Thoughts about Bell, Bohm, Christian, et al.

[Heinera]

Again there is no "nice", or "bad" in epr-simple, that you keep using those words reveals severe misunderstandings. A 0 value means the particle was not detected. What exactly is "bad" about that? What physical principle allows you to claim that all particles must be detected? Does QM say all particles must be detected?

The proof in Bell's original paper doesn't apply to models where some particles are not detected. For that you need other proofs and inequalities, e.g. the CH-inequality. I have tested your epr-simple against that inequality, and it solidly fails to violate the inequality.

[minkwe]

Again there is no "nice", or "bad" in epr-simple, that you keep using those words reveals severe misunderstandings. A 0 value means the particle was not detected. What exactly is "bad" about that? What physical principle allows you to claim that all particles must be detected? Does QM say all particles must be detected?

The proof in Bell's original paper doesn't apply to models where some particles are not detected. For that you need other proofs and inequalities, e.g. the CH-inequality. I have tested your epr-simple against that inequality, and it solidly fails to violate the inequality.

epr-simple is a CHSH-test simulation. To suggest you "tested" against CH-inequality tells me you do not know what you are doing. Did you rewrite the simulation for the CH-inequality. Please do provide a link to the code, if your claims are true.

Maybe you can help Schmelzer, to explain where the loophole is in the calculation of


from epr-simple. Where exactly do you guys claim particles are being eliminated using settings ("a", and "b") in the calculation of that correlation? Which particle should have been included in that calculation, but has now been eliminated because it produced a zero outcome?

The proof in Bell's original paper doesn't apply to models where some particles are not detected. For that you need other proofs and inequalities, e.g. the CH-inequality.

Does the proof in Bells original paper apply to models where none of the expectation values P(a,b), P(a,c), P(b,c) are counterfactual?

If you think it does, then please derive Bell's inequality from

[Heinera]

epr-simple is a CHSH-test simulation. To suggest you "tested" against CH-inequality tells me you do not know what you are doing.

There is no such thing as a "CHSH-test simulation". There is a simulation; the tests are performed afterwards based on the outcomes of the simulation. One now has a choice of performing one or more of several different tests (and for your model, with particles not detected, the CHSH-test doesn't even apply).. To suggest otherwise shows you are clueless about everything concerning Bell's theorem.

Please do provide a link to the code, if your claims are true.

No, since I believe in learning by doing, I will not do that. It is a trivial task for you to modify your code yourself, in order to check against CH.

[Heinera]

If you think it does, then please derive Bell's inequality from

You still don't get it. One can't derive Bell's theorem from experimental data. The only thing you have shown is that experimental data can violate Bell's inequality. Thank you (but we already knew that).

[minkwe]

There is no such thing as a "CHSH-test simulation".
...
No, since I believe in learning by doing, I will not do that. It is a trivial task for you to modify your code yourself, in order to check against CH.

There is, I wrote the code. I know what I did. In the CHSH-test there are 2 detectors per station, one labelled +1 and another labelled -1. In a CH-test, there is a single detector per station. Again, you do not know what you are talking about. You don't provide the code because you did not test anything against the CH inequality as you claimed, pants-on-fire!

[minkwe]

If you think it does, then please derive Bell's inequality from

You still don't get it. One can't derive Bell's theorem from experimental data. The only thing you have shown is that experimental data can violate Bell's inequality. Thank you (but we already knew that).

No you don't get it. I'm not asking you to derive Bell's inequality from experimental data. I'm asking you to derive it without using counterfactual terms since you continue to believe that the inequality should apply to experiments where no counterfactual terms are present.

BTW, you ignored my question about the loophole. In epr-simple, P(a,b), P(a,c) and P(b,c) are calculated as prescribed for experimental data:


See equation (1) of https://en.wikipedia.org/wiki/Bell_test_experiments

Now where is the loophole? Please show exactly where in that calculation you and schmelzer claim results are being rejected based on "a" and "b". Or have all the shouting "loophole! loophole!" by you guys simply been empty noises. Please explain which results would have been included in that calculation but were left out by epr-simple because of 0 outcomes.

