= Solution
<Bounded completeness of an arithmetic theory> means that every bounded <arithmetic> <first-order sentence> true in the standard <natural numbers> is provable in $S$. If $S$ is <first-order consistent> and proves a false <bounded arithmetic sentence> $B$, <bounded completeness of an arithmetic theory> also proves its true bounded <negation>, a <contradiction>. Conversely, if every bounded theorem is true, $S$ cannot be inconsistent, because inconsistency would prove the false <bounded arithmetic sentence> $0=1$. Thus
$$
\boxed{\operatorname{Con}(S)\iff\Sigma_0\text{-soundness}.}
$$
If $S$ is <one-consistent> and proves $\exists xB(x)$ with bounded matrix, that <existential sentence> cannot be false: otherwise every $\neg B(\bar n)$ would be a true <bounded arithmetic sentence> and hence provable, contrary to <one-consistency>. Thus $S$ has <Sigma-1 soundness>.
Conversely, <Sigma-1 soundness> excludes inconsistency. If $S\vdash\exists xB(x)$, it supplies a standard witness $n$ with $B(n)$ true. <Bounded completeness of an arithmetic theory> proves $B(\bar n)$, so <first-order consistency> forbids a <formal proof> of $\neg B(\bar n)$. Hence there cannot simultaneously be refutations of all standard instances. Therefore
$$
\boxed{1\text{-consistency}\iff\Sigma_1\text{-soundness}.}
$$
Finite blocks of existential <logical variables> are treated as finite <tuples> of witnesses, with the same argument.
Back to article page
