So I have to show: $\displaystyle \Sigma$ is complete $\displaystyle \iff$ $\displaystyle \Sigma$ has a unique satisfying truth assignment. So my proof has to go in both directions...

To show the "$\displaystyle \Leftarrow$", we can try proving its contrapositive:

"If $\displaystyle \Sigma$ is not complete then it has no unique truth assignment."

If there is a truth assignment $\displaystyle f_{\Sigma}$ then for an arbitrary A the following is one of the options that must be true:

$\displaystyle f_{\Sigma}(A)=0$ then $\displaystyle f_{\Sigma}(\neg A) = 1$.

And in order for that to be true we need $\displaystyle \Sigma$ to be complete (because if we have $\displaystyle f_{\Sigma}(A)=0$, then $\displaystyle A \notin \Sigma$ so $\displaystyle \neg A \in \Sigma$ so $\displaystyle f_{\Sigma}(\neg A) = 1$).

Suppose $\displaystyle \Sigma$ is not complete, then if $\displaystyle f_{\Sigma}(A)=0$ then $\displaystyle f_{\Sigma}(\neg A) \neq 1$. So there's no truth assignment. Is this right?

To show the "$\displaystyle \Rightarrow$", similarly I'll try proving its contrapositive:

"If $\displaystyle \Sigma$ doesn't have a unique satisfying truth assignment, then it is not complete."

Suppose $\displaystyle \Sigma$ doesn't have a unique satisfying truth assignment. Then how could we prove that $\displaystyle \Sigma$ is therefore not complete? Here's where I'm stuck...

P.S: By the truth assignment $\displaystyle f_{\Sigma}(A): form(L) \to \{ 0,1 \}$ I meant:

$\displaystyle f_{\Sigma}(A)= \left\{\begin{matrix}1 \ if A \in \Sigma \\ 0 \ if \ A \notin \Sigma\end{matrix}\right.$