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Jun 30, 2011 at 15:40 comment added Gyorgy Sereny Thank you for your answer. As a matter of fact, I take it for granted, that the completeness of a theory is a syntactic property, that is, one formulated in terms of provability rather than in terms of semantic consequence: T is complete just in case $T\vdash\sigma$ or $T\vdash\lnot\sigma$ for all sentences $\sigma$. In this case, the decidability of complete recursively axiomatized theories can be shown without the Completeness Theorem.
Jun 26, 2011 at 22:28 comment added Dave Marker @Gyorgy--I am using "completeness in two senses. We want to look at "complete theory" $T$, i.e., one where $T\models \phi$ or $T\models\neg\phi$ for all sentences $\phi$. But we are also using the Completeness Theorem to argue that for any $T$ the set of logical consequences of $T$ is recursively enumerable in $T$.
Jun 26, 2011 at 17:45 comment added Gyorgy Sereny You write: "Completeness comes into play when proving the decidability of complete recursively axiomatized theories ...". Don't you use here the notion of "completeness" in another sense (namely as a characteristics of a theory as opposed to one that appears in the Completeness Theorem, which characterizes a logic)?
Jun 26, 2011 at 1:39 history edited Dave Marker CC BY-SA 3.0
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Jun 26, 2011 at 0:15 history answered Dave Marker CC BY-SA 3.0