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This question already has an answer here:

(current answers neglect the fact that the set of all concepts( $C_{U}$) is a subset of U as all of them are physically encoded( symbolically represented by the physical events themselves(brains, computers et al)) ( note $\oplus$ is the mutually exclusive or, A $\vdash$ S means S can be derived(proven) as a theorem of axiom system A) Given U, $(\exists u \leftrightarrow u \in U)$ (also suppose that U is the physical universe/multiverse, etc. [which it trivially is , as all concepts are physical phenomena {neurotransmitters, computers et al.}] this is obviously assuming platonism is wrong, or can still be classified as physical, i.e. another universe in the multiverse (trivially)) Also given A $\subseteq$ U s.t A is a set of initial conditions and some equations describing the progression of physical events s.t. these equations are the constraints on the strings of the formal language associated with the axiom system (this trivially exists as given the ordering/index over the axioms, st.t obviously one may write a code/equation that transforms the string of Ax.n. to Ax.n+1. then given the initial conditions have the index associated with them one may use the code/equation inductively to reach any Ax.m. or one can prove its equivalence to the existence cause and effect. Then ( $(A \vdash S) \leftrightarrow (\exists S \leftrightarrow S \in U )) \leftrightarrow ( (\neg(A \vdash S)) \leftrightarrow (\neg(\exists S))) $, that is there does not exist a single statement that can not be derived from A, i.e. it is an axiomatization of everything( the physicsl universe/ answer to hilberts sixth problem). However by Godels incompletness theorem ( as obviously it can prove basic arithmetic truths ) $( ((\exists \phi (A \vdash \phi \wedge A \vdash \neg\phi)) \leftrightarrow ( \exists \phi( \exists \phi \wedge \neg\exists \phi )) \leftrightarrow (\exists \phi(( \phi \in U )\wedge (\phi \notin U)))) \vee((\exists \phi( \phi \notin U)) \leftrightarrow (\exists \phi(( \phi \in U )\wedge (\phi \notin U))))) $ .

That is the set is both incomplete and inconsistent, this is a contradiction as U by definition is both complete and consistent so trivially either our ideas of mathematical formalism have no bijective relation to complete models of physical reality and there exists a negative solution to Hilbert's sixth problem or existence and non existence are equivalent. This would trivially also disprove the existence of a "theory of everything" and a single unifying equation as one could use the equation as the constraints on the strings in the language of the axiom system, of which the equation is the only axiom, and (algebraic/differential et al.) manipulations of the equation are theorems.

Is there a way resolve this paradox?

This paper fully presents the paradox. https://www.academia.edu/11102734/On_undecidable_physical_statements_in_current_mathematical_formalism (and also illustrates its importance and current undecidable nature showing it is different from previously answered questions.)

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marked as duplicate by user10851, John Rennie, Mark Mitchison, Martin, ACuriousMind Feb 3 '15 at 12:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ First, this could use better formatting. Second, your question sounds like this: "There is a theorem that says that every X that is Y has problem Z. Suppose we have an X that is Y. Is it true that X has the problem Z?" $\endgroup$ – ACuriousMind Feb 2 '15 at 21:05
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    $\begingroup$ So you are stating it is a trivial tautology? $\endgroup$ – kb jnknlknlkn Feb 2 '15 at 21:29
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    $\begingroup$ Yes, I believe it is a tautology to say that an axiom system solving Hilbert's problem would be incomplete. It is a totally undecidable question whether the statements that are true but cannot be proven should bother us, though, since Gödels proof constructs a very silly example of inconsistency (roughly of the type "I am false"). $\endgroup$ – ACuriousMind Feb 2 '15 at 21:39
  • $\begingroup$ Related: physics.stackexchange.com/q/14939/2451 , physics.stackexchange.com/q/87239/2451 , and links therein. $\endgroup$ – Qmechanic Feb 2 '15 at 21:46
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    $\begingroup$ The Godel numbering in the formal proof is not particularly silly nor trivial merely because it is recursive, also the statements are not necessarily true nor false and could feasibly be some unknown or incomprehensible value or property. $\endgroup$ – kb jnknlknlkn Feb 2 '15 at 21:57
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The existence of undecidable statements does not mean that we cannot hypothetically manage, adding suitable consistent axioms, to determine the truth of all physically relevant statements. Suppose you have a theory $L$ with a meaningful undecidable statement $A$, such that both $L+A$ and $L+\bar{A}$ (I use bar for negation) are consistent. Then you can choose either $L+A$ or $L+\bar{A}$ as your new theory (depending on what is more relevant, physically).

