This may be a too obvious question, but then in completeness theorem, direction is only in one direction: if a theory is consistent, then it has a model. Can we make it stronger and say that a theory has a model if and only if a theory is consistent?
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"A consistent theory has a model" is also known as Henkin's Theorem, or The Henkin Theorem. – DanielWainfleet Jul 23 '18 at 02:08
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@DanielWainfleet I've never heard it called that - just "the completeness theorem" (of which it's an immediate corollary). Is it referred to separately as such? – Noah Schweber Jul 23 '18 at 02:50
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@NoahSchweber. I just recall that name from the hand-written lecture notes from a course in Model Theory, decades ago. – DanielWainfleet Jul 23 '18 at 02:55
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Yes; the other half you're asking about is the soundness theorem, which states that if $T$ proves $\varphi$ then $\varphi$ is true in every model of $T$. In particular, if $T$ proves $\perp$ then $T$ has no model, and so by the contrapositive any satisfiable theory is consistent.
The proof of the soundness theorem is much simpler than that of the completeness theorem: we just show that each of the clauses of our proof system match up appropriately with the definition of satisfaction.
Noah Schweber
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