[Heinera]

There is no such thing as a "CHSH-test simulation".
...
No, since I believe in learning by doing, I will not do that. It is a trivial task for you to modify your code yourself, in order to check against CH.

There is, I wrote the code. I know what I did. In the CHSH-test there are 2 detectors per station, one labelled +1 and another labelled -1. In a CH-test, there is a single detector per station. Again, you do not know what you are talking about. You don't provide the code because you did not test anything against the CH inequality as you claimed, pants-on-fire!

You should read up on both CHSH and CH. The two tests both assume the same experimental setup. Clueless.

[Heinera]

You still don't get it. One can't derive Bell's theorem from experimental data. The only thing you have shown is that experimental data can violate Bell's inequality. Thank you (but we already knew that).

No you don't get it. I'm not asking you to derive Bell's inequality from experimental data. I'm asking you to derive it without using counterfactual terms since you continue to believe that the inequality should apply to experiments where no counterfactual terms are present.

The inequality does not apply to experiments, only to LHV models. That's why it is observed to be violated in experiments. Got it?

[minkwe]

There is no such thing as a "CHSH-test simulation".
...
No, since I believe in learning by doing, I will not do that. It is a trivial task for you to modify your code yourself, in order to check against CH.

There is, I wrote the code. I know what I did. In the CHSH-test there are 2 detectors per station, one labelled +1 and another labelled -1. In a CH-test, there is a single detector per station. Again, you do not know what you are talking about. You don't provide the code because you did not test anything against the CH inequality as you claimed, pants-on-fire!

You should read up on both CHSH and CH . The two tests both assume the same experimental setup. Clueless.

You are the clueless one. Anybody else can read up on the CHSH and CH (https://en.wikipedia.org/wiki/Bell_test_experiments) and discover the difference between "single-channel" and "double-channel" and if they were not utterly stupid, they might even notice the diagrams on Wikipedia illustrating the different experimental setups used for them. If you had the slightest clue, you would know that.

Now, where is the code? Where is the loophole in the calculation of



by epr-simple???

[minkwe]

The inequality does not apply to experiments, only to LHV models. That's why it is observed to be violated in experiments. Got it?

Tell that to Weihs, Zeillinger, Aspect, etc, etc who have been spending millions on Bell-test experiments. You are clueless indeed. You've probably forgotten 99 % of everything you learnt from the last several pages of this thread, and gone back to your empty claims that the inequality applies to LHV.

[minkwe]

Bell names it differently (and I think more adequate) "predetermined". Whatever,
...
I have misunderstood your understanding of CFD, but in this case I would not care about your understanding of CFD anymore, because the second part of Bell's proof needs only (1) and (2), and, in case this does not contain your CFD, your CFD is not necessary for the proof and your irrelevant hobby.

Sorry, this is incoherent. When Bell says on page 406:

"it follows that 'c' is another unit vector",
P(a,b) - P(a,c) = ...

What do you think he is doing? There are two particles and three angles "a", "b", "c". Where do you think the "c" result comes from, if it is not counterfactual? I've explained this clearly already. You claimed my understanding of CFD was not mainstream, yet you can't even bring yourself to state what my understanding of CFD is, and how it differs from the mainstream one. In fact, you haven't even presented a mainstream definition of CFD. You rambled about predetermination, which is a different concept from CFD. How do you expect anyone to have an insightful discussion with you, if you make claims out of hot air, about things you don't even understand.

Feel free to find loopholes in this derivation of predetermination, (1) and (2), and replace the "has been derived" by "falsely claims to have derived" if you like, then we can argue about this. But common sense tells me that the use of words like "follows" and "implies" describe something what at least the author considers as a derivation.

