Understanding bell's theorem: why hidden variables imply a linear relationship?

[lugita15]

Agreed.

Great! Here's my next question. If the relative frequency of something is the same for three different sets, then it's also the same for their union. So if we let s be the union of p, q, and r (the sets from my previous post), then the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for p and s. Do you agree with that?

 

[billschnieder]

Great! Here's my next question. If the relative frequency of something is the same for three different sets, then it's also the same for their union. So if we let s be the union of p, q, and r (the sets from my previous post), then the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for p and s. Do you agree with that?

Yes. if p, q, and r are identically prepared, then of course the union of (p + q + r) is also identically prepared to any of the component sets p, q and r.

 

[lugita15]

Yes. if p, q, and r are identically prepared, then of course the union of (p + q + r) is also identically prepared to any of the component sets p, q and r.

OK, so we're now agreed on the following 3 statements:
1. The relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for p and s
2. The relative frequency of getting (1,-1) or (-1,1) for f(0,30) is the same for q and s
3. The relative frequency of getting (1,-1) or (-1,1) for f(-30,30) is the same for r and s

So now my next question is, do you agree that
the relative frequency of getting (1,-1) or (-1,1) for f(-30,30) in s is less than or equal to the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) in s, plus the relative frequency of getting (1,-1) or (-1,1) for f(0,30) in s?

 

[DevilsAvocado]

[Bill, I’ll get back to your replies, but first the “Holy Grail” that could finally solve the elongated “Bill-Bell debates”! ;]

I'm glad (although again surprised) to hear that we're agreed on that. In future discussion, let's assume that we're dealing with sufficiently large sets of photons, so that we don't need to worry about whether they're identically prepared.

Fine with me. I was just trying to be very clear so there was no chance for misunderstanding.

And I’m extremely glad! This is what I’ve been shouting about for a couple of weeks now! ()

So now my next question is, do you agree that
the relative frequency of getting (1,-1) or (-1,1) for f(-30,30) in s is less than or equal to the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) in s, plus the relative frequency of getting (1,-1) or (-1,1) for f(0,30) in s?

WTF!? <-- oral cavity state vectors

I can’t believe we FINALLY gotten to the “Holy Grail”!?

Beautiful Lugita! Let’s just add Bell’s inequality for complete clarity:

N(+30°, -30°) ≤ N(+30°, 0°) + N(0°, -30°)​

Yay!

 

[billschnieder]

OK, so we're now agreed on the following 3 statements:
1. The relative frequency of getting (1,-1) or (-1,1) for f(-30,0) is the same for p and s
2. The relative frequency of getting (1,-1) or (-1,1) for f(0,30) is the same for q and s
3. The relative frequency of getting (1,-1) or (-1,1) for f(-30,30) is the same for r and s

I agree to each of those statements considered independently, but I do not agree to all of them considered together. In other words I would not agree if the s in 1, 2 and 3 were the exact same set, rather than 3 bigger arbitrarily larger identically prepared sets. In other words I do not agree that the operations for measuring f(-30,0), and f(-30,30) on the same set as f(0,30) on the exact same set commute.

 

[DrChinese]

I am baffled as to why you would even be comparing the first two streams with the third. Are you suggesting that stream 3 is simply a linear combination of the first two?

It (3) is a physical combination of 2 different streams. It is polarization entangled. Yet, according to your world view, all of the member pairs are predetermined with known polarization. Their statistics can never result in perfect correlation. So that does not match experiment, which has them giving entangled state statistics.

So you should assert that the result of 3) is a linear combination of 1) and 2), yes. There will be approximately twice as many photons detected and matched. You should assert there is no physically entangled state, since it is all predetermined. After all, each stream's pairs originates in a different crystal after undergoing down conversion. In terms of a photon's time, they arrive far apart.

 

[billschnieder]

It (3) is a physical combination of 2 different streams. It is polarization entangled.

I don't understand how you take 2 different streams, each not polarization entangled and then simply combine them linearly and get a polarization entangled stream.

Yet, according to your world view, all of the member pairs are predetermined with known polarization.

I do not believe you understand my world view yet, as I've explained already multiple times. I'm talking apples and you are talking oranges.

So you should assert that the result of 3) is a linear combination of 1) and 2), yes.

I've made no such assertion, I'm still trying to understand what you mean by 1, 2 and 3 and how linear combination of 1 and 2 results in 3.

