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by **FrediFizzx** » Sun Apr 11, 2021 11:53 am

Ok, so back to the "proof" paper.

https://arxiv.org/abs/2012.00719

I promised Gill that I would shred it to pieces so here we go. The first mistake they make is right before eq. (1) which is the same mistake Bell and the person on stackexchange make. B = -A only half of the time; the other half of the time B = A. So, the "proof" is only good half of the time? However, it doesn't matter because their eq. (1) is equal to zero. The integrand has two unspecified functions as a product. All we know is that they are either +1 or -1 so the product is always +1 or -1. So, the integral divided by 2pi is always +1 or -1. Which means that they are summing a bunch of +1's and -1's which converges to zero as N goes to infinity. Here is the equation in question,

$\rho_N(\theta)~=~-\frac1N\!\sum\limits_{n=1}^N\,\,\frac1{2\pi}\!\!\!\int\limits_{u \in (-\pi,\pi]}A_n(u)A_n(u -\theta)\,{\mathrm d} u.$

Here is a simple Mathematica evaluation of it.

So, we will stop there for now to see if they want to present a new eq. (1) that might make some sense as far as producing a negative cosine curve.
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by **Joy Christian** » Sun Apr 11, 2021 2:05 pm

.
It is a ridiculous attempt by Gill to patch up the failed theorem of Bell by using pseudo mathematics.
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by **FrediFizzx** » Sun Apr 11, 2021 5:04 pm

Another path to zero would be to include when B = A,

$\rho_N(\theta)~=~-\frac1N\!\sum\limits_{n=1}^N\,\,\frac1{2\pi}\!\!\!\int\limits_{u \in (-\pi,\pi]}A_n(u)A_n(u -\theta)\,{\mathrm d} u$
$+\,\frac1N\!\sum\limits_{n=1}^N\,\,\frac1{2\pi}\!\!\!\int\limits_{u \in (-\pi,\pi]}A_n(u)A_n(u -\theta)\,{\mathrm d} u = 0.$
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by **FrediFizzx** » Mon Apr 12, 2021 6:21 am

Moving on here, their eq.(4) is also zero,

Which is absurd because A_n(u) is supposed to be +/-1's. I guess there is no point in continuing with the shredding. It is like they just threw this together and didn't calculate anything out.

So, this "proof" is shot down big time. Try again.
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by **FrediFizzx** » Wed Apr 14, 2021 9:37 am

There is a clue in this on how to fix eq. (4).

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by **FrediFizzx** » Thu Apr 15, 2021 1:12 pm

This is all similar to the nonsense Bell did in his first paper eq. (2).

$P({\bf a, b}) = \int\,d\lambda\,\rho(\lambda)\,A({\bf a}, \lambda)\,B({\bf b}, \lambda)$

Then Bell says,

You have two unspecified functions as a product and all we know is that they are +/-1. So, the product is going to be either +1 or -1. Taking $\rho(\lambda)=1$ we will have,

So, you are averaging a bunch of +lambdas with a bunch of -lambdas which of course is equal to zero. Then we have,

$P({\bf a, b}) = \int\,d\lambda\,\rho(\lambda)\,A({\bf a}, \lambda)\,B({\bf b}, \lambda) = 0$

There is one way around this dilemma and that is to do what Joy did. So, Bell was wrong because it is possible!
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by **FrediFizzx** » Fri Apr 16, 2021 3:49 pm

Anyways, it is pretty obvious that something is not quite right with Bell's formulation. The product calculation has to be done before the +/-1's are generated or else it is complete nonsense.
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by **FrediFizzx** » Sat Apr 17, 2021 7:09 am

Of course Joy's formulation is a correct way of doing local measurement functions for EPR.

