# Independent statements that cannot be weakened

Let $T$ be a theory and let $\phi,\psi$ be statements that are independent of $T$. Say that $\psi$ is a $T$-weakening of $\phi$ if $T$ proves $\phi \Rightarrow \psi$ but cannot prove $\psi \Rightarrow \phi$, and say that $\phi$ is $T$-basic if there is no $T$-weakening of $\phi$.

If $T$ is at least as strong as Peano arithmetic, do $T$-basic sentences always exist? Is $\phi$ $T$-basic when $T$ is ZFC minus infinity and $\phi$ is the axiom of infinity?

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For any theory which is essentially incomplete -- meaning there is no effective complete theory extending it -- there are no basic sentences. Because if $\phi$ is independent, so is $\lnot \phi$, and if $\phi$ has no weakening over $T$ then $\lnot \phi$ has no consistent strengthening over $T$, so $T + \lnot \phi$ is a complete extension of $T$.
Comment, part 1 :Why is it that if $\phi$ has no weakening over $T$ then $\lnot \phi$ has no consistent strengthening over $T$ ? – Ewan Delanoy Oct 2 '11 at 14:03
Comment, part 2 :I suppose this follows from the contraposite, so that if $\lnot\phi$ has a consistent strenthening over $T$, which means that there is a $\psi$ that is independent from $T+\lnot\phi$, then we should be able to construct a $T$-weakening of $\phi$ from $\psi$. – Ewan Delanoy Oct 2 '11 at 14:03
Comment, part 3 : All we know is that $T$ cannot prove $\lnot\phi \Rightarrow \psi$ or $\lnot\phi \Rightarrow \lnot\psi$. By contraposition, we see that $T$ cannot prove $\lnot\psi \Rightarrow \phi$ or $\psi \Rightarrow \phi$. So $\psi$ and $\not\psi$ are candidates for the weakening we are looking for, but we are not done yet : it is unclear (at least to me) if $\phi \Rightarrow \psi$ or $\phi \Rightarrow \lnot\psi$. – Ewan Delanoy Oct 2 '11 at 14:06
A strengthening $\psi$ of $\lnot \phi$ would have the properties that $T \vdash \psi \to \lnot \phi$ and $T \not \vdash \lnot \phi \to \psi$. Take contrapositives of both: $T \vdash \phi \to \lnot \psi$ and $T \not \vdash \lnot \psi \to \phi$. So $\lnot \psi$ is a weakening of $\phi$. (This is really just the fact that, in a Boolean algebra, $a < b$ if and only if $b^c < a^c$.) Also $T \vdash \lnot \psi$ if and only if $T + \psi$ is inconsistent. – Carl Mummert Oct 2 '11 at 16:05