 

[billschnieder]

So now my next question is, do you agree that
the relative frequency of getting (1,-1) or (-1,1) for f(-30,30) in s is less than or equal to the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) in s, plus the relative frequency of getting (1,-1) or (-1,1) for f(0,30) in s?

If the set "s" is the exact same set of photons in all three measurements, then I agree that the relative frequency of getting (1,-1) or (-1,1) for f(-30,30) in s is less than or equal to the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) in s, plus the relative frequency of getting (1,-1) or (-1,1) for f(0,30) in s.

Do you now see where the disagreement is? (after considering my answer to this question and the one answered in my last response to you) If you do see the point of disagreement but do not understand what I mean, I can explain further, but only after it is clear to you on exactly which point we disagree.

 

[Nugatory]

I agree to each of those statements considered independently, but I do not agree to all of them considered together.

Lugita15, that was a very good try and I'm impressed that you got as close as you did... But it looks as if once again Lucy has snatched the football away at the last moment.

 

[DrChinese]

Lugita15, that was a very good try and I'm impressed that you got as close as you did... But it looks as if once again Lucy has snatched the football away at the last moment.

We all know we are never going to change Bill's* mind. Every experimental form of entanglement is essentially a refutation of every local realistic viewpoint. If you simply deny this, as Bill does, well... here we are.


*As he is not likely to change mine.

 

[DrChinese]

I don't understand how you take 2 different streams, each not polarization entangled and then simply combine them linearly and get a polarization entangled stream.

In your world, you can't. In the real world, you get Type I polarization entangled pairs.

http://arxiv.org/abs/quant-ph/0205171

See Fig. 2. Note that here they use a single pump laser. But you could accomplish the same thing with 2 phase locked pumps too if you combine the outputs so the source is indeterminate.

 

[Nugatory]

I agree to each of those statements considered independently, but I do not agree to all of them considered together. In other words I would not agree if the s in 1, 2 and 3 were the exact same set, rather than 3 bigger arbitrarily larger identically prepared sets. In other words I do not agree that the operations for measuring f(-30,0), and f(-30,30) on the same set as f(0,30) on the exact same set commute.

So f I understand you properly, you're challenging the promiscuous transfer of measurement results from one set to another set in which the same measurements have not been made?

I mentioned much further up in the thread that there is an assumption embedded in the construction of the Bell experiments (but not in the theorem itself), and that this assumption is essential to accepting the experiments as falsification of the inequality. That assumption is fair sampling, which along with counterfactual definiteness allows us to transfer the results from measuring properties in one set to another set in which we do not measure the those properties.

If a property is counterfactually definite and I am measuring it within one statistically representative subset of a larger population, I can conclude that my measurements of that property are applicable to (by fair sampling) any other statistically representative subset of that population, even if I don't measure the property on that subset (by counterfactual definiteness).

 

[stevendaryl]

We all know we are never going to change Bill's* mind. Every experimental form of entanglement is essentially a refutation of every local realistic viewpoint. If you simply deny this, as Bill does, well... here we are.


*As he is not likely to change mine.

I like the attempt to get past disagreements about the meaningfulness or validity of abstract arguments by making it into an Amazing-Randy style wager:

The Bell side makes a bet, that there is no way to simulate EPR-style correlations without nonlocal communication, using a combination of deterministic devices plus random number generators.

The challenge is to design a "pair generator" that will produce a sequence of pairs of "secret messages", together with a box, "Alice's detector" and "Bob's detector" that will receive a secret message, together with a real-number input from Alice or Bob, and will output either +1 or -1.

The challenge proceeds as follows: We pick a number of rounds, say 100. Each round proceeds as follows: On round number n,

1. The pair generator creates a pair of secret messages m_{A,n} and m_{B,n}, and sends m_{A,n} to Alice's detector, and m_{B,n}to Bob's detector.
2. Alice rolls a 6-sided die. If the result is 1 or 2, she picks \alpha_n = 0°. If the result is 3 or 4, she picks \alpha_n = 120°. If the result is 5 or 6, she picks \alpha_n = 240°. She inputs the value into her detector.
3. The detector produces an output, either A_n =+1 or -1, which is only seen by Alice. She records her choice of \alpha_n and the output A_n from the detector.
4. Bob similarly chooses \beta_n from the set { 0°, 120°, 240° }.
5. Bob's detector produces an output, either +1 or -1, which is only seen by Bob. He records his choice of \beta_n and the output, B_n from his detector.