I haven't written anything about "predetermination" in this thread. Now you want to change the topic from CFD to predetermination, and you are challenging me to find loopholes in the "derivation" of predetermination? More confusion.
I've said it before. It is impossible to have an insightful discussion with you, I've tried.

If Alice measures first, and obtains +1, and Bob later, then, yes, he would obtain -1.

Alice and Bob have already measured. You know that Alice measured along "a". You do not know what axis Bob measured along. Alice obtained +1. Is it or is it not true that According to QM, if Bob's measurement axis was along "a" he MUST have obtained -1.

True.

Good! You answer "True", when you don't yet know along what axis Bob measured. Now I reveal that indeed Bob had measured along axis "b" not "a". In other words, the only thing that has changed is you now know the axis along which Bob measured and it is not "a". Would you change your answer from "True"? If not, then you have agreed that an experiment which was not performed (ie, Bob measuring along "a") has a single definite result -1.

http://arxiv.org/pdf/1007.4281.pdf: [CFD] is the assumption that a measurement that was not performed had a single definite result.

Get it?

why don't you explain why the following definitions are not mainstream enough for you

Why should I? They are.
...
No necessity. I see a difference in the conclusions - the mainstream says that there is no CFD in QM, you seem to think there is.

I just proved to you that any such claim is false, and you admitted it yourself. Please scroll up 5 lines and read my question and your answer. And if you continue to see difference in conclusions, explain why the definition is wrong, or why it should not apply to argument you just agreed to. If you can't substantiate claims, don't make them.

It's not my job to find out where exactly you err, that's your problem.

It's your job to substantiate your accusation that I err. Failure to substantiate claims you make points to dishonesty and deceit, especially when you don't even understood the opposing argument.

Maybe you misunderstand the definition, whatever else, I don't really care.

Seriously?! Now it is "maybe I misunderstand". You mean you don't even know if I misunderstand it or not, yet you were quick to claim that I'm in error?! And you don't care? You obviously care enough to make false claims you can't substantiate about an argument you don't understand, simply because it doesn't fit your made-up mind.

I see the difference, have mentioned it to avoid misunderstandings (say, I say something about mainstream CFD, you interpret it as about your CFD), that's all.

You still haven't specified exactly what this difference is between "my CFD" and mainstream CFD. I've asked you many times. You don't even know that there is a difference yet you continue to make claims about differences.

Again there is no "nice", or "bad" in epr-simple, that you keep using those words reveals severe misunderstandings. A 0 value means the particle was not detected. What exactly is "bad" about that? What physical principle allows you to claim that all particles must be detected? Does QM say all particles must be detected?

Bell's proof requires that all particles must be detected, without this assumption the theorem cannot be proved. That's the detector efficiency loophole.

So anything which Bell did not think about is "invalid" or "bad"? All of a sudden, Bell has become your god. If Bell forgets to consider that particles may not all be detected, then anytime a particle is not detected, the experiment is "invalid". If Bell forgets to consider hidden instrument parameters in his derivation, then instruments are forbidden from having hidden parameters. If Bell forgets to consider time-delays in particle detections, then when time delays are present, the experiment is "invalid". Yet when it is pointed out to you that Bell required counterfactual results in the derivation. You insist in thinking experiments which do not contain any counterfactual results are "nice" and "good" and "valid". This is a religion not a science, and you are a bishop in the church of Bell.

Please think before you write, what rejection are you talking about in experiments expectation value is calculated as

Of course, about the outcomes with a 0 value, which means "A 0 value means the particle was not detected".

The expectation values are calculated in exactly the same way in all CHSH-test experiments, and in epr-simple, using the above equation. You claim that some experiments are rejected because they have zero outcome. The outcomes containing zero terms are N{+0}, N{0+}, N{-0}, N{0-}, which are not present in the above expression. What difference does it make to the expectation value if I reject the results with zero outcomes, if none of them are relevant for the calculation above???

The sum is over those a, b, which have values , .
If you sum over those which have, additionally, also , resp. for B, you get a sum over a different subset of experiments, and even if the experimental outcomes themself are the same for all choicse in above sums, the sums will be different.

