Okay then, you’re wrong and we have something to disagree about! ;)
Do you disagree when I say that if the momentum is defined, then the position is not defined?
Certainly, that’s the point! It’s not defined, so it’s not within [-1, 1], and it’s not not within [-1, 1]! We can ‘shrink’ things down by just considering an approximate momentum eigenstate, where we know the momentum to be within an interval ∆p, and consequently, know the position to be within an interval ∆x. Now we just cut the ∆x interval in two, intervals ∆x[sub]1[/sub] and ∆x[sub]2[/sub], such that the two taken together are equal to ∆x. We know that the momentum is in ∆p, and the position is in ∆x[sub]1[/sub] or ∆x[sub]2[/sub]. However, the position is not in either x[sub]1[/sub] or ∆x[sub]2[/sub] – this extends beyond the limits given by the uncertainty principle. So it’s not the case that the momentum is in ∆p and the position in x[sub]1[/sub] – since it’s false that the position is in x[sub]1[/sub] --, and it’s not the case that the momentum is in ∆p and the position is in ∆x[sub]2[/sub] – since it’s false that the position is in ∆x[sub]2[/sub].
I have to dash now, so I’m posting this without checking, so any mistakes are of course only due to haste…
Wait sorry, I’m having double-negation trouble here.
Because I am retarded, please answer the following true/false question for me:
True or False: If the momentum of a particle is defined, then position is undefined.
ETA: I think I see now you’re agreeing–i.e., answering true. On that assumption, I’ll formulate my argument again in the next post.
p = “the particle is moving right.”
q = “the particle is in the interval [1, -1]”
r = “the particle is not in the interval [1, -1]”
- Assumption: For all x and y, if a particle’s position is undefined, then any statement of the form “the particle is not in the interval [x, y]” is true.
- Assumption: If a particles’ momentum is defined, then its position is undefined.
- Assumption: If a particle is moving right, then its momentum is defined.
- Suppose p is true.
- Then the momentum of the particle is defined.
- So the position of the particle is undefined.
- Therefore r is true.
- So if p is true, then r is true.
Are you with me so far, or is there something in the above that we should stop and discuss?
This appears to me to be an error.
Since the position is undefined, there is no position that it has. This means that it’s not within [-1, 1]. (And, importantly, not “not not.” Just not.) There’s no interval that it’s within, because it has no position at all.
That 1/0 is undefined doesn’t mean that it’s not 2 and not not 2. It just means it’s not 2, and not 3, and not 4, and not etc for each number.
Rephrase it in terms of information about the particle’s location. If you have perfect information about its momentum, you must have zero information about its location. Knowing that there’s an interval where it can’t be – i.e. knowing that it’s not within [-1, 1] – means having some information about its location – which violates the uncertainty principle. In other words, if you have perfect knowledge of its momentum, it can’t be true that it’s not in [-1, 1] – it’s not not in [-1, 1].
How about the slightly rephrased version I gave, where you don’t have perfect momentum information? It’s really the same thing – one has to keep in mind that what we’re really talking about here are subspaces of the particle’s Hilbert space, so ‘and’ and ‘or’ can be literally taken to be their meet and join. Thus, it’s true that the particle is in the join of ∆x[sub]1[/sub] and ∆x[sub]2[/sub], but crucially, this does not mean it is in either ∆x[sub]1[/sub] or ∆x[sub]2[/sub] – so if we take ∆x[sub]1[/sub] = [-1, 1] and ∆x[sub]2[/sub] = ![-1, 1], since it is not in either, it is not in [-1, 1], and also not in its complement, i.e. it is not not in [-1, 1]. This is unfamiliar, but that doesn’t mean it’s not true.
Indeed, this can be regarded as the reason why quantum mechanics is so weird, because if the logic it is founded on were distributive and hence, equal to propositional logic, QM would reduce to ordinary probability theory – which it doesn’t.