After many rounds, Alice and Bob each have a list of pairs. They put their lists together to compute joint probabilities as follows:

P(\alpha, \beta, A, B) = \dfrac{N_{\alpha, \beta, A, B}}{N_{\alpha,\beta}}

where N_{\alpha, \beta, A, B} is the number of rounds in which Alice chose \alpha_n = \alpha, and Bob chose \beta_n = \beta, and A_n = A and B_n = B, and where N_{\alpha, \beta} is the number of rounds in which Alice chose \alpha_n = \alpha and Bob chose \beta_n = \beta.

The bet is that there is no way to design the "pair generator" and the "detectors" so that the simulated joint probability distribution P(\alpha, \beta, A, B) agrees with the quantum spin-1/2 EPR predictions:

P(\alpha, \beta, +1, +1) = P(\alpha, \beta, -1, -1) =\frac{1}{2} sin^2(\frac{1}{2} (\beta - \alpha))

P(\alpha, \beta, +1, -1) = P(\alpha, \beta, -1, +1) =\frac{1}{2} cos^2(\frac{1}{2} (\beta - \alpha))

Where the "local" comes in is the assumption that Bob's detector is not allowed to use Alice's input, and vice-versa, and that the pair generator is not allowed to use either input. If you violate these locality restrictions, it's easy to get the QM results.

 

[lugita15]

I agree to each of those statements considered independently, but I do not agree to all of them considered together. In other words I would not agree if the s in 1, 2 and 3 were the exact same set, rather than 3 bigger arbitrarily larger identically prepared sets. I

I definitely intend s to be the same set in all three statements. After all, s is just the union of p, q, and r. So you're saying that if I use the same set s, then my 3 statements aren't all true. Well, if they're not all true, then at least one of them is wrong, say the first one. So are you saying it's possible for the relative frequency of something to be the same for three different sets, but not to be the same for their union?

 

[billschnieder]

I definitely intend s to be the same set in all three statements.

I'm saying a relationship between 3 properties within a single set p, is the same relationship between 3 properties within another single q, and the same relationship between 3 properties within another single set r, and the same relationship between 3 properties within a bigger set s=(p+q+r). But I'm also saying that same relationship does not necessarily apply between one property from "p" and a second property from "q" and a third property from "r", even if p, q and r are each components of "s".

For example, Let us assume that we have a pair of perfectly correlated photons hidden properties which predetermine outcomes at three angles a,b,c for each photon and each of the outcomes can be +1 or -1.We can calculate the sum of the paired product of outcomes ab and ac as follows that

ab + ac = a(b + c) = ab(1+bc)
since b = 1/b, and taking the absolute value of both sides of this equation we get

|ab + ac| = |ab (1+bc) | ≤ 1 + bc or |ab + ac| - bc ≤ 1, a Bell inequality.

Clearly, we have derived this as a condition which applies to 3 outcomes a,b,c from a single pair of photons, without any other assumption than that the 3 outcomes exist. In fact, we might have started with an abstract assumption that we have an arbitrary set of 3 variables a,b,c each with values +1 or -1, without any physical meaning assigned to the set, and still obtain the inequalities. Furthermore, if we have a very large number of such photons, because individually they obey the inequality, collectively, set of photons also obeys the inequality
|<ab> + <ac>| - <bc> ≤ 1, another Bell inequality, or if you prefer |C(a,b) + C(a,c)| - C(b,c) ≤ 1, Bell's original form.

Do you agree?

 

[billschnieder]

Continuing ...

The terms C(a,b), C(a,c) and C(b,c) in the above inequality are each from the exact same set of photons, not three different sets. And the reason why it MUST be so, is because in order to derive the inequality, we started with two terms "ab + ac", and we factored out "a(b+c)", then we factored out b=1/b again to get "ab(1+bc)" and in this way we "created" the "bc" term by stitching together part from "ab" and part from "ac".

So what is the problem? You may ask.

Suppose instead we have 3 different pairs of photons each with predetermined outcomes at 3 angles, say (a1, b1, c1) for pair 1, (a2,b2,c2) for pair 2 and (a3, b3, c3) for pair 3. Obviously, each of those pairs will independently satisfy the inequality since we could simply say (a=a1,b=b1,c=c1) etc. Similarly, we could have 3 sets, each with a large number of photon pairs and each one will independently satisfy the inequalities.

