Deeper than primes

[epix]

[latex]$p \lor p$[/latex]​

is not equivalent to

[latex]$\lnot(p \leftrightarrow p)$[/latex]​

no matter how you try to slice it. When p is true, the first statement is true, and the second statement is false. This rules out even one-way implication. To summarize your wrongness and inability to use logical implication, feel free to verify that the following is false for any true p:

[latex]$$(p \lor p) \rightarrow \lnot(p \leftrightarrow p)$$[/latex]​

Once again, Doron is disproven by basic logic.

The material implication is still contingent (neither true nor false) even if q=p.


Due to the material implication equivalency,

(~(p v p)) v (~(p ↔ p))

must be contingent as well, and that's the case. So, go figure . . .

Edit: Oops. It's not contingent for p=true.

 

[epix]

If you get out of your you find that there is no problem to get 0 "≤ local\non-local number ≤ ∞" it terms of "0 ≤ x ≤ ∞"

Doron, we must check the circuitry and actually compute something to arrive at a humble number. Before pondering the philosophical aspects of infinity, we should concern ourselves with the finite but sufficiently large to create the intermediate step.

The universe may glitter with far more stars than even Carl Sagan imagined when he rhapsodized about billions upon billions. A new study suggests there are a mind-blowing 300 sextillion of them, or three times as many as scientists previously calculated. That is a 3 followed by 23 zeros. Or 3 trillion times 100 billion.

http://news.yahoo.com/s/ap/20101202/ap_on_sc/us_sci_starry_night

What do you think? Is stars = 3*10²³ a number large enough?

That's lots of stars and that's a lot planetary systems. The question is how many planets are out there in the universe that harbor civilizations virtually identical to ours, like +/- 5000 earth years of the cultural development difference.

The formula that computes the number of such civilizations in the universe is actually the formula that tells us how many numbers divide into stars = 3*10²³ including number 1 and the large number itself. But I couldn't find the formula, so I don't know if my number agrees with yours. How many positive integers divide into stars, Doron? Those civilizations are non-local to our solar system and you deal with locality and non-locality on a daily bases, so you surely know how to go about it.

 

[doronshadmi]

Therefore(2): False statements do not exist.

No, the result of False statment does not exist, for example: all negative natural numbers.

 

[doronshadmi]

[latex]$p \lor p$[/latex]​

is not equivalent to

[latex]$\lnot(p \leftrightarrow p)$[/latex]​

no matter how you try to slice it. When p is true, the first statement is true, and the second statement is false. This rules out even one-way implication. To summarize your wrongness and inability to use logical implication, feel free to verify that the following is false for any true p:

[latex]$$(p \lor p) \rightarrow \lnot(p \leftrightarrow p)$$[/latex]​

Once again, Doron is disproven by basic logic.

You are still run after the case where p=p or q=q.

But my case is this:

If (("the term "are" is used in the case of p and q") AND ("there is full influence of some logical connective on the result")) then ("p ≠ q").

In order to understand the meaning of "full influance" please look at http://www.internationalskeptics.com/forums/showpost.php?p=6650358&postcount=13259.

If you get it, you see that your p=p or q=q limited case is not the considered case (which is not less than p ≠ q).

 

[doronshadmi]

In order to be clearer about OR connective, it is actually not limited to it.

It is actually related to any binary connective, as follows:

No binary connective is fully expressible if the propositions are p=p or q=q, because given

F F ---> expressible
F T ---> not expressible
T F ---> not expressible
T T ---> expressible

only F F or T T are expressible if p=p or q=q, where F T or T F are not expressible.

In other words, full expression of binary connectives is possible only if p≠q, such that:

p≠q
----
F F ---> expressible
F T ---> expressible
T F ---> expressible
T T ---> expressible

 

[doronshadmi]

Again any way you slice it Doron it is nonsensical gibberish.

Once again this is nonsensical gibberish entirely due to your propensity to be sloppy, inconsistent or simply lazy in your assertions. You assert the “inability of x to be OR("=" is T ,"<" is T)” apparently and incorrectly inferring that TRUE OR TRUE must be FALSE. Then you use an “XOR” truth table as evidence for your incorrect assertion of “OR”. The facts of the matter are that "=" and “<" are mutually exclusive (a concept you have yet to demonstrate and inkling of) as well as that no one on this thread (other than you) has ever claimed that they can be both TRUE for some “x” or some “x”, “y”. So go ahead slap around you own strawman (and what you assert below as part of your “notions”) as much as you like, but please let use know when you are done amusing yourself by beating yourself up.

Just more nonsensical gibberish.
Well you are right about the contrary part as this assertion is simply contrary to itself. Which only serves to demonstrate that you lack of consistency is in fact deliberate. Though not that it actually matters much, you do certainly revel in your lack of consistency and rigor as well as specifically base major portions of your purported notions upon such lacking. The problem is Doron that as always, intentional or not, that lacking remains simply and entirely yours.