Oh and, of course, ‘the particle is not in [-1, 1]’ is also simply not true – experimentally, that might just be where it is found, which is what I was getting at earlier. Yes, the two are inequivalent propositions, but nevertheless, I don’t know what it would mean to consider ‘the particle is not in [-1, 1]’ to be true if experiment could very well find it just there.
It’s my understanding that the uncertainty here doesn’t amount to us simply not knowing its location–the idea, IIUC, is that there simply is no location.
Isn’t that correct?
You said that knowing it’s not within [-1, 1] would mean we have information about its location. But if there is no location to know about, then it’s impossible to have information about it. Knowing it’s not within such-and-such interval doesn’t give us information about its location, because there’s no location to have information about.
In my previous post, with the assumptions and derivations etc, where in the derivation do you want to stop me? Right at assumption one? Or do you grant that and want to stop me elsewhere? Or do you buy every step of it and think the derivation isn’t relevant to our problem?
I don’t know what these deltas mean, and I don’t know what meets and joins are, and I don’t know what a Hilbert space is, so I can’t say much about this. But if it’s not in [-1, 1] and it’s also not not in [-1, 1], then it is not anywhere. What argument is there to say that it is somewhere? Is it just that an experiment may find it to be somewhere? But to repeat myself, “The particle is here” and “An experiment will find a particle here” are non-equivalent propositions. They don’t mean the same thing. If it’s not in [-1, 1] and also not not in [-1, 1], and also an experiment will find it somewhere, then what we have to say is that the particle is not anywhere, and then upon the occasion of the experiment, the particle will be somewhere. No PL-contradiction here.
BTW, is the “undefinedness” of position that we’re talking about basically the same thing as the “undefinedness” of division by zero? I thought that it was. But if it is, then notice this: “1/0 is not 2” is a true statement exactly because 1/0 is undefined. In the very same way, “Particle is not in [1, -1]” is true exactly because the particle’s position is undefined.
What’s the difference between the rules of Euclidean geometry and the rules of Boolean propositional logic? The rules of Euclidean geometry are definitionally true… for Euclidean geometry. There is no Euclidean space which does not satisfy the rules of Euclidean geometry. Of course, one may well be interested in other, related but in ways markedly different, accounts of “spaces”, and once one moves to such a different context, one needn’t shackle oneself to using the rules of Euclidean geometry anymore. They may well not apply. In that sense, one can find a counterexample, though perhaps “counterexample” is not the best word to describe the idea that “Not every rule-system that talks of things like points and lines is a rule-sytem which includes the rules of Euclidean geometry”.
The rules of Boolean algebra are definitionally true… for Boolean algebra. There is no model of Boolean algebra which does not satisfy the rules of Boolean algebra. Of course, one may well be interested in other, related but in ways markedly different, accounts of “propositional structures”, and once one moves to such a different context, one needn’t shackle oneself to using the rules of Boolean logic anymore. They may well not apply. In that sense, one can find a counterexample, though perhaps “counterexample” is not the best word to describe the idea that “Not every rule-system that talks of things like conjunctions and disjunctions is a rule-system which includes the rules of Boolean algebra”.
Are you proposing that there is a fundamental difference between the status of the abstract rules of Euclidean geometry and the abstract rules of Boolean algebra?
I’ll admit I was a little iffy about saying that–but it’s my understanding that it is possible to do propositional logic with no axioms, and impossible to do geometry without axioms. Am I wrong about that, though?
Which is basically all I’m saying, but my claim is that this means if you think you’ve got a tautology in boolean algebra which is false, then you must be mistaken. You’ve either misinterpreted a proposition, misapplied an operator, or you’re simply outright not doing boolean algebra.
The arguments I’ve seen that quantum mechanics somehow “violates” PL (not just “can be modeled by other logics” but positively violates) always seem to me to turn on a misinterpretation or equivocation on a propositional variable. So for example see our discussion of the supposed violation of distributivity. What I’m trying to show is that if you interpret the propositional variables consistently, you find that the supposed contradiction disappears. The view that there’s a contradiction arises from a mistaken tendency to glide back and forth (in this discussion and others I’ve seen) from “there’s a particle here” to “observations will show a particle here” and back–and to treat these both as instances of the same proposition.
