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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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up vote 7 down vote accepted

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 $T$.

The more technical way to say this is that the Lindenbaum algebra of an essentially incomplete theory is always atomless. Any theory which interprets Peano arithmetic (or even interprets some weaker arithmetical theory) is essentially incomplete.

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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
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