However if we calculate the "ab" term from the first photon pair, the "ac" term from the second photon pair and "bc" from the third pair (ie, a1b1, a2c2, b3c3), is it reasonable to expect the inequality |ab + ac| - bc ≤ 1 to be obeyed?

Similarly if we calculate the "ab" correlation from the first set of photons, the "ac" correlation from the second set of photons and the "bc" correlation from the third set of photons (ie a1b1, a2c2, b3c3), is it reasonable to expect the inequality |C(a,b) + C(a,c)| - C(b,c) ≤ 1 to be obeyed?

I say the answer to the above two questions is NO. Let us see why, by verifying the maximum value of left hand side |a1b1 + a2c2| - b3c3
Clearly, the LHS is maximum when b3c3 = -1, a1b1 = a2c2 = ±1
Therefore the correct inequality for three different pairs is
|a1b1 + a2c2| - b3c3 ≤ 3 NOT |a1b1 + a2c2| - b3c3 ≤ 1

And, the correct inequality for 3 different sets is
|<a1b1> + <a2c2>| - <b3c3> ≤ 3 NOT |<a1b1> + <a2c2>| - <b3c3> ≤ 1

In other words, the relationship we derived from properties within a single pair of photons does not apply between properties from 3 different pairs of photons.
Do you agree?

 

[billschnieder]

Some may wonder why we got a maximum of 3 for different sets and 1 for the same set. I thought I should clarify again. For three sets we have |a1b1 + a2c2| - b3c3 , with the maximum of 3 obtained when b3c3 = -1, a1b1 = a2c2 = +1.

This can happen if b3 = -1, c3 = 1, a1 = 1, b1 = 1, a2 = 1, c2 = 1 which is possible within three different sets since b3 is allowed to be different from b1. However in a single set, it is impossible for the "b" used to calculate "bc" term to be different from the "b" used to calculate the "ab" term.

 

[Nugatory]

But I'm also saying that same relationship does not necessarily apply between one property from "p" and a second property from "q" and a third property from "r", even if p, q and r are each components of "s".

Of course it does not necessarily apply. But it does apply if we make two additional assumptions, namely counterfactual definiteness and fair sampling within statistically representative sets. Which of these assumptions do you reject?

 

[rlduncan]

Bill, your #346 and #347 posts are excellent summaries of your argument. They tie in nicely with your post #270 which reveal that Bell-Type inequalities cannot apply to both scenarios.

 

[lugita15]

I'm saying a relationship between 3 properties within a single set p, is the same relationship between 3 properties within another single q, and the same relationship between 3 properties within another single set r, and the same relationship between 3 properties within a bigger set s=(p+q+r). But I'm also saying that same relationship does not necessarily apply between one property from "p" and a second property from "q" and a third property from "r", even if p, q and r are each components of "s".

I'm not sure what properties you're talking about. Do you mean relative frequencies?

Anyway, let me just ask you this. For each of the four sets p, q, r, and s, we can calculate the following relative frequencies:
1. The relative frequency of getting (1,-1) or (-1,1) for f(-30,0)
2. The relative frequency of getting (1,-1) or (-1,1) for f(0,30)
3. The relative frequency of getting (1,-1) or (-1,1) for f(-30,30)

Now you've already agreed that all of these relative frequencies are the same for p, q, and r. But you've disagreed that all of these relative frequencies are the same for p, q, r, and s. Well then, you must believe that at least one of these relative frequencies can different for s than for p, q, and r. In other words, you must believe that the relative frequency of something can be different for the union of three sets even though it's the same for all three sets.

So first of all, do you in fact believe that? If so, I think I can prove you wrong with a very simple argument.

 

[billschnieder]

Of course it does not necessarily apply. But it does apply if we make two additional assumptions, namely counterfactual definiteness and fair sampling within statistically representative sets. Which of these assumptions do you reject?

I partly agree and partly disagree. If we start talking about measurements of the predermined outcomes, then since we have 3 outcomes but only 2 identical photons, we can only measure a pair of properties, and the other two pairs are counterfactual. In other words, CFD is NOT an additional assumption, to derive Bell's inequality for a single pair.
For three different pairs, we can measure each correlation term from a different pair and CFD does not come in at all. It would not even make since to "assume CFD" in this case. What would that even mean?