And I’m claiming that QM is a model of Boolean algebra.
Agreed–but that does not make what you’re working with isnt a model of Boolean algebra. It’s just that there are much better models to work with. (Just as a description of a predicate logic gives eo ipso a model for a propositional logic–but will nevertheless be a better model to work with (than Propositional logic would be) in many contexts.)
Well yeah, you’re damn right it’s not the best word to describe it. 
Given what “counterexample” means, it’s entirely wrong to use the word to describe this idea, isn’t it?
It would be like calling “Socrates is a man and all men are mortal therefore Socrates is mortal” a “counterexample” to propositional logic.
But calling it a “counterexample” is exactly the kind of thing, I think, that leads people (in internet forums to be sure, and I’m fairly sure I’ve seen Physicists say this as well, and well you know there’s always a Philosopher (at least one) will have said ‘X’ for any ‘X’…) to say wildly false stuff like ‘Quantum mechanics violates propositional logic’ or ‘quantum mechanics shows that some statements are neither true nor false’ and ‘qm violates the law of noncontradiction’ etc.
It’s impossible to violate PL because any statement that seems to violate it can not be a statement of PL. The rules of PL guarantee this. I THINK
Yet another post.
It sounds almost naive to say this, but the more I think about it, the more I think it’s right:
Just look at the truth table. You can see from it that distributivity simply can’t be violated.
Here’s what I mean.
IF you’re saying something of the apparent form:
(p & (q v r)) & --((p & q) v (p & r))
and you REALLY mean ‘&’, ‘v’ and ‘–’ by these symbols
and you are REALLY interpreting p, q and r consistently (i.e. you mean the same thing by each letter each time you use it)
and you REALLY mean each of p, q and r to be understood as being either true or false and not both*
then by doing all of the above, you are limiting yourself to the rules that apply to propositional logic. So then, draw up the truth table. All the truth table does is draw out the implications of all of the above. And what you’ll find is that the truth table proves you must be wrong. The statement can’t be true.
Or perhaps you don’t really mean some of the things you said you really mean above. But in that case, you’re not actually saying “(p & (q v r)) & --((p & q) v (p & r))”. You’re saying something else, in some other formalism. You’re not talking about distributivity at all, it turns out.** You’ve just got a kind of “homophone” for distributivity in whatever formalism you’re using.
Axioms are just another word for “rules”, in the end. It’s certainly the case that people can say that they’re doing propositional logic with no extra assumed axioms, but all that means is that they’ve already taken the axioms of their propositional logic as so implicit that they’re not willing to consider any rule system without them; they’ve been baked in. But you could do that with anything; you could do the same with the rules of Euclidean geometry if you liked.
Does “x & (~x v y) = x & y”, say, hold without the assumption of any axioms? Well, one could certainly play a game where “x & (~x v y)” and “x & y” aren’t considered equal (Why use the same symbols for that game as for the Boolean game? Well, it might be very similar in many respects, even if different in others. Why use the same + for addition of natural numbers, of integers, of vectors, of bits modulo 2, etc., when they all act quite differently?). Their equality doesn’t follow immediately, from no rules at all, such that one is led inexorably from the one to the other. It’s only under the implicit assumption of some rules/axioms that “x & (~x v y) = x & y” is necessarily the case. One could just as well study some other rules governing &, if one liked.