The Gödel undecidable statement for example does not seem very relevant (in my opinion) for a physical theory, and also the fact that the consistency of the theory is not provable within it (since you can, in principle, use forcing/impredicativity to prove consistency and find truth complexities of the logical system, even if it would mean to consider mathematical entities that are not "physical").

As a concrete example: would physics care if the generalized continuum hypothesis is true or false? $GCH$ is undecidable in $ZFC$, and $ZFC$, $ZFC+GCH$, $ZFC+\overline{GCH}$ are consistent theories. I suppose it is not so important which one to choose for physics, but surely you need set theory. The axiom of choice instead is more relevant, since there is a model of $ZF+DC$ (dependable choice) where all sets of reals are measurable (but on the other hand you have hard times defining distributions), and that may be more relevant physically (maybe...I don't know).

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  • $\begingroup$ The incompleteness theorem still applies to (L + A $\oplus L + \neg A )$ so it doesn't matter how many undecidable statements you add as axioms (provided you somehow know their truth value). As to the "arbitrary" relation between (relevance of) any theorem to physics in general (not to get to much in to philosophy) if it isn't relevant then obviously the axiom system does not abide by the constraints defined upon it (somehow physical reality and its models transcend logic) then it is no longer an axiom system thus it is no longer an "axiomatization of physics" $\endgroup$ – kb jnknlknlkn Feb 3 '15 at 1:41
  • $\begingroup$ Also as any arbitrary equation can be transformed into (reduced to) an axiom system then if we assume a theory of everything exists and it is in the form of some arbitrary single equation we take as an axiom (or multiple, it doesn't matter) then as reality is not bound by the laws equivalent to this axiom(or any axiom system) then it is thus not a theory of everything, nor even a model of reality. $\endgroup$ – kb jnknlknlkn Feb 3 '15 at 1:41
  • $\begingroup$ first of all, it is not true that an equation reduces to an axiom system; a physical theory (or model of reality) is formulated within a given logical model, if else you cannot do inferences and therefore improve your knowledge. what I am saying is that, even there will always be undecidable statements, you can in principle impose on your system the truth of the physically relevant ones and check the consistency of the theory by means of the logical tools as forcing. so, in my opinion, you are not prevented by undecidability to get a complete axiomatization of physics $\endgroup$ – yuggib Feb 3 '15 at 7:37
  • $\begingroup$ and a theory of everything would anyways be a model, i.e. a logical/mathematical system (with a huge uncountable truth complexity) such that every physically relevant statement is provable. obviously we are very far from that... $\endgroup$ – yuggib Feb 3 '15 at 7:39
  • $\begingroup$ If a physical theory is formulated within a logical model then those axioms and rules of inference from which the equations and physical models would be derived would be an axiomatization of physics. Furthermore (assuming there is a single unifying equation )a single equation must be assumed to be "true" (an axiom) and various algebraic or mathematical manipulations would be theorems or lemmas and those mathematical rules used to manipulate the equation and derive those theorems (occurrences of variables certain derivatives, inequalities etc.) $\endgroup$ – kb jnknlknlkn Feb 3 '15 at 23:22
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I think it is interesting to pose this question in the setting of a very simple hypothetical universe. Suppose that the universe consisted, for example, exclusively of perfectly hard spheres that move according to classical physics and collide with each other elastically. (Really, the specific rules of our toy universe don't matter for what I will say, as long as they are simple and unambiguous).

