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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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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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    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.2011-10-02