The sense in which a model of a propositional theory necessarily validates the rules of Boolean algebra is only the sense in which one has baked into their account of what a “model of a propositional theory” is, at some level or another, the rules of Boolean algebra as a hardcoded requirement. [E.g., often one’s definition of a model of a propositional theory involves the requirement that propositions be interpreted as elements of the 2-element Boolean algebra. Well, of course the laws of Boolean algebra will be validated by every interpretation of propositional logic if you demand that every interpretation of propositional logic be a Boolean algebra (and a particular one at that). If I wanted any rule to be considered inviolable, I could pull the same trick: demand the models include that rule as part of their structure. “(B -> A) -> A = B -> A” could be just as much an inescapable logical principle, on a suitable account of what a “model” must be (e.g., propositions must be interpreted as natural numbers, and A -> B must be interpreted as max(A + 1, B)]
That’s right. But the same applies to Euclidean geometry… if you think you’ve got a theorem in Euclidean geometry which is false, then you must be mistaken. You’ve either misapplied the rules of Euclidean geometry or you’re simply outright not doing Euclidean geometry. (You may have thought you were doing Euclidean geometry, in that you thought the criteria you were directly concerned with for judging “validity” of results were criteria which validated all the rules of Euclidean geometry, only to discover that this was not so. But this is just one particular way in which you may not have been doing Euclidean geometry.)
I agree. I’m not thrilled with those arguments for similar reasons as you are. But in the same way, I’m not thrilled to say that any mathematical theory (which is to say, any abstract system of rules) can be empirically violated. Boolean algebra, non-Boolean algebra, Euclidean geometry, anything… All that can fail is for a particular system of rules’ consequences, interpreted in a particular fashion, to be empirically correct claims about a particular phenomenon. Which only means that system of rules under that interpretation does not model that phenomenon; it may well model that same phenomenon under another interpretation, or model other phenomena, or such things. It is, to me, meaningless to speak of a system of rules as being empirically violated; it’s only particular ways of using those rules to do things which may fail to do what one wants it to do.
I am happy with this argument of yours, that Boolean logic correctly applies to this situation and distributive holds and so on, suitably interpreted. But I would also be happy with treating “There is a particular here” and “Observations will show a particle here” as the same proposition and stating that distributivity fails and that quantum logic were more applicable, if it seemed a productive way of viewing things, as well. The same situation may well be modelled by apparently conflicting theories, under different interpretations.
One could say the surface of the Earth is perfectly well modelled by Euclidean geometry, and there are no triangles whose angles add up to 180 degrees, because geodesics across the surface of the Earth aren’t actually straight lines; they bend quite noticeably. Or one could say the surface of the Earth displays a non-Euclidean geometry, and a North pole and two points on the equator induce a triangle with angles of more than 180 degrees, counting geodesics across the Earth as straight lines. And both of these are correct; they just involve different choices for how one might link up abstract statements in the language of geometry with statements about the surface of the Earth. They seem to contradict each other, but they are only using similar terminology with different interpretations.
As indeed it is, if one interprets the propositional terminology in a certain way into terms of QM. At the same time, QM is also a model of non-distributive logic, if one interprets the propositional terminology in a different way into terms of QM. Whether one, the other, both, or neither of these interpretations (that is, bridges between different rule systems) is actually useful or improves understanding of the rule-systems involved is a matter to be judged.
I’m not sure what you mean by this. Can I claim that anything isn’t a model of anything else? For example, I would like to say alphanumeric sequences and their concatenation do not comprise a model for the theory of a commutative binary operation (insofar as “cat” + “dog” = “catdog” is different from “dog” + “cat” = “dogcat”), where we interpret the binary operation of the theory as the concatenation operation on those sequences. Can I say that? Or would I have to deal with the objection that “Well, no, that IS still a model of commutativity, you’re just looking at it wrong; if you picked a different interpretation, you would find that commutativity still holds…”. Because, I agree, I could find some other way to interpret + in the context of alphanumeric sequences where it was commutative. But when I say that this is not a model of commutativity, I mean that this particular interpretation does not model commutativity.
And in that same way, I could certainly pick ways of interpreting the language of Boolean algebra into a context which did not validate the rules of Boolean algebras. For example, if I have two distinct elements, 0 and 1, such that 0 * 1 = 1 but 1 * 0 = 0, and choose to interpret the & of Boolean algebra as this * operator (and make up some interpretation for ~ and all the rest of it). This would be an interpretational structure which was not a model of Boolean algebra. You could still say “Well, sure, but there’s some other thing around which is a model of Boolean algebra; the equation 0 * 1 = 1 has a truth value, as does the equation 1 * 0 = 0, and those truth values can be conjoined, and that conjunction is commutative”, but that’s not what I’m talking about. That’s some other model of Boolean algebra (plus, that involves the presumption that these equations have been externally assigned truth values that live in a Boolean algebra, which is not in fact necessarily the case…); it’s still the case that my original interpretation was an example of something which is not a model of Boolean algebra.
