Suppose we have a proof system for classical first-order logic, where $\vDash$ denotes model-theoretic consequence and $\vdash$ denotes proof-theoretic consequence.
We can distinguish two forms of completeness of the proof system. Call them weak and strong, respectively*:
$$\text{for all sentences $\phi$, if $\vDash\phi$, then $\vdash\phi$}\tag{weak}$$ $$\text{for all sets of sentences $\Gamma$ and sentences $\phi$, if $\Gamma\vDash\phi$, then $\Gamma\vdash\phi$}\tag{strong}$$
Both of these statements are true of classical first-order logic with any of the standard proof systems. It is also obvious that strong completeness implies weak completeness, by taking $\Gamma=\emptyset$.
The weak statement of completeness also comes in two other equivalent forms: (1) every sentence $\phi$ is either satisfiable or refutable, and (2) if $\phi$ is consistent, then $\phi$ is satisfiable. Similarly, the strong statement of completeness is equivalent to (3) if $\Gamma$ is consistent, then $\Gamma$ is satisfiable. (My understanding is that in his 1930 dissertation Gödel proved a statement even stronger than (3), namely (4): if $\Gamma$ is consistent, then $\Gamma$ is satisfiable in a countable domain.)
Does weak completeness imply strong completeness? I get stuck when I try to prove it does.
Here is how I tried to prove that (weak) implies (strong). My thought was to show (weak) implies (3). Suppose $\Gamma$ is consistent. Then for all sentences $\phi$, $\Gamma\nvdash\phi\land\lnot\phi$. What you'd like to do at this point is use the syntactic deduction theorem** to say $\nvdash\Gamma\rightarrow(\phi\land\lnot\phi)$, and then use (weak) to conclude $\nvDash\Gamma\rightarrow(\phi\land\lnot\phi)$. From here it would follow that $\Gamma$ is satisfiable. But of course pushing $\Gamma$ into the antecedent of a conditional doesn't make sense for arbitrary sets of sentences $\Gamma$ (instead of a finite set, which can be pushed into the antecedent as a conjunction).
This suggests some sort of compactness trick would help, and that weak completeness does not imply strong completeness in general. Is this right?
*If there are more standard names for these properties, please tell me.
**Which says $\Gamma\cup\left\{\phi\right\}\vdash\psi$ if and only if $\Gamma\vdash\phi\rightarrow\psi$.