When you get out of your predilection for inconstancy and lack of rigor you find that “0 "≤ local\non-local number ≤ ∞" it terms of "0 ≤ x ≤ ∞"” is just gibberish and the meaning you thought it entailed was simply your own apparently self-imposed “fog” of confusion.

You simply can't comprehend the simple notion of OR=XOR+AND, such that OR("=" is T ,"<" is T) --> T is the AND part, which is related to non-local numbers, and OR("=" is F ,"<" is T) --> T , OR("=" is T ,"<" is F) --> T is the XOR part, which is related to local numbers.

 

[laca]

Let's not wander too far, Doron. I understand you have limitless supply of nonsensical gibberish, but you have still not pointed out the error in the following:

[latex]$$$\begin{eqnarray}
\exists z \, \forall x \, (x \in z) \rightarrow (x \ne x) \equiv \\
\exists z \, \forall x \, (\lnot (x \in z)) \vee (x \ne x) \equiv \\
\exists z \, \forall x \, (\lnot (x \in z)) \vee \lnot (x = x) \equiv \\
\exists z \, \forall x \, (x \notin z) \vee \lnot (x = x) \equiv \\
\exists z \, \forall x \, (\lnot (x = x)) \vee (x \notin z) \equiv \\
\exists z \, \forall x \, (x = x) \rightarrow (x \notin z)
\end{eqnarray}
$$$[/latex]​

Since I seem to have struck some kind of nerve and got on ignore, I must rely on others to not let Doron off the hook on this one.

 

[zooterkin]

Let's not wander too far, Doron. I understand you have limitless supply of nonsensical gibberish, but you have still not pointed out the error in the following:

[latex]$$$\begin{eqnarray}
\exists z \, \forall x \, (x \in z) \rightarrow (x \ne x) \equiv \\
\exists z \, \forall x \, (\lnot (x \in z)) \vee (x \ne x) \equiv \\
\exists z \, \forall x \, (\lnot (x \in z)) \vee \lnot (x = x) \equiv \\
\exists z \, \forall x \, (x \notin z) \vee \lnot (x = x) \equiv \\
\exists z \, \forall x \, (\lnot (x = x)) \vee (x \notin z) \equiv \\
\exists z \, \forall x \, (x = x) \rightarrow (x \notin z)
\end{eqnarray}
$$$[/latex]​

Since I seem to have struck some kind of nerve and got on ignore, I must rely on others to not let Doron off the hook on this one.

QFD

 

[doronshadmi]

QFD

zooterkin, jsfisher propositions are not relevant to the considered case, which is:

P:

[latex]$$$
\exists z \, \forall (x \ne x) \, (x \ne x) \rightarrow ((x \ne x) \in z)$$$[/latex]​

In p case z exists as empty set only if the non-existing belongs to it, but then p case does not cover the case of the existing that does not belong to z.


q:

[latex]$$$
\exists z \, \forall (x = x) \, (x = x) \rightarrow ((x = x) \notin z)$$$[/latex]​

In q case z exists as empty set only if the existing does not belong to it, but then z is one of the exiting things that are used to define the existence of z, which is a circular reasoning.


As can be seen p≠q, so this case is closed.

 

[laca]

zooterkin, jsfisher propositions are not relevant to the considered case, which is:

No, Doron. You're trying to sneak out of it. You clearly addressed that case and even said which steps are wrong. Now, please show why.

P:

[latex]$$$
\exists z \, \forall (x \ne x) \, (x \ne x) \rightarrow ((x \ne x) \in z)$$$[/latex]​

Sorry, but that proposition doesn't have the right syntax. It is nonsensical.

q:

[latex]$$$
\exists z \, \forall (x = x) \, (x = x) \rightarrow ((x = x) \notin z)$$$[/latex]​

Sorry, but that's nonsensical as well. Want to try again?

As can be seen p≠q, so this case is closed.

You just wish. It's not closed until you address the issue either by formally showing where the proof went wrong, by admitting the proof is correct or by admitting you don't know what the hell you're talking about.

 

[epix]

Sorry, but that proposition doesn't have the right syntax. It is nonsensical.

I wonder what the expression (x=x)(x=x) turns out to be. It resembles the closest the implied multiplication sometimes used in algebra. So feel free to interpret it that way until Doron again climbs the stairs to confer with the demons in the attic.

 

[jsfisher]

I wonder what the expression (x=x)(x=x) turns out to be. It resembles the closest the implied multiplication sometimes used in algebra. So feel free to interpret it that way until Doron again climbs the stairs to confer with the demons in the attic.

It really doesn't matter what Doron thinks it means. He's again trying to discredit Mathematics by forcing his twisted concepts on everything. His expression is gibberish. It is not what we were discussing; it is not worth discussing on its own; it is gibberish.

Admittedly, following the spiral of some of Doron's more insane posts can be amusing, but here is no point humoring this excursion into the absurd just so he can rant about yet another definition, that of the equality relation, being wrong.

 

[HatRack]

You are still run after the case where p=p or q=q.