.

by **FrediFizzx** » Sat Apr 17, 2021 3:20 pm

Actually this,

$P({\bf a, b}) = \int\,d\lambda\,\rho(\lambda)\,A({\bf a}, \lambda)\,B({\bf b}, \lambda) = 0$

will have a range of -1 to +1 when the product is +1 or -1 with the high probability centered on zero instead of just being zero. Same with the ones in Gill's version of Gull's "proof". Well..., they do converge to zero at infinity.
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by **gill1109** » Sat Apr 24, 2021 8:35 am

FrediFizzx wrote:Actually this,

$P({\bf a, b}) = \int\,d\lambda\,\rho(\lambda)\,A({\bf a}, \lambda)\,B({\bf b}, \lambda) = 0$

will have a range of -1 to +1 when the product is +1 or -1 with the high probability centered on zero instead of just being zero. Same with the ones in Gill's version of Gull's "proof". Well..., they do converge to zero at infinity.

As you should know Fred, it is easy to have P(a, b) take all values between -1 and +1 as a and b vary throughout their ranges. Bell gave the example of how to generate the triangle wave by a suitable choice of the set Lambda of all possible lambda, probability measure rho(lambda)d lambda on the set Lambda, and functions A and B with values in {-1, +1}.

by **FrediFizzx** » Sat Apr 24, 2021 10:49 am

gill1109 wrote:
FrediFizzx wrote:Actually this,

$P({\bf a, b}) = \int\,d\lambda\,\rho(\lambda)\,A({\bf a}, \lambda)\,B({\bf b}, \lambda) = 0$

will have a range of -1 to +1 when the product is +1 or -1 with the high probability centered on zero instead of just being zero. Same with the ones in Gill's version of Gull's "proof". Well..., they do converge to zero at infinity.

As you should know Fred, it is easy to have P(a, b) take all values between -1 and +1 as a and b vary throughout their ranges. Bell gave the example of how to generate the triangle wave by a suitable choice of the set Lambda of all possible lambda, probability measure rho(lambda)d lambda on the set Lambda, and functions A and B with values in {-1, +1}.

Ah, he's back with more nonsense. I said, "when the product is +1 or -1...". The a and b vectors are out of the RHS of the equation so it doesn't matter what they are.

$P({\bf a, b}) = \int\,d\lambda\,\rho(\lambda)\,\pm1 = 0$

That is a legitimate substitution which produces nonsense. Without specifying actual functions for A and B, you end up with nonsense. Same with Gull's nonsense. And we already know that Bell's example is nonsense.
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by **gill1109** » Tue May 04, 2021 10:53 pm

FrediFizzx wrote:
gill1109 wrote:
FrediFizzx wrote:Actually this,

$P({\bf a, b}) = \int\,d\lambda\,\rho(\lambda)\,A({\bf a}, \lambda)\,B({\bf b}, \lambda) = 0$

will have a range of -1 to +1 when the product is +1 or -1 with the high probability centered on zero instead of just being zero. Same with the ones in Gill's version of Gull's "proof". Well..., they do converge to zero at infinity.

As you should know Fred, it is easy to have P(a, b) take all values between -1 and +1 as a and b vary throughout their ranges. Bell gave the example of how to generate the triangle wave by a suitable choice of the set Lambda of all possible lambda, probability measure rho(lambda)d lambda on the set Lambda, and functions A and B with values in {-1, +1}.

Ah, he's back with more nonsense. I said, "when the product is +1 or -1...". The a and b vectors are out of the RHS of the equation so it doesn't matter what they are.

$P({\bf a, b}) = \int\,d\lambda\,\rho(\lambda)\,\pm1 = 0$

That is a legitimate substitution which produces nonsense. Without specifying actual functions for A and B, you end up with nonsense. Same with Gull's nonsense. And we already know that Bell's example is nonsense.
.

Sorry Fred, that was an illegitimate substitution which you just did.

