[latex]$$$ \exists x : ( x \, \wedge \, \neg x ) $$$[/latex]

or

[latex]$$$ \exists (x \epsilon A) : (x \epsilon A) \wedge \neg (x \epsilon A) $$$[/latex]

Yesterday you had another story.

Notions are developed during interaction with other notions.

But don't you worry, it does not hold in your case because you already know everything so you do not need any evolution.