This makes no sense. None of the expectation values contain N{+0}, N{0+}, N{-0}, N{0-} terms, so it boggles my mind why you would think the zero outcomes some how magically change the calculated expectation values. Expectation values are always calculated from a list of pairs of outcomes. The point of the demonstration was to show what happens when you do not use counterfactual outcomes, and what happens when you use counterfactual outcomes. But as I suspected, detractors will miss the point and go down a rabbit hole.

[Heinera]

Anybody else can read up on the CHSH and CH (https://en.wikipedia.org/wiki/Bell_test_experiments) and discover the difference between "single-channel" and "double-channel" and if they were not utterly stupid, they might even notice the diagrams on Wikipedia illustrating the different experimental setups used for them.

If they also read the Wikipedia article on the CHSH inequality, they will find the statement "In their 1974 paper,[7] Clauser and Horne show that the CHSH inequality can be derived from the CH74 one. As they tell us, in a two-channel experiment the CH74 single-channel test is still applicable...".

So, the CH-inequality for your simulation is



for stations , settings , and where are the coincidence counts, and the singles counts, is the total number of trials where the detector setting was , and is the number of trials where the detector setting was (regardless of the setting on the other side). (I have used the same notation as Christensen et. al.)

Now, care to compute it?

[Joy Christian]

So, the CH-inequality for your simulation is



for stations , settings , and where are the coincidence counts, and the singles counts, is the total number of trials where the detector setting was , and is the number of trials where the detector setting was x (regardless of the setting on the other side). (I have used the same notation as Christensen et. al.)

Now, care to compute it?

Not so fast. There is a conceptual mistake in what is written above, and without correcting it one cannot "care to compute" CH. I am not telling what the mistake is.

[Heinera]

Not so fast. There is a conceptual mistake in what is written above, and without correcting it one cannot "care to compute" CH. I am not telling what the mistake is.

You really know how to make a useful contribution. And there is no "conceptual mistake".

[Joy Christian]

...there is no "conceptual mistake".

You are better off correcting your mistake before minkwe pounces on you again.

[Schmelzer]

Bell names it differently (and I think more adequate) "predetermined". Whatever,
...
I have misunderstood your understanding of CFD, but in this case I would not care about your understanding of CFD anymore, because the second part of Bell's proof needs only (1) and (2), and, in case this does not contain your CFD, your CFD is not necessary for the proof and your irrelevant hobby.

Sorry, this is incoherent. When Bell says on page 406:

"it follows that 'c' is another unit vector",
P(a,b) - P(a,c) = ...

What do you think he is doing? There are two particles and three angles "a", "b", "c". Where do you think the "c" result comes from, if it is not counterfactual?

It is, and I do not claim that Bell is not using CFD. My claim was that he uses a more adequate name "predetermined". And, if your notion of CFD (which seems to differ from the mainstream way, which does not claim that CFD exists in QM, but admits a definiteness only after a preparational measurement.) differs from the notion used in the word "predetermined" and the formulas (1), (2), I do not have to care about this strange notion.

You claimed my understanding of CFD was not mainstream, yet you can't even bring yourself to state what my understanding of CFD is, and how it differs from the mainstream one. In fact, you haven't even presented a mainstream definition of CFD. You rambled about predetermination, which is a different concept from CFD.

So, we see that your concept of CFD differs from what Bell has used ("predetermined"). I see differences too, these differences being in favour of using "predetermined", but essentially, if one uses the mainstream understanding of CFD, these are minor, irrelevant differences.

Now you want to change the topic from CFD to predetermination, and you are challenging me to find loopholes in the "derivation" of predetermination? More confusion.