No matter what p denotes or what q denotes, it is ALWAYS the case that p=p and q=q. Your inability to understand this simple fact is entirely your failure and no one else's.

 

[The Man]

You simply can't comprehend the simple notion of OR=XOR+AND, such that OR("=" is T ,"<" is T) --> T is the AND part, which is related to non-local numbers, and OR("=" is F ,"<" is T) --> T , OR("=" is T ,"<" is F) --> T is the XOR part, which is related to local numbers.

Doron "You simply can't comprehend the simple" fact that "=" and "<" are mutually exclusive for any “x” or “x”, “y”. So it doesn’t matter whether your use “OR”, “XOR”, “AND”, or whatever you like to pretend “XOR+AND” means. Your assertion is once again irrelevant and apparently your argument is once again simply with yourself.

 

[The Man]

I find Doron’s latest tact rather curious. Most of his notions center around his consideration of a line being indivisible, (one of his “atoms”). So his "line" passing through some partition can not be bifurcated (into at least two rays) by that partition but contradictorily must 'belong to AND not belong to' both sides of that partition in its (Doron imposed "line") entirety. This is the 'foggy' property of his “non-local”, well, whatever. That its indivisibility (that he imposes) results in him not being able to specificaly assert that some part is on one side of the partition, while the other part is, well, on the other side or that the entire whatever is entirely in neither side of the partition. The curious part is that now Doron apparently just wants to partition “OR” into a modification of “OR” (“XOR”) “+” “AND” reversing his own (self imposed) pervious modification of “OR”. Seems a rather long way to go against one’s self. Especially when considering that it still doesn’t gain him the ability to determine what part of his “line” is on what side of the partition or that the entirety of his "line" is not contained by (does not belong to) either side of the partition, while some part of a line (including a ray) is so entirely contained.

 

[epix]

I can understand this

Rd|-x-xx-x-----|-x-xx-x-----|

or this

HH|-x-x-x-x-x-x|-x-x-x-x-x-x|

but

[latex]$$$
\exists z \, \forall (x \ne x) \, (x \ne x) \rightarrow ((x \ne x) \in z)$$$[/latex]​

that's the weirdest drum tab one can imagine to come across. What's that arrow mean? Flam on foot hi-hat?

 

[doronshadmi]

No matter what p denotes or what q denotes, it is ALWAYS the case that p=p and q=q. Your inability to understand this simple fact is entirely your failure and no one else's.

It is ALWAYS your limited p=p or q=q cases, so?

In other words, you can't comprehend http://www.internationalskeptics.com/forums/showpost.php?p=6655853&postcount=13285 .

 

[doronshadmi]

EDIT:

The curious part is that now Doron apparently just wants to partition “OR” into a modification of “OR” (“XOR”) “+” “AND” reversing his own (self imposed) pervious modification of “OR”. Seems a rather long way to go against one’s self.

This is the beauty of OR connective, it is used to demonstrate XOR+AND connectives under a one framework, where the XOR aspect of OR is used for locality, such that a given object is understood by (partial (in the case of a line)) OR (complete (in the case of line or point)) membership w.r.t a given domain.

Pay attention that p="partial" OR q="complete" are used here, such that p≠q and therefore there is no contradiction exactly because p≠q.

The AND aspect of OR is a novel notion if the OR T T inputs are related to an indivisible object that is simultaneously at AND not at a given domain.

If it is comprehended as an indivisible object > a given domain , then its size (the property of length>0) w.r.t a given domain (being simultaneously at AND not at a given domain) provides its consistent non-local existence.

Lines or line segments (if not understood in terms of collections) have the ability of being simultaneously at AND not at a given domain, where a point can't be simultaneously at AND not at a given domain.

As long as a line or a line segment is understood only in terms of collection, the simultaneity of the AND aspect of OR is understood as a contradiction, because OR is understood only of terms of p≠q (it is not understood in therms of the same existing thing that its size > a given domain) in order to be considered as a consistent result.

Especially when considering that it still doesn’t gain him the ability to determine what part of his “line” is on what side of the partition

Being simultaneously at AND not at a given domain is possible only if the considered object has size > 0 and it is not partitioned (it is a non-local atom).

If you get it in terms of partitioning, then its non-locality is not considered anymore and you get things only in terms of collections.

At this very moment there is an unresolved dis-communication between us.

 

[epix]

Doron, we must check the circuitry and actually compute something to arrive at a humble number...
...How many positive integers divide into stars, Doron?

All quiet on the Western front...
See what happens when you start shuffling stuff around at will and there is no end to it? You lose your basic math instinct and then you can't figure out a simple problem.

Seeing the carnage, I think you should start learning to walk again with the help of Pythagorean Theorem-related exercises. Try this one.

See? You still have some of it, Doron. Now proceed to Exercise No. 4...

 

[zooterkin]

It is ALWAYS your limited p=p or q=q cases, so?

In other words, you can't comprehend http://www.internationalskeptics.com/forums/showpost.php?p=6655853&postcount=13285 .

Says the man who can't even understand what X=X means...