It’s easy to fail to be a model of Boolean algebra; it’s as easy as failing to be a model of any other theory.
Absolutely, distributivity cannot be violated… by an interpretation in which everything is assigned either a value 0 or 1, and & means the function such that 0 & 0 = 0, 0 & 1 = 0, etc., and so on, and all the rest of it.
All that means is that distributivity is not violated in the standard 2 element Boolean algebra. The standard 2 element Boolean algebra is, in fact, a Boolean algebra. No one denies that.
Now, you might say “The only thing ‘&’ means is & in the standard 2 element Boolean algebra. Every other use of the language of ‘&’ is just a kind of misguided homophone; people should really be saying something else instead.” And you are free to take that perspective. But what’s wrong with using such homophones, so long as one understands what’s the same and what’s different?
We use the same symbol + and the same language of addition to speak of many very different things; we use the same language because they are related by a web of family resemblances, but yet, we also understand that they aren’t all the same in all respects. Addition of natural numbers acts very differently from addition of integers, which acts very differently from addition of bits modulo 2, which acts very differently from addition as concatenation of English strings, which acts very differently from myriad other uses of the language of +.
Is it an inviolable theorem that addition is commutative? Yes, it is inviolable… for those interpretations of addition on which it is required to be commutative. But it is also perfectly plausible and natural to speak of “cat” + “dog” = “catdog”. Is it in an inviolable theorem that x + y = 0 entails both x = 0 and y = 0? This is inviolable for basic counting: if two bags together are empty, then they are each individually empty. But this property of natural number addition does not generalize to integer addition. And so on and so on…
Is there any property that is guaranteed to hold for any use of “+”, without any further rules being assumed? No; the only way a property can be guaranteed to hold is by some rule or combination of rules doing the guaranteeing. There’s no mathematical force of guarantee beyond the force of the rules we agree to follow (and the rules we agree upon might change depending on the context and what seems useful to us to discuss). And this is just as true for the so-called “logical operations” as it is for any other operation of mathematics.
So many typos… so little editing time.
None of this means I think “quantum logic” is necessarily the best way to think about quantum mechanics. I don’t have a dog in that fight (at least, not right now). This is all more targeted to the idea that the rules of Boolean logic are inviolable in a way which the rules of other mathematical theories (e.g., Euclidean geometry) are not. I’m happy to say either that all are inviolable or that all are violable (whichever one I say, I’ll mean the same thing; I just want to emphasize that there’s no fundamental difference between the one rule-system and the other. Of course, one often defines the Euclidean geometry rule-system as implicitly including all the rules from some ancillary rule-system for “logic”. It’s only the fact that we are led to use some systems of rules much more often in this ancillary fashion than others that leads us to supposing a spurious distinction).
I prefer quantum logic to frame discussions about quantum systems simply because it is more useful; also, if one were to force classical logic upon quantum mechanics, then one would effectively just do away with quantum mechanics, and reduce it to classical probability theory. Which one of course may well do, and some even want to (see hidden variable theories), but I don’t think is a very natural way to view things – in principle, you could also do differential geometry using nothing but number theory (by means of a Gödel numbering), but really, why would you want to do that?
And just to make it clear once more, I’m not claiming that quantum mechanics means that propositional logic is wrong – this doesn’t follow anymore than it follows from the fact that one body of water added to another body of water gives one body of water that 1 + 1 = 2 is wrong.
But it’s a simple fact that different descriptions are more appropriate to different circumstances; otherwise, it would make no sense to talk about different mathematical fields, different physical theories, or different universal Turing machines (which end up all computing the same things) at all. So to the extend there’s different mathematical formalisms, it is useful to talk about different parts of reality allowing different descriptions.