As far as I can tell, nothing prevents us from modeling this toy universe as rigorously as we wish. But would our model be a consistent, complete theory? And if so, what of Gödel's theorem?

Gödel's theorem applies to formal systems, in which strings of symbols may represent statements, proofs, and so on. The toy universe is obviously not in itself a formal system. Nor does knowing the mathematical rules of the toy universe immediately tell us what the formal system is. So what is our formal system, and more importantly, what will be the semantics by which we map its strings onto facts about the toy universe? And under what conditions will be consider the system to be a "complete" description of the toy universe? We have to agree on precise answers to those questions. Otherwise, I believe the answer to the main question can only be "it depends what you mean."

If we lived in that toy universe, would we be justified in saying that we had a "theory of everything"?

It is true that part of the mathematics used in the theory would be arithmetic, and therefore there would be undecidable statements about arithmetic. No surprise there. As yuggib says, those are not physically relevant. We don't care that our "toy universe theory of everything" can't tell us whether some Diophantine equation has solutions or not, if it doesn't bear on the physics of the toy universe. What if it did? What if we concocted some horribly complicated "computer" out of hard spheres, in such a way that its behavior mapped onto the Diophantine equation and could not be predicted? Would that constitute "incompleteness" of the model?

Well, I leave that up to you to decide. Personally, I consider a mathematical model of a physical system "complete" if it allows us to do computations to any desired precision in principle. So if we lived in a hard-sphere universe, I would say we possessed a "theory of everything." And furthermore, that any a priori argument precluding a "theory of everything" for the real universe should explain in what essential way it differs from the toy one.

Of course, one is free to choose a different definition of "completeness" and say that a model will only be complete if the detailed outcomes of specific configurations/systems/situations are provable. In that case, there will never be a "theory of everything" for even a universe as simple as a Turing machine!

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  • $\begingroup$ I mean complete in the sense of mathematical logic, i.e. the maximal consistent set of sentences that model physical reality. For instance in your toy universe one would take the basic equations of classical physics as axioms and manipulations according to the rules of inference (take the rules of inference as those of algebra (associativity etc) those of mathematical analysis and all other relevant mathematical constructs) as theorems and lemmas etc. As the theory is complete those manipulations would be representative of physical phenomena and would be true in the universe. $\endgroup$ – kb jnknlknlkn Feb 3 '15 at 23:50
  • $\begingroup$ I suppose then the question is whether it is worth attempting to formalize the negation of the existence of a "theory of everything" using these facts in a paper or will it merely be a useless tautology. $\endgroup$ – kb jnknlknlkn Feb 4 '15 at 1:34
  • $\begingroup$ Well, the computer would probably then just take forever to calculate solutions to the equation, and the statement of Gödel's theorem, applied to that computer, would just become a statement of the "Does it finish?" problem. $\endgroup$ – Jerry Schirmer Feb 8 '15 at 21:00
  • $\begingroup$ @ Jerry Schirmer The undecidable statements arising from the halting problem applied to the algorithm equivalent to the rules of implication and the axiom system for U={x|∃x} with input equivalent to a theorem to prove using those rules of implication applied to those axioms(that is the algorithm)is equivalent to the undecidable statements godels incompleteness theorem proves are undecidable (assuming consistency ) and the question also proves they result in the same paradox (however trivially regardless of the reducibility of one to the other (equivalence) one can prove also if... $\endgroup$ – kb jnknlknlkn Feb 11 '15 at 22:53
  • $\begingroup$ ...($A_{i}$) is equivalent to saying that the algorithm A halts given input i( trivially this means the halting problem is decidable for a given input i) and A is the algorithm constructed of the axiom system for U with the rules of implication and i is a theorem it is to prove using the rules of implication then obviously $(\exists i(\neg A_{i})) \leftrightarrow (\exists i( i \notin U )) \leftrightarrow (\exists i( \neg\exists i))$ which is the same paradox as in the question $\endgroup$ – kb jnknlknlkn Feb 11 '15 at 22:54

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