Bell’s example is merely that: an example. Particular functions A and B, which happen to lead to the saw-tooth function (aka triangle wave) as correlation; not the negative cosine. What’s the problem with that? Try using Mathematica to check Bell’s maths. Or do an R Monte Carlo simulation. You’ll find that Bell’s maths is correct (the published papers not surprisingly do contain the odd typo, it’s always obvious how to fix them). Bell does not claim that this is a correct model from the point of view of physics when we are in the EPR-B situation. Obviously, it isn’t. That’s the point which Bell wanted to make.

by **FrediFizzx** » Wed May 05, 2021 2:05 am

gill1109 wrote:
FrediFizzx wrote:
gill1109 wrote:
FrediFizzx wrote:Actually this,

$P({\bf a, b}) = \int\,d\lambda\,\rho(\lambda)\,A({\bf a}, \lambda)\,B({\bf b}, \lambda) = 0$

will have a range of -1 to +1 when the product is +1 or -1 with the high probability centered on zero instead of just being zero. Same with the ones in Gill's version of Gull's "proof". Well..., they do converge to zero at infinity.

As you should know Fred, it is easy to have P(a, b) take all values between -1 and +1 as a and b vary throughout their ranges. Bell gave the example of how to generate the triangle wave by a suitable choice of the set Lambda of all possible lambda, probability measure rho(lambda)d lambda on the set Lambda, and functions A and B with values in {-1, +1}.

Ah, he's back with more nonsense. I said, "when the product is +1 or -1...". The a and b vectors are out of the RHS of the equation so it doesn't matter what they are.

$P({\bf a, b}) = \int\,d\lambda\,\rho(\lambda)\,\pm1 = 0$

That is a legitimate substitution which produces nonsense. Without specifying actual functions for A and B, you end up with nonsense. Same with Gull's nonsense. And we already know that Bell's example is nonsense.
.

Sorry Fred, that was an illegitimate substitution which you just did. ...

How could it be illegitimate? Don't make nonsensical claims like that without backing it up. Are A and B equal to +1 or -1 or not? Of course they are. So their product is always going to be +1 or -1. The formulation produces nonsense just like I said. You have to figure out a way to keep the vectors a and b in the calculation while producing the +/-1's.
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by **FrediFizzx** » Fri May 07, 2021 4:26 pm

Here is how you get around the nonsense with the +/-1 substitution. You have to specify actual measurement functions for A and B. Note that the limit replacement functions drop out of the product calculation.

PlotArray1 is a partial list of the A and B outcome pairs. Note that the limit replacement function on the A and B functions give the 4 different outcome pairs.

Bell loses again! :mrgreen:

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by **Justo** » Fri May 07, 2021 4:49 pm

I am just curious. Bell is supposed to have shown that $A(a,\lambda)$ and $B(b,\lambda)$ do not exist such that
$\int A(a,\lambda)B(b,\lambda)\rho(\lambda) d\lambda=-\cos(b-a)$
You say that is false. The question is how do you mathematically define the functions A and B that proves Bell is wrong?

by **FrediFizzx** » Fri May 07, 2021 4:51 pm

Justo wrote:I am just curious. Bell is supposed to have shown that $A(a,\lambda)$ and $B(b,\lambda)$ do not exist such that
$\int A(a,\lambda)B(b,\lambda)\rho(\lambda) d\lambda=-\cos(b-a)$
You say that is false. The question is how do you mathematically define the functions A and B that proves Bell is wrong?

I just posted it above your post or you can read Joy's papers. :roll:

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by **gill1109** » Sun May 09, 2021 6:49 am

FrediFizzx wrote:
gill1109 wrote:Sorry Fred, that was an illegitimate substitution which you just did. ...

How could it be illegitimate? Don't make nonsensical claims like that without backing it up. Are A and B equal to +1 or -1 or not? Of course they are. So their product is always going to be +1 or -1. The formulation produces nonsense just like I said. You have to figure out a way to keep the vectors a and b in the calculation while producing the +/-1's.
.