Knowledge that the particle is not in some location certainly constitutes information about the particle’s location. All information about a particle’s location can be framed in this form: if it’s not anywhere else, it’s gotta be here – if in my rephrased example it is not anywhere in the plane except for the interval ∆x (where the ∆ just is shorthand for ‘interval’), then it is within ∆x – and indeed, that’s where experiment will find it.
Yes, basically. Consider the equivalent formulation:
p: The particle is moving to the right
q: The particle is in [-1, 1]
r: The particle is in [-∞, -1) U (1, ∞]
Where U is the join of both intervals. This is the same as Putnam’s version, just written differently. By your own reasoning – that definite momentum means indefinite position and that hence, the particle can’t be in any interval – both q and r are false taken in conjunction with p, and we observe a lack of distributivity.
As for the ‘truth table’ argument, truth tables form a decision procedure only for Boolean logic (and some sufficiently similar things), so you’re kinda putting in what you want to get out – the problem is, whatever r or q mean on their own, p ^ q and p ^ r are unambiguously false, so it’s not right to assign truth values to either, and then check things through in the standard way. The reason for this is that p and r (or q) refer to incompatible observables – i.e. observables that don’t necessarily have a definite value on their own, but rather, whose values depend on the measurement of the other observable. This is precisely where quantum and classical mechanics differ.
It’s just an analogy to how propositional logic can be regarded as talking about sets: a property of some system A can be phrased in the form ‘A is in B’, where B is some appropriate set – say, the set of all rectangular things, so the property we’re talking about is that ‘A is rectangular’. These properties can be combined by using the set-theoretic operations of meet and join – the meet of two sets is the set of all elements that are in both sets, while their join is the set of all elements that are in either set. These sets form a structure ordered by inclusion (i.e. some set A > B if B is included in A), and that structure is a Boolean lattice.
Conversely, properties of quantum systems are given by projectors, operators in Hilbert space (the space of states of a quantum system – its detailed properties are not important here), that yield 1 if the system has the property, 0 if it doesn’t, similar to how the membership function of a set yields 1 if an element is in the set, 0 if it isn’t. Projectors are called that because they project to a subspace of the system’s Hilbert space: if the projection yields 1, the system is in the associated subspace. The lattice of subspaces thus is for a quantum system the equivalent of the lattice of sets, and that lattice happens not to be Boolean, but rather just what’s called an orthocomplemented lattice, basically Boolean sans distributivity.
Basically, yes. What we mean when we say ‘the particle is in [-1, 1]’ is that an experiment will find it there; but a similar proposition is not true of either [-1, 0] or [0, 1].
I’m not saying they do, but if ‘the particle is in [-1, 1]’ being true does not entail that an experiment will find it within [-1, 1] – and conversely, if ‘the particle is in [-1, 1]’ not being true does not entail that experiment won’t find it within [-1, 1] --, then I’m not sure what the truth or falsity of the proposition means, operationally.
I don’t think the two are really in the same category, so I don’t think the same notion of ‘undefinedness’ can be applied. For instance, you can easily find extensions of number theory where 1/0 is perfectly well defined, pretty much without anything else going haywire; while if you wish to insist that the position of a particle is well defined, then you are going to have to wrestle with severely limitative theorems (the no-go theorems of Bell and Kochen-Specker) that constrain the allowed theories to things most people don’t find very natural.
I should perhaps elaborate on that, as I think this gets to the heart of the problem and I did a poor job explaining it. The way quantum mechanics works, it is unambiguously false to say that a particle has a definite momentum and a definite position – I think we’re agreed on that. So the problem is that position and momentum are not independent in their truth values, meaning that you can’t assign a truth value to a proposition like ‘the particle is in [-1, 1]’ on its own, disregarding its momentum state – doing so leads to contradictions, in particular violations of uncertainty. So if you’re doing truth tables, you can’t just take proposition r and assign it a value of either true or false – it is only in conjunction with propositions regarding the particle’s momentum that a definite truth value can be assigned. This is why distributivity fails: (p ^ r) and (p ^ q) are unambiguously false, as they are in violation of uncertainty; but p ^ (r v q) is true (if p is), as it just amounts to saying ‘the particle’s momentum and position relate to each other in a way allowed by the uncertainty principle’.
