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Let $T$ be a theory, let $L$ be its language, let $A$ be its set of axioms and let $P_0 \in L$ be a property.

$P_0$ could be :

  • Consequence of $A$
  • The negation of a consequence of $A$
  • Independent of $A$

Assume that the property $P_1$ : "$P_0$ is independent of $A$" is also in $L$. Can $P_1$ be itself independent of $A$ ?

For instance, assume $P$ versus $NP$ is independent of the axioms (let's say ZFC), could it be the case that this independence is itself independent of the axioms ? This could explain why nobody already solved this problem.

More generally, assume that $P_{i+1}$ : "$P_i$ is independent of $A$" is also in $L$. Does it exist a property $P_0$ such that $\forall i.\ P_i$ is independent of $A$ ?

If $A$ is empty, any property is independent of the axioms, in particular all the $P_i$ are independent of the axioms whatever is $P_0$.

If $A$ is the axioms of Presburger arithmetic, none of the property is independent of the axioms, but I am not sure that we can express "$P_i$ independent of $A$" in the language of the theory.

What happens when A is ZFC, Peano axioms, Euclide axioms, or any other usual theory ?

Asaf Karagila
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François
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  • But, if $P_0$ is independent of $A$, then $P_1$ is true. – ajotatxe Aug 18 '15 at 08:30
  • @ajotatxe $P_1$ is true, but does a proof of $P_1$ can be derived from the axioms ? – François Aug 18 '15 at 08:32
  • If $A$ is empty, then every property is independent of $A$. But for every property $P$, the independence of $P$ can be easily shown using the fact that $A$ is empty. – ajotatxe Aug 18 '15 at 08:40
  • @ajotatxe You can't prove anything but you can prove something, this is not consistent. I agree with the first part, but I am not sure you can use the emptiness of A in a proof, it's not an axiom. – François Aug 18 '15 at 08:47

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