I simply want to stay close to the text, and this text uses predetermination and derives it using the EPR argument. So, no, I don't want to change the topic, you have changed it by introducing a notion not mentioned in Bell's paper, and introducing it in a strange, confusing version, with strange implications like suggestions that we have CFD in QM. It is your point that there is somthing wrong with Bell's paper, not my. So you have to challenge Bell's paper, which mentions predetermination, but not your personal version of CFD which makes QM a theory with CFD.

Good! You answer "True", when you don't yet know along what axis Bob measured. Now I reveal that indeed Bob had measured along axis "b" not "a". In other words, the only thing that has changed is you now know the axis along which Bob measured and it is not "a". Would you change your answer from "True"? If not, then you have agreed that an experiment which was not performed (ie, Bob measuring along "a") has a single definite result -1.

http://arxiv.org/pdf/1007.4281.pdf: [CFD] is the assumption that a measurement that was not performed had a single definite result.

Get it?

Yes. This is the exceptional case, where the measurement of Alice creates the state of Bob as an eigenstate. There is a very small class states and corresponding measurements where QM predefines the result of a measurement. The most important example is simply a repetition of the same measurement. There are other cases, the simplest one being some nontrivial unitary evolution, so that after this the state becomes a different one, and the measurement which measures it too. So, in these exceptional cases, QM tells us that there is a unique outcome of this experiment.

So, we have CFD in QM in some exceptional, artificially constructed cases. In general, we have no CFD in QM.

You still haven't specified exactly what this difference is between "my CFD" and mainstream CFD. I've asked you many times. You don't even know that there is a difference yet you continue to make claims about differences.

I see continuing suggestions that we have CFD in QM, which are nonsense, and supported only by some trivial exceptional cases, which have nothing to do with the general case. In the particular case of the Bell experiment, out of the full S^2 of possible directions we have 1 direction (ok, 2, the opposite too) where QM gives us CFD. In above cases, there is not much conterfactual, given that there have been definite experiments with definite outcomes.

If Bell forgets to consider that particles may not all be detected, then anytime a particle is not detected, the experiment is "invalid".

As if it would matter how you name it. Name it "invisible pink unicorn" if you like.

Moreover, it does not matter at all if Bell has forgotten about this possibility or not. The detection loophole is a loophole of particular experimental tests of BI, and the aim of the paper was not to consider particular tests.

If Bell forgets to consider hidden instrument parameters in his derivation, then instruments are forbidden from having hidden parameters.

They are not forbidden, the space of possible values of is arbitrary, thus, can contain instrument hidden parameters too. What matters is that there is predetermination, derived from the EPR criterion, which does not allow them to depend on a and b.

If Bell forgets to consider time-delays in particle detections, then when time delays are present, the experiment is "invalid".

Bell's theorem is not about experimental realizations, thus, he has not "forgotten" such things, they are simply not part of what is considered in the paper.

Yet when it is pointed out to you that Bell required counterfactual results in the derivation. You insist in thinking experiments which do not contain any counterfactual results are "nice" and "good" and "valid". This is a religion not a science, and you are a bishop in the church of Bell.

Name them as you like, I couldn't care less. Of course, once Bell has derived that the experimental outcomes have to be predefined, he can use CFD after this in the derivation.

Of course, about the outcomes with a 0 value, which means "A 0 value means the particle was not detected".

The expectation values are calculated in exactly the same way in all CHSH-test experiments, and in epr-simple, using the above equation. You claim that some experiments are rejected because they have zero outcome. The outcomes containing zero terms are N{+0}, N{0+}, N{-0}, N{0-}, which are not present in the above expression. What difference does it make to the expectation value if I reject the results with zero outcomes, if none of them are relevant for the calculation above???

It is the difference that if there are such detection failures, we have the detector efficiency loophole open, thus, the experiment does not allow to rule out local realism. And the simulation does not present a counterexample to Bell's theorem. In above cases, the experiment as well as the simulation do not reach their aim.

The sum is over those a, b, which have values , .
If you sum over those which have, additionally, also , resp. for B, you get a sum over a different subset of experiments, and even if the experimental outcomes themself are the same for all choicse in above sums, the sums will be different.

