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Godel's incompleteness theorem basically says that a set of axioms cannot prove everything. But you can add those unprovable truths to your set of axioms to expand it. Suppose you keep expanding your set of axioms until it contained all truths (which I think isn't possible in finite time). Or suppose somehow, you had a complete set of axioms. Would it necessarily be inconsistent, because if it wasn't, then it would be consistent and compete?

Croolsby
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    Adding the statements as an axiom does not help because other statements must appear that are not provable, unless the theory is inconsistent, in which case, every statement can be proven. Weak enough systems, as the Pressburg-arithmetic, can escape from Goedel's result. In fact, this arithmetic is both consistent and complete. – Peter Jan 27 '20 at 10:26

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Goedel's first incompleteness theorem states that every theory that is strong enough (containing the peano axioms) is either inconsistent or incomplete. If it is is inconsistent, it can prove everything ! If it is consistent, there are necessarily statements that cannoe be proven within this theory. This does not mean that it can prove nothing.

Goedel's second incompleteness theorem states that no system that is strong enough (containing the peano axioms) can prove its own consistency.

Goedel's completeness theorem states that a statement is provable within a theory if and only if it is true under every interpretation within this theory.

Peter
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    A mathematically irrelevant historical detail almost always overlooked in the incompleteness theorems terminology is that Gödel only showed a complete Peano-prover is $\omega$-inconsistent; Rosser extended this to the familiar requirement that the theory be inconsistent. – J.G. Jan 27 '20 at 10:38
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Yes, you could have a system $S$ in which all propositions that are true in $S$ are axioms of $S$. The system $S$ is then, by definition, complete.

If $S$ is powerful enough to represent all the truths of the arithmetic of natural numbers then we know there are an infinite number of such true statements, so $S$ must have an infinite number of axioms.

But then you cannot be sure that $S$ is consistent because there is no effective decision procedure (a procedure that always halts eventually) that can show that $p$ and $\lnot p$ are not both axioms of $S$.

gandalf61
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