# Given a φ independent of PA which is true in the standard model, will always (PA+ ¬φ) be Σn-unsound for some n?

Given a φ independent of PA which is true in the standard model, will always (PA+ ¬φ) be Σn-unsound for some n? This is a follow up from a previous question: Given a φ independent of PA which is true in the standard model, will always (PA+ ¬φ) be ω-inconsistent?.

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Isn't this tautological? $\lnot\varphi$ is false, and provable in PA$+\lnot\varphi$. But $\lnot\varphi$ is (equivalent to) a $\Sigma_n$ formula for some $n$. – Andrés E. Caicedo Nov 3 '12 at 2:05
I see your point, but what if ¬φ is equivalent to a PIn formula? (I might be asking something obvious, I apologize if that is the case, I am just a physicist) – Wolphram jonny Nov 3 '12 at 3:34
A $\Pi^0_n$ formula is also $\Sigma^0_{k}$ for every $k > n$ (and $\Pi^0_k$ and $\Delta^0_k$). So every formula is $\Sigma^0_n$ for infinitely many $n$. – Carl Mummert Nov 3 '12 at 12:08

Yes. Every formula "is" $\Sigma^0_n$ for sufficiently large $n$, and the rest follows directly from the definition of a theory being $\Sigma^0_n$ unsound.
There are two ways to read the above statement. Some people would define the arithmetical hierarchy so that each formula has infinitely many rankings. But even if we only give a formula its lowest ranking, every formula $\phi$ will be logically equivalent to a formula $\psi$ that is $\Sigma^0_n$ for some $n$; just add dummy quantifiers to the prenex form of $\phi$ until it starts with an existential quantifier. Then any theory that proves $\phi$ also proves $\psi$. This is why, for most purposes, we can look at the arithmetical hierarchy "modulo logical equivalence" or even modulo provable equivalence in our theory.