There is however an additional assumption that can be made, which you may call "fair sampling" although it is more subtle than you describe. We may assume for the three separate photon pairs that a1=a2 and b1=b3 and c2=c3. Extending this to three separate sets of photon pairs, this means the 6 lists of outcomes from the separate sets can be reduced to 3 pairs of identical lists [a1:a2, b1:b3, c2:c3], in which not only the number of +1s and -1s must be the same in each pair, but also the pattern of switching between +1 and -1 in each list. Let me explain another way. If you place the six lists side by side, each row should obey a1=a2 and b1=b3 and c2=c3 OR it should be possible to sort them (while keeping pairs from a set together) so that the six lists obey a1=a2 and b1=b3 and c2=c3. Only then will the 3 photon pairs in each row obey Bell's inequality and consequently the correlations from the three sets will obey the inequality.

Obviously this is more subtle than the regular "fair sampling".

 

[billschnieder]

I'm not sure what properties you're talking about.

You did not say if you understood the argument in #345 and #346. The properties are the correlations.

 

[billschnieder]

Anyway, let me just ask you this. For each of the four sets p, q, r, and s, we can calculate the following relative frequencies:
1. The relative frequency of getting (1,-1) or (-1,1) for f(-30,0)
2. The relative frequency of getting (1,-1) or (-1,1) for f(0,30)
3. The relative frequency of getting (1,-1) or (-1,1) for f(-30,30)

Now you've already agreed that all of these relative frequencies are the same for p, q, and r. But you've disagreed that all of these relative frequencies are the same for p, q, r, and s.

I'm not sure you are getting all the subtleties yet, that is why I ask you to confirm if you understand the argument in #345 and #346. You started by saying p, q, and r are completely distinct sets, and s = (p + q + r). And then you go on to talk of measuring f(-30,0), f(0,30) and f(-30, 30) in s and the results of these measurements being the same as the results from completely distinct sets p, q and r.

The reason I disagree with this is because if you define "p" as the set of photons from "s" which
did not match for f(-30,0), and "q" as the set of photons in "s" which did not match for f(0,30), and "r" as the set of photons which did not match for f(-30,30), then it is impossible for "p", "q" and "r" to be completely distinct sets, contrary to our starting assumption. That is why the relationship between three measurements on a single set "s" is not the same relationship between three measurements from three "independent" sets "p", "q", and "r". If you are admitting that "p", "q", and "r" are not independent/distinct/separate sets, then we are not going anywhere towards answering the core issue.

 

[lugita15]

You started by saying p, q, and r are completely distinct sets, and s = (p + q + r).

Yes, and I'm still saying that. (Although I'd use the union sign rather than the plus sign.)

And then you go on to talk of measuring f(-30,0), f(0,30) and f(-30, 30) in s and the results of these measurements being the same as the results from completely distinct sets p, q and r.

I'm sorry if I was imprecise. By "getting" I didn't mean measuring. Let me rephrase:
1. The relative frequency of f(-30,0) being equal to (1,-1) or (-1,1)
2. The relative frequency of f(0,30) being equal to (1,-1) or (-1,1)
3. The relative frequency of f(-30,30) being equal to (1,-1) or (-1,1)

The reason I disagree with this is because if you define "p" as the set of photons from "s" which
did not match for f(-30,0), and "q" as the set of photons in "s" which did not match for f(0,30), and "r" as the set of photons which did not match for f(-30,30), then it is impossible for "p", "q" and "r" to be completely distinct sets, contrary to our starting assumption.

No, that's not how I'm defining them at all. I'm simply defining them to be a set of photon pairs measured at (-30,0), a set of photon pairs measured at (0,30), and a set of photon pairs measured at (-30,30), respectively.

If you are admitting that "p", "q", and "r" are not independent/distinct/separate sets, then we are not going anywhere towards answering the core issue.

No, p, q, and r are definitely three completely separate sets, and s is their union.

 

[billschnieder]

Please confirm that you understand and/or agree with the argument in #345 and #346. AFTER you've done that, I will explain using a similar argument why your inequality based on set "s" does not necessarily apply to terms from three distinct sets p, q and r.

 

[lugita15]

Please confirm that you understand and/or agree with the argument in #345 and #346.

I apologize, but I'm not that familiar with Bell's original proof, so I can't really comment much on the validity of what you're saying in these posts. So if you don't mind, rather than making analogies with another Bell's theorem proof, can we just stick to the line of reasoning that I've been presenting to you?

