> On 3 Dec 2018, at 16:02, agrayson2...@gmail.com wrote: > > > > On Monday, December 3, 2018 at 2:42:26 PM UTC, Bruno Marchal wrote: > >> On 2 Dec 2018, at 15:00, agrays...@gmail.com <javascript:> wrote: >> >> >> >> On Sunday, December 2, 2018 at 12:11:50 PM UTC, Bruno Marchal wrote: >> >>> On 30 Nov 2018, at 12:13, agrays...@gmail.com <> wrote: >>> >>> >>> >>> On Friday, November 30, 2018 at 12:34:13 AM UTC, Brent wrote: >>> What can be inferred always depends on what you take as premises. If you >>> start from the Hilbert space formulation of QM or an equivalent formulation >>> and you premise that there is a probability interpretation of a state, >>> then Gleason's theorem tells you that the Born rule provides the unique >>> probability values. >>> >>> Brent >>> >>> So to get Born's Rule, Bruno would have to assume a huge amount IN ADDITION >>> TO ARITHMETIC. I don't buy it. AG >> >> On the contrary, mechanism assumes less than any other theory. And Mechanism >> is roughly the idea that the brain does not invoke magical things. >> >> The theory of everything, with mechanism assumed at the metalevel, assume >> only S K, S≠K, and the axioms >> >> 1) If x = y and x = z, then y = z >> 2) If x = y then xz = yz >> 3) If x = y then zx = zy >> 4) Kxy = x >> 5) Sxyz = xz(yz) >> >> I doubt that you will find an easier theory. >> (Exercice: prove that x = x) >> >> Bruno >> >> But you haven't replied to my objection. In addition to logic and the axioms >> of arithmetic, you must ALSO assume such a thing as probability exists to >> even approach QM. What you have above won't cut it, IMO. AG > > > I do not assume any probabilities in the ontology. I justify them through the > phenomenology. Here I was just making clear that I assume only the 5 rules > and axioms above. There is three inference rules: > > 1) If x = y and x = z, then y = z > 2) If x = y then xz = yz > 3) If x = y then zx = zy > > And two axioms: > > 4) Kxy = x > 5) Sxyz = xz(yz) > > Are the variables restricted to natural numbers, that is, the positive > integers?
A combinatory algebra is any set with some law or operation verifying the axioms above. That can be birds (Smullyan) or whatever you want, as long as the formula above are satisfied. And they can be numbers indeed. I will probably prove, soon, that if you take the set N (the natural numbers), and fix some universal machinery phi_i (that is an enumeration of the partial computable functions), then by defining (i j) by phi_i(j) we make N into a combinatory algebra. So any universal system/machine/number automatically endow N with a structure of (partial) combinatory algebra. There are other more sophisticated models though. phi_i(j) denotes the result of applying the ith program (of the fixed enumeration of all programs in the fixed universal machinery) on the number j. Kxy abbreviates ((K x) y) where (K x) denotes what is noted usually K(x), that is K applied on x. > What are these axioms, explicitly? They are axioms defining the combinatory algebra, which captures the Turing Universal system. Like the axioms of group theory defines the group. Not sure of the meaning of your question, as it is hard to be more explicit than by giving axioms. > And No, I don't believe there's enough here to infer de Broglie matter waves, > or the quantum interference pattern for, say, the double slit. AG Being skeptical is good … as long as your skepticism does not prevent you to verify the proof and perhaps to justify your disbelief with some specific argument. If not, I will think that your disbelief is based on some personal opinion (like believing in some ontological matter that we would have to assume). In that case we are no more doing science. And, yes, the axiom above capture basically the whole of computer science, although we can add many axioms, to get more particular theories (Turing complete or not). Of course, by incompleteness, to get the whole theoretical computer truth, we would need a (non recursively enumerable) infinity of axioms. Such truth are beyond *all* theories. Note that I could use the numbers instead of combintaors; using Robinson Arithmetic instead: 0 ≠ s(x) (0 is not the successor of a number) s(x) = s(y) -> x = y (different numbers have different successors) x = 0 v Ey(x = s(y)) (except for 0, all numbers have a predecessor) x+0 = x (if you add zero to a number, you get that number) x+s(y) = s(x+y) (if you add a number x to the successor of a number y, you get the successor of x added to y) x*0=0 (if you multiply a number by 0, you get 0) x*s(y)=(x*y)+x (if you multiply a number x by the successor of y, you get the number x added to the multiplication of the number x with y) OK? That theory, amazingly, is also Turing complete, despite being very weak (it cannot prove that 0 + x = x, for example). Bruno > > > Nothing else is assumed, except mechanism and as much as needed mathematics > at the meta level, like in all theories. > > I don’t expect you to believe this immediately. I just present the result, > hoping you will study the proof. > > Bruno > > > > >> >> >> >> >> >>> >>> On 11/29/2018 10:23 AM, agrays...@gmail.com <> wrote: >>>> Regardless of rules of arithmetic and mathematical logic, I simply don't >>>> believe that something like Born's Rule can be inferred without actually >>>> observing a quantum interference pattern. AG >>> >>> >>> -- >>> You received this message because you are subscribed to the Google Groups >>> "Everything List" group. >>> To unsubscribe from this group and stop receiving emails from it, send an >>> email to everything-li...@googlegroups.com <>. >>> To post to this group, send email to everyth...@googlegroups.com <>. >>> Visit this group at https://groups.google.com/group/everything-list >>> <https://groups.google.com/group/everything-list>. >>> For more options, visit https://groups.google.com/d/optout >>> <https://groups.google.com/d/optout>. >> >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Everything List" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to everything-li...@googlegroups.com <javascript:>. >> To post to this group, send email to everyth...@googlegroups.com >> <javascript:>. >> Visit this group at https://groups.google.com/group/everything-list >> <https://groups.google.com/group/everything-list>. >> For more options, visit https://groups.google.com/d/optout >> <https://groups.google.com/d/optout>. > > > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to everything-list+unsubscr...@googlegroups.com > <mailto:everything-list+unsubscr...@googlegroups.com>. > To post to this group, send email to everything-list@googlegroups.com > <mailto:everything-list@googlegroups.com>. > Visit this group at https://groups.google.com/group/everything-list > <https://groups.google.com/group/everything-list>. > For more options, visit https://groups.google.com/d/optout > <https://groups.google.com/d/optout>. -- You received this message because you are subscribed to the Google Groups "Everything List" group. 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