Show that $Σ$ is complete if, and only if, it is inconsistent or satisfactory for exactly one assessment Let $Σ$ be a set of propositional formulas and $α$ be any propositional formula. 
We will say that a set of formulas $Σ$ is complete if for all formula $α$ we have that $Σ |= α Σ |= ¬α$. Show that it is complete if, and only if, it is inconsistent or satisfactory for exactly one assessment.
I’m thinking of using the fact that $Σ$ is satisfactory if there’s a valuation σ so that $σ(Σ)=1$. And in another case, $Σ$ is inconsistent. But I can't figure out how to do the whole exercise. 
Thank you in advance!
 A: First suppose that $\Sigma$ is complete. If there is a propositional variable $P$ such that $\Sigma \models P$ and $\Sigma \models \neg P$, then $\Sigma$ is inconsistent. If this is not the case, then we can define a valuation as follows:
$$
\sigma(P) = \begin{cases}
1 & \Sigma \models P \\
0 & \Sigma \models \neg P
\end{cases}
$$
Completeness of $\Sigma$, together with the fact that we never have $\Sigma \models P$ and $\Sigma \models \neg P$ at the same time guarantee that $\sigma$ is a well-defined valuation. Clearly $\Sigma$ is satisfied by $\sigma$. If $\sigma'$ is any other valuation, then there must be $P$ such that $\sigma(P) \neq \sigma'(P)$, so $\sigma'$ cannot satisfy $\Sigma$. We thus see that $\sigma$ is the only valuation satisfying $\Sigma$.
For the other direction, if $\Sigma$ is inconsistent then $\Sigma \models \alpha$ for any formula $\alpha$, so in particular $\Sigma$ is complete. If $\Sigma$ has only one satisfactory valuation $\sigma$, then for any formula $\alpha$ we have $\Sigma \models \alpha$ if and only if $\sigma(\alpha) = 1$. So $\Sigma \models \alpha$ precisely when $\sigma(\alpha) = 1$ and $\Sigma \models \neg \alpha$ precisely when $\sigma(\alpha) = 0$. One of these must always happen, so $\Sigma$ is complete.

I just wanted to note that to me it is a bit strange to consider inconsistent $\Sigma$ to be complete. I know this is the definition you are working with, but to me it seems to make more sense to define completeness only for consistent $\Sigma$. I just mention this, because there are other texts out there that use this convention.
