# System of Axioms that violate Godel's Incompleteness Theorem

As I understand it, Godel's incompleteness theorem will only apply to any system for which the set of all statements in the system can be mapped to a set unique elements in that system (The Godel numbers of the statements).

If this understanding is correct, would it not therefore be possible to design a system such that the cardinality of the set of all possible statements within a system was greater than the cardinality of the set of all possible elements in a system which would then not be bound by the incompleteness theorem? (Perhaps through some infinite set of operators or if not that then through some other means)

• Why the downvotes? If there is something wrong with the question please at least leave a comment so I can understand what needs to change. – DaggerOfMesogrecia Apr 10 '17 at 16:38
• I don't know why the downvotes are there; but simply put : 1- there are many "natural" (i.e. not constructed for this purpose) systems that are not subject to Gödel's theorem's hypotheses (for instance Presburger's arithmetic, Tarski's geometry, etc.); 2- you cannot bind the cardinality of the "system" if you want there to be an infinite "system" : by the compactness theorem, if a first order theory has an infinite model, then it has models of all infinite cardinalities. – Max Apr 10 '17 at 16:56
• Gödel's Theorem applies to any system that allows one to algorithmically verify if a certain statement is an axiom, to algorithmically verify whether a statement follows by a rule of inference from some given statements, and that is powerful enough to express basic arithmetic of the natural numbers ... – Hagen von Eitzen Apr 10 '17 at 16:59
• Unrelated to the question, but @Max, that's not entirely true. By including lots of constants in your language (say $\omega_\alpha$ many) and saying in your theory that they are pairwise different, you can make sure that all models have cardinality at least $\omega_\alpha$. But you're right of course when one restricts attention to countable theories. – martin.koeberl Apr 10 '17 at 17:09
• @Mauro ALLEGRANZA I'm not proposing limiting the "amount of arithmetic" I'm proposing increasing the expressiveness of a system such that it's statements can no longer be represented as elements within the system. – DaggerOfMesogrecia Apr 10 '17 at 17:24

The most obvious is to take the language of arithmetic $L=\{+,\times, 0, 1, <\}$ and add a whole ton of junk to it. For example, we could add uncountably many constant symbols $c_i$ ($i\in I$, $I$ appropriately large). The resulting language $L'$ would be bigger than the standard model $\mathbb{N}$ of arithmetic, so there would be no way of representing each $L'$-sentence by an element of $\mathbb{N}$ as in Goedel numbering. And there are lots of similar tricks we can play.
• The theories we get from this language-expansion may still be incomplete, for essentially the same reason as PA! For example, let $L'$ be as above, and consider $PA$ as a theory in this language (so we add no new axioms governing the $c_i$s). Then $Con(PA)$ is still unprovable in this setting. Moreover, even if we add some axioms to $PA$, as long as we do so in a "nice" way we'll still hit up against incompleteness. E.g. let $PA'$ be $PA$ together with "$c_i\not=c_j$" for all $i\not=j$ (note that any model of $PA'$ must be uncountable!). Then every finite fragment of $PA'$ is interpretable in $PA$, and in fact $PA$ proves the consistency of each finite fragment of $PA'$ appropriately construed; so no finite fragment of $PA'$ proves $Con(PA)$ by Goedel. But this in turn means that $PA'$ doesn't prove $Con(PA)$, by Goedel's completeness theorem. So at the end of the day, you still have to add some interesting axioms after expanding your language, and you have to do so in a sufficiently complicated way as to get around Goedel; so really this approach, although it does appear to block Goedel coding, isn't the way to go about things.