This makes no sense. None of the expectation values contain N{+0}, N{0+}, N{-0}, N{0-} terms, so it boggles my mind why you would think the zero outcomes some how magically change the calculated expectation values.

Why should I care what boggles your mind? The sum is over a different set of outcomes, one symmetric to permutations between a,b,c,d, the other one asymmetric. We need symmtry in the proof of the theorem, and it is easy to construct counterexamples with asymmetric sets.

Expectation values are always calculated from a list of pairs of outcomes. The point of the demonstration was to show what happens when you do not use counterfactual outcomes, and what happens when you use counterfactual outcomes. But as I suspected, detractors will miss the point and go down a rabbit hole.

It happened exactly what I would have predicted based on Bell's theorem. So, indeed, I miss the point why you think that this somehow supports your beliefs.

[minkwe]

You should read up on both CHSH and CH. The two tests both assume the same experimental setup.

So you finally saw the diagrams of the two different experimental setups. Or are you still clueless that they two use the same experimental setup.

So, the CH-inequality for your simulation is



Now, care to compute it?

Please derive this inequality, step by step let us see how clueless you are. And don't forget to post the code in which you claim to have tested with epr-simple? I suspect it is simply pants-on-fire, and you did not test anything. But we shall see. First, show the derivation of the above inequality, step by step.

[Heinera]

Please derive this inequality, step by step let us see how clueless you are. And don't forget to post the code in which you claim to have tested with epr-simple? I suspect it is simply pants-on-fire, and you did not test anything. But we shall see. First, show the derivation of the above inequality, step by step.

Since others have derived it before me, I will just point you to the original CH74 paper, and the first pages of Christensen et.al. There are more authoritative sources than Wikipedia, you see.

[Heinera]

You should read up on both CHSH and CH. The two tests both assume the same experimental setup.

So you finally saw the diagrams of the two different experimental setups. Or are you still clueless that they two use the same experimental setup.

That should be "The two tests can both assume the same experimental setup." The point is unchanged, however, that it is no problem to test your simulation against the CH-inequality.

[minkwe]

It is, and I do not claim that Bell is not using CFD. My claim was that he uses a more adequate name "predetermined". And, if your notion of CFD (which seems to differ from the mainstream way, which does not claim that CFD exists in QM, but admits a definiteness only after a preparational measurement.) differs from the notion used in the word "predetermined" and the formulas (1), (2), I do not have to care about this strange notion.

Look, you define CFD, before you can see what the definition applies to. It is nonsensical to define CFD in a way that applies to QM arguments and then turn around and claim because your so-called "mainstream" does not belief CFD applies to QM, it must be so. Even though I've shown you and you agreed that it applies.

Similarly, it is nonsensical to make statements like:

Bell uses the CFD in his proof. But the result is about averages, and these averages can be measured without measuring P(a,b), P(a,c) and P(b,c) on the same particles. So, it is the proof which uses CFD, but the BI[Bell inequalities] is not about them.

Think: you agree that Bell used CFD to derive Bell inequalities and without CFD he could not have derived the Bell's inequalities. Yet you shoot yourself in the foot by at the same time claiming that the Bell inequalities are not about the counterfactual averages which were used to derive them, but some different ones which you are unable to derive the inequalities from.

So, we see that your concept of CFD differs from what Bell has used ("predetermined"). I see differences too, these differences being in favour of using "predetermined", but essentially, if one uses the mainstream understanding of CFD, these are minor, irrelevant differences.

You think the difference between my concept of CFD and Bell's concept of CFD is that I use CFD while bell uses "predetermined" but the differences are minor and irrelevant anyway, blah blah blah, mainstream, blah, .. blah .. . Again, you are not making any sense at all.

So when you write that CFD is actually "predetermined", and that there is no CFD in QM you are probably also claiming that there is nothing "predetermined" in QM. Find a good dictionary and look up the meaning of "prediction", and start planning how you will convince yourself that a theory can predict anything, if "predetermination" is absent in it.

