Hi guys I need to prove a Finite Support Theorem which states that $K$ is finitely definable iff $K$ has finite support. Unfortunately I succeeded in proving only the first part of if and only if.
$Introduction:$ We say that $S\subseteq \{p_i : i\in\mathbb{N}\}$ is a support for a set of assignments $K$, if for every two assignments $v$ and $v'$ that agree on all propositional variables in $S$ $\Rightarrow$ $v$$\in$$K$$\iff$$v'$$\in$$K$
$K$ is finitely definable if there exists a finite number of formulas in $\Sigma$ s.t. $Ass(\Sigma) = K$
$Proof \ attempt:$
Let $K$ be a set of truth assignments.
$\Rightarrow$ Suppose $K$ is finitely definable, I can just go over all the formulas in $\Sigma$ because it is finite and construct a finite set $S$ from a propositional variables in $\Sigma$, with some formality it's not a problem to prove that $S$ is a finite support for $K$.
$\Leftarrow$ Suppose $S$ is a finite support of $K$. I need to show that $K$ is finitely definable. In this direction I'm struggling to find a formal proof. Maybe I need somehow to construct $\Sigma$ from the a finite support I have, but I don't know how to use the given property of assignments... Anyway I'm a little lost here. Would be thankful for some help.
$\Leftarrow$: Suppose $K$ has finite support, for some finite $S\subseteq \{p_i\mid i\in \Bbb N\}$ such that for all assignments $v,v'$, if $v|S = v'|S$ then $v\in K\iff v'\in K$. (Note, abuse of notation: strictly speaking, $v|S$ should really be $v\,|\,\{i\mid p_i\in S\}$.)
Assume that assignments are functions $v\colon \Bbb N\to 2 = \{0,1\}$.
Let $K|S = \{v|S\mid v\in K\}$, the set of all restrictions of assignments in $K$ to $S$, so $K|S$ is finite. (Note, another abuse of notation: of course $K$ isn't a function, so don't take $K|S$ literally.)
For each partial assignment $s\in K|S$ and each $p_i\in S$, let $$ p_i^{(s)} = \begin{cases} p_i &\text{if $s_i = 1$},\\ \neg\, p_i &\text{if $s_i = 0$.}\\ \end{cases}$$
For each $s\in K|S$, let $\phi^{(s)}$ be the formula $$ \phi^{(s)} = \bigwedge_{p\in S} p^{(s)}, $$ so $\phi^{(s)}$ is true under any assignment $v\supseteq s$.
Finally, define the formula $\phi$ by $$ \phi = \bigvee_{s\in K|S} \phi^{(s)}. $$
We claim that $K = Assign(\{\phi\})$. Clearly, every $v\in K$ satisfies $\phi$, as $v$ satisfies $\phi^{(v|S)}$. For the reverse inclusion, suppose $v$ is an assignment satisfying $\phi$. Then for some $s\in K|S$, $v$ satisfies $\phi^{(s)}$. By definition of $K|S$, there is $v'\in K$ such that $s= v'|S$. By definition of $\phi^{(s)}$ and the fact that $v$ satisfies it, we must have $v|S = v'|S$. Hence, because $K$ has finite support with respect to $S$ and $v'\in K$, we must have $v\in K$. So $K$ is finitely definable.