Whenever you’re talking about the truth of q or r on their own, you’re essentially throwing out what’s quantum about quantum mechanics, which is precisely the idea of complementary observables.
On the one hand, you (HMHW) are saying you don’t mean to argue that propositional logic can in some sense be falsified.
But on the other hand, you’re offering a series of what you claim to be inconsistent triads of propositions.
I don’t think you can do both of those things.
Your latest triad is this:
p=The particle is moving to the right
q=The particle is in [1, -1]
r=The particle is in [-inf, -1] U [1, inf]
You said q and r are both false, so distributivity fails.
But if q and r are both false, then p & (q v r) is false, and so distributivity doesn’t fail.
But if the particle’s location is undefined, then it has no location. (Do you agree?) And if it has no location, then it’s not true that “if it’s not anywhere else, it’s gotta be here.”
That’s fine (but what is meant by “p ^ q”?) but none of this means you’ll ever find an inconsistent triad. Yet you keep saying various pqr triads are inconsistent on account of the fact that they don’t satisfy distributivity.
Got it. I learned those as “intersection” and “union.”
Got it. And as far as I can tell, the “or” (and possibly the “and”) of quantum logic simply doesn’t mean the same thing as the “or” and the “and” of boolean logic. It’s a different operation which we give the same name for some reason. What’s the reason? What makes the “or” and “and” of quantum logic "or"ish and "and"ish? (If I know this, I may have a better sense as to why people are tempted to claim that quantum phenomena don’t just model a different logic, but actually “violate” boolean logic.)
Indistinguishable, sorry for the anti-climax here but for the most part I simply agree with you, and also I simply think I was in a muddle about the distinction I was trying to draw between Euclidean geometry and Propositional logic.
Right: For a long time, people thought that by doing physics, they were doing something that extends Euclidean geometry. They discovered, however, that this wasn’t true–the geometry they were doing (when doing physics) was (or anyway, had to become) non-Euclidean.
If you interpret “point,” “line,” etc in the traditional ways, then the actual world turns out not to be a model for Euclidean geometry. So people might say the world “violates” Euclidean geometry in that sense.
But, if you interpret A, B, C, &, v, –>, etc in the traditional ways, then the actual world doesn’t turn out not to be a model for Propositional Logic. Propositional Logic works just fine on the traditional interpretation. And I’m saying that the actual world can’t turn out not to be a model for Propositional logic. I guess my argument’s not a very good one, though–because it basically boils down to a faith-y assertion that we’ll never find a set of observations which require us to assign both True and False to any single proposition. It’s not exactly “faith” though–it’s just a strong sense that “propositions don’t work that way.” If something is supposed to take on more than one value, it’s simply not a proposition. In other words, we’re not dealing with true and false anymore.
I suppose I’d finally admit that the world has violated propositional logic (in the sense modeled above) if it turned out that there are not truths and falsehoods. But how the heck could anyone believe that? By believing it, they’re already assigning the value “true” to it… etc etc.
We can treat the possible states of affairs expressed by those two phrases as equivalent (just as, when I say A = “My cat is fat” I’m saying A represents any state of affairs in which my cat is fat–I’m treating all those disparate states as equivalent). But since they (the two phrases) don’t have the same truth conditions, this equivalence isn’t truth-equivalence. It’s some other kind of equivalence. Which is great, and could be the best way to do the science. But it’s not a “violation” of propositional logic. (BUT I can see thinking that QM violates propositional logic if it turns out there is no place whatsoever for truth-equivalence as a basis for delineation of “propositions.”)
Not to ignore anything else in your post, but in responding I found myself mostly repeating myself so I’ll leave it at that for now.