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Does the following situation occur?

$T$ is strong enough to interpret PA. There is a sentence $\sigma$ such that $Con(T+\sigma)$ is equivalent to $Con(T+\neg \sigma)$ over $T$, and $Con(T)$ does not imply $Con(T+\sigma)$.

Heuristically, making a decision about $\sigma$ either way requires a jump in strength.

Follow-up question. Can we have $T$ and $\sigma$ such that $Con(T) < Con(T+\neg \sigma) < Con(T+\sigma)$? (The less-than means that the right side implies the left but not vice versa.)

mbsq
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    If each one of $T + \sigma$, $T + \neg \sigma$ has higher consistency strength than $T$, then neither one of $Con(T + \sigma)$ and $Con(T + \neg \sigma)$ can imply the other. The reason is same as before. – hot_queen Nov 01 '15 at 20:54

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So long as $T$ is reasonable, $T$ will prove $$Con(T)\implies\forall\sigma [Con(T+\sigma)\vee Con(T+\neg\sigma)].$$ So if $T$ proves $Con(T+\sigma)\iff Con(T+\neg\sigma)$, then $T$ proves $Con(T)\implies Con(T+\sigma)$ and $Con(T)\implies Con(T+\neg\sigma)$. So the answer is no, that can't happen.

Noah Schweber
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