Can I infer $\Gamma\cup\{\alpha\}$ is inconsistent from $\Gamma\nvdash\alpha$? My reason for thinking so is this:
A set of wffs $\Gamma$ is inconsistent just in case: $\Gamma\vdash\alpha$ and $\Gamma\vdash\lnot\alpha$.
So, if $\Gamma$ is not inconsistent and $\Gamma\vdash\alpha$ then it's not the case that $\Gamma\vdash\lnot\alpha$, or $\Gamma\nvdash\lnot\alpha$.
So then, to make an inconsistent set out of $\Gamma$, we can add $\lnot\alpha$ to the set, $\Gamma\cup\{\lnot\alpha\}$.
So, if $\Gamma\nvdash\lnot\alpha$, then $\Gamma\cup\{\lnot\alpha\}$.
I suspect that this doesn't follow. Since I assumed that $\Gamma\vdash\alpha$ for the derivation to work. What I am wondering though is what exactly I can infer from $\Gamma\nvdash\alpha$. Like can I infer that $\Gamma\vdash\lnot\alpha$?
For context I am trying to prove that if we suppose that every consistent set of wffs is satisfiable (ACT), then FOL completeness (CT), i.e. If $\Gamma\vDash\alpha$, then $\Gamma\vdash\alpha$ follows. My strategy is by way of reductio. So I suppose, CT is not true. That is, I suppose that $\Gamma\vDash\alpha$ for some set of wffs $\Gamma$ and wff $\alpha$, but $\Gamma\nvdash\alpha$. My initial question then regards whether I can derive from this assumption that $\Gamma\cup\{\alpha\}$ is inconsistent. If so, then I think I can use the contrapositive of ACT to derive a contradiction.
Thanks for any and all help.
 A: You certainly can't infer $\Gamma \vdash \lnot \alpha$ from $\Gamma \not\vdash \alpha$. E.g., $ \mathrm{FOL} \not\vdash \forall x, y.x = y$ and $ \mathrm{FOL} \not\vdash \exists x, y.x \neq y$, since both of these sentences have counter-models.
What you can infer from $\Gamma \not\vdash \alpha$, is that $\Gamma \cup \{\lnot \alpha\}$ is consistent. (Because if it were inconsistent, then $\Gamma, \lnot\alpha \vdash \beta$ for any $\beta$, and in particular, $\Gamma, \lnot\alpha  \vdash \alpha$, but that is equivalent to $\Gamma \vdash \lnot \alpha \Rightarrow \alpha$, which is equivalent to $\Gamma \vdash \alpha$, contradicting $\Gamma \not\vdash \alpha$). This gives you the contrapositive of CT, if you are given that every consistent set of sentences is satisfiable: if $\Gamma \not\vdash \alpha$, then $\Gamma \cup \{\lnot\alpha\}$ has a model, so $\Gamma \not\models \alpha$.
A: Did you mean to say "So, if $\Gamma\nvdash\lnot\alpha$, then $\Gamma\cup\{\lnot\alpha\}$ is inconsistent"? If so, the answer is no. You also need that $\Gamma\vdash\alpha$. But then you can deduce anything at all, because ($\Gamma\vdash\alpha$ and $\Gamma\nvdash\alpha$) is a logical contradiction.
