How exactly are consistency and satisfiability related in first order logic? A set $\Gamma$ is consistent if there is $\psi$ such that $\Gamma \not\vdash \psi$.
A set $\Gamma$ is satisfiable if there is a model such that $\Gamma \vDash \psi$ for any $\psi \in \Gamma$.
Intuitively, I understand that for a set to be consistent it must not contain any contradictions, otherwise we can prove anything from it. 
I know that if a set is consistent then it is satisfiable, from the completeness theorem, but is it true the other way around? Does satisfiability imply consistency when talking about sets of sentences? 
I would think not, since satisfiability depends on the model chosen, but I haven't been able to come across any proof or theorem regarding this, or maybe I haven't understood this correctly.
 A: First of all, your definition of satisfiability is wrong.  It is not that $\Gamma \vDash \psi$ for every $\psi \in \Gamma$ (which is of course trivially true for any $\Gamma$!), but rather that $\Gamma$ has a model, i.e. that there is an interpretation $I$ such that $I \vDash \psi$ for every $\psi \in \Gamma$
Now, if you say that consistency is $\Gamma \not\vdash \psi$, then by the $\vdash$ you must be referring to some derivational system $S$, so it depends on that derivational system. 
For example, for the derivational system that contains no derivational rules (the 'null' system $N$), we have that $\{P, \neg P\} \not \vdash_N Q$, and thus relative to $N$, $\{P, \neg P\}$ is consistent .. but it is clearly not satisfiable.
On the other hand, for the derivational system $B$, whose only rule of derivation is Hokus Ponens (which is: $\vdash \phi$, i.e. you can infer any statement from nothing), we have don't have any $\psi$ such that $\{ \} \not \vdash_B \psi$, and thus relative to $B$, $\{ \}$ is inconsistent, and yet clearly it is satisfiable.
OK, but those are weird systems.  As long as we are dealing with a system $S$ that is both sound and complete, consistency and satisfiability are the same thing. Here is why. First, let's establish that for any sound and complete $S$:
Lemma: $\Gamma \vdash_S \bot$ iff for any $\psi$: $\Gamma \vdash \psi$
Proof:
'if': Trivial. Just pick $\psi = \bot$
'only if': We know that $\bot$ implies any $\psi$, and so since $S$ is complete, it can derive $\psi$ from $\bot$. Hence, if $\Gamma \vdash_S \bot$, then $\Gamma \vdash_S \psi$ for any $\psi$
For any $\Gamma$:
$\Gamma$ is consistent iff
There is some $\psi$ such that $\Gamma \not \vdash_S \psi$ iff (Lemma)
$\Gamma \not \vdash_S \bot$ iff (soundness and completeness of $S$)
$\Gamma \not \vDash \bot$ iff 
There is no interpretation $I$ such that $I \vDash \psi$ for all $\psi \in \Gamma$ and $I \not \vdash \bot$ iff ($I \not \vdash \bot$ is trivially true)
There is no interpretation $I$ such that $I \vDash \psi$ for all $\psi \in \Gamma$ iff
$\Gamma$ is satisfiable