Dear Fred

A and B are not identically equal to -1 or identically equal to +1. For given a and b their product still depends on lambda. As lambda varies, their product can take both values. Not at the same time. But for some lambda, their product can be +1. For other lambda, their product can be -1. There are no other possibilities.

For each a, b and lambda, A(a, lambda) equals +1 or it equals -1, and B(b, lambda) equals +1 or it equals -1.

Joy Christian does not have a counterexample to Bell’s theorem … because Bell’s theorem is a simple, true, theorem of elementary pure mathematics. Which can be proved in a myriad of different ways.

If you use the CHSH inequality to prove the theorem you don’t even need any calculus. Since there are only two a’s in the game and only two b’s, everything depends on a single list of 16 probabilities adding up to +1. Because (in obvious notation) there only 16 possible values of the quadruple (A1, A2, B1, B2). As lambda varies it takes values in the set {-1, +1}^4. That’s a set of 16 elements. It’s easy to show that A1 B1 - A1 B2 - A2 B1 - A2 B2 always equals -2 or +2. Its mean value therefore lies between -2 and +2 (inclusive).

Bell wrote that Bohr would not have been impressed by Bell’s theorem. Bohr did not subscribe to local realism.

by **Joy Christian** » Sun May 09, 2021 8:12 am

gill1109 wrote:
Joy Christian does not have a counterexample to Bell’s theorem … because Bell’s theorem is a simple, true, theorem of elementary pure mathematics.

Both parts of the above statement have been proven wrong many times before. For the latest demonstrations of why they are wrong, please see my latest paper, published in IEEE Access.

PS: The above paper of mine has gone through two rounds of peer-review involving 14 reviewers. That means that Gill lives in his own fantasy world and is not believed by the community.
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by **FrediFizzx** » Sun May 09, 2021 8:31 am

gill1109 wrote:
FrediFizzx wrote:
gill1109 wrote:Sorry Fred, that was an illegitimate substitution which you just did. ...

How could it be illegitimate? Don't make nonsensical claims like that without backing it up. Are A and B equal to +1 or -1 or not? Of course they are. So their product is always going to be +1 or -1. The formulation produces nonsense just like I said. You have to figure out a way to keep the vectors a and b in the calculation while producing the +/-1's.
.
... A and B are not identically equal to -1 or identically equal to +1. ...

Egads! More freakin' nonsense! This proves that you and Bell are wrong.

PlotArray1 is a partial list of the A and B outcome pairs. Note that the limit replacement function on the A and B functions give the 4 different outcome pairs.

Bell loses again! :mrgreen:

by **FrediFizzx** » Sun May 09, 2021 2:40 pm

gill1109 wrote:
FrediFizzx wrote:
gill1109 wrote:Sorry Fred, that was an illegitimate substitution which you just did. ...

How could it be illegitimate? Don't make nonsensical claims like that without backing it up. Are A and B equal to +1 or -1 or not? Of course they are. So their product is always going to be +1 or -1. The formulation produces nonsense just like I said. You have to figure out a way to keep the vectors a and b in the calculation while producing the +/-1's.
.
...
If you use the CHSH inequality to prove the theorem you don’t even need any calculus. Since there are only two a’s in the game and only two b’s, everything depends on a single list of 16 probabilities adding up to +1. Because (in obvious notation) there only 16 possible values of the quadruple (A1, A2, B1, B2). As lambda varies it takes values in the set {-1, +1}^4. That’s a set of 16 elements. It’s easy to show that A1 B1 - A1 B2 - A2 B1 - A2 B2 always equals -2 or +2. Its mean value therefore lies between -2 and +2 (inclusive)...

It is really hard to believe that any mathematician would believe such nonsense. The A1 in the first pair is impossible to be the same as the A1 in the second pair, etc. So, you could have,

A1 B1= +1, A1 B2= -1, A2 B1= -1, A2 B2= -1 --> +1+1+1+1 = 4

Bell was wrong, get over it and move on!!!!!
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