Oh by the way here you are admitting the opposite:

There is a very small class states and corresponding measurements where QM predefines the result of a measurement.

So what would you call somebody who knowing, this continues to claim there is no CFD (which they claim is the same as "predetermined") in QM. I would say they are either liars, or clueless.

I simply want to stay close to the text, and this text uses predetermination and derives it using the EPR argument. So, no, I don't want to change the topic, you have changed it by introducing a notion not mentioned in Bell's paper, and introducing it in a strange, confusing version, with strange implications like suggestions that we have CFD in QM. It is your point that there is somthing wrong with Bell's paper, not my. So you have to challenge Bell's paper, which mentions predetermination, but not your personal version of CFD which makes QM a theory with CFD.

I cited mainstream definitions of CFD, despite claiming they weren't mainstream you weren't able to provide a mainstream definition. After repeated challenge from me you finally admitted that the definitions I provided were mainstream. Then I provided an argument which Bell himself made "according to QM". Despite repeated claims from you that there is no CFD in QM, you finally admitted that indeed the argument which Bell made uses CFD. So sorry, I haven't changed any subject or introduced any strange concept. I simply used the mainstream views, combined with your own brain, and Bell's claims to show you that you are inconsistent. If that sounds confusing to you, it is because you are trying to suppress a clear logical argument. It can be bewildering if you find suddenly that the idea you've invested all of your life on, is bogus. Facts and logic can do that to a person. And when those facts come, the are always "strange" and "confusing". Those who are wise learn, and cut their losses. The unwise, dig-in, and deceive themselves that "I must be right because I've spent so much time and money on this idea".

Good! You answer "True", when you don't yet know along what axis Bob measured. Now I reveal that indeed Bob had measured along axis "b" not "a". In other words, the only thing that has changed is you now know the axis along which Bob measured and it is not "a". Would you change your answer from "True"? If not, then you have agreed that an experiment which was not performed (ie, Bob measuring along "a") has a single definite result -1.

http://arxiv.org/pdf/1007.4281.pdf: [CFD] is the assumption that a measurement that was not performed had a single definite result.

Get it?

Yes. This is the exceptional case

I see some progress, from claiming, "there is no CFD in QM" to, "well maybe there is in this one exception". Did you forget your previous claims:

Why is that not counterfactual definiteness in Quantum Mechanics?

Because there is no counterfactual definiteness in QM. Counterfactual definiteness requires that the outcome of unperformed experiments exists, is well-defined even before the measurement.

2. The counterfactual definiteness assumption invoked by Bell applies to QM as well. Bell himself uses it on page 1 to make a QM argument.

No. In the minimal interpretation there is no CFD.

I suppose you now admit that there is CFD in QM. You see, this "strange" and "confusing" thing called logic says I can disproof your claim that "there is no CFD in QM", by showing just a single example of CFD in QM.

Yes. This is the exceptional case ...
So, we have CFD in QM in some exceptional, artificially constructed cases.

I rest my case QED. I take it you believe Bell's argument argument on page 1 of the paper that is an exceptional, artificially constructed case which is not relevant to his paper, or the inequalities he derived?

where the measurement of Alice creates the state of Bob as an eigenstate.

Alice could have measured after Bob, yet according to you her measurement created Bob's eigenstate, even though Bob's particle had since ceased to exist. Utter garbage.

You still haven't specified exactly what this difference is between "my CFD" and mainstream CFD. I've asked you many times. You don't even know that there is a difference yet you continue to make claims about differences.

I see continuing suggestions that we have CFD in QM, which are nonsense

Even though you have admitted that we indeed have CFD in QM, you still claim it is nonsense to suggest that we have CFD in QM. Sounds like the left brain is fighting with the right brain.

and supported only by some trivial exceptional cases, which have nothing to do with the general case.

trivial cases like the one in Bell's paper , which is used to claim that CFD should be rejected .