AFTER you've done that, I will explain using a similar argument why your inequality based on set "s" does not necessarily apply to terms from three distinct sets p, q and r.

Can you please just present your argument directly, without relying on posts #345 and #346? And before you get to why the inequality doesn't apply, can you back up and answer my question about whether the relative frequency of something can be the same in three disjoint sets and yet different for their union?

 

[billschnieder]

I apologize, but I'm not that familiar with Bell's original proof, so I can't really comment much on the validity of what you're saying in these posts.

The argument is self contained. You do not need any information outside of those posts to understand the argument. If you do not understand that argument, you very likely won't understand the core issue.

 

[DevilsAvocado]

However if we calculate the "ab" term from the first photon pair, the "ac" term from the second photon pair and "bc" from the third pair (ie, a1b1, a2c2, b3c3), is it reasonable to expect the inequality |ab + ac| - bc ≤ 1 to be obeyed?

Similarly if we calculate the "ab" correlation from the first set of photons, the "ac" correlation from the second set of photons and the "bc" correlation from the third set of photons (ie a1b1, a2c2, b3c3), is it reasonable to expect the inequality |C(a,b) + C(a,c)| - C(b,c) ≤ 1 to be obeyed?

I say the answer to the above two questions is NO. Let us see why, by verifying the maximum value of left hand side |a1b1 + a2c2| - b3c3
Clearly, the LHS is maximum when b3c3 = -1, a1b1 = a2c2 = ±1
Therefore the correct inequality for three different pairs is
|a1b1 + a2c2| - b3c3 ≤ 3 NOT |a1b1 + a2c2| - b3c3 ≤ 1

And, the correct inequality for 3 different sets is
|<a1b1> + <a2c2>| - <b3c3> ≤ 3 NOT |<a1b1> + <a2c2>| - <b3c3> ≤ 1

In other words, the relationship we derived from properties within a single pair of photons does not apply between properties from 3 different pairs of photons.
Do you agree?

No.

Because you are making the assumption that somehow the LHV knows on beforehand what the polarizer settings will be, which of course is non-locality.

Then only (best) feasible way to implement your “three-stage rocket” is to create a “random generator” that will assign 1 of 3 possible values/predetermined outcome, i.e. for (+30°, 0°) and (0°, -30°) and (+30°, -30°). The first two values/predetermined outcomes gives the same result/correlation, which gives you only 2 possible choices. Looking good, huh?

Now, you’re the “Bayesian Grandmaster Flash” and can easily do the calculations for this. However, I skip this part and tell you right away that there’s no way in h*ll your “random generator” (that delivers the right answer 50% of the time for 2 possibilities x 3 pairs/set) can beat QM that get it right 100% of the time, for every pair/set.

No way.

Sorry Bill, good try but it doesn’t work.


[I’ll get back on your first replies soon/later]

 

[lugita15]

The argument is self contained. You do not need any information outside of those posts to understand the argument. If you do not understand that argument, you very likely won't understand the core issue.

I apologize, but I've been examining in great detail the inequality we've been talking about. On the other hand, if you give me an analysis of a completely different inequality, and ask me my opinion of that analysis, I can't readily spot whether or where you're making errors. So again, can you phrase your argument in terms of the concepts and terminology we've been discussing all this time?

 

[billschnieder]

can you back up and answer my question about whether the relative frequency of something can be the same in three disjoint sets and yet different for their union?

But I've already answered this question. I've already agreed that if the relative frequency of "something" is the same in three disjoint sets, then the relative frequency of that same "something" will be the same in the union of those three disjoint sets. The part you are not understanding, which I disagree with is if you take a single set, and measure the relative frequency of three different but logically related things in that same set, and write an inequality which relates those three things, that inequality which you have now written, does not necessarily apply to the three disjoint sets. For this to be obvious, you will have to write down clearly how you arrived at the inequality from the single set, by writing down clearly the starting equation. I'm talking about the statement

do you agree that
the relative frequency of getting (1,-1) or (-1,1) for f(-30,30) in s is less than or equal to the relative frequency of getting (1,-1) or (-1,1) for f(-30,0) in s, plus the relative frequency of getting (1,-1) or (-1,1) for f(0,30) in s?

I agreed to this question already. However, to make things clear, please can you prove this inequality by deriving it, it will be evident later why this is important and since you brought it up, let it be on you to derive it from a single set "s".