In logic via algebra (page $93$), Halmos tries to prove strong completeness ( if $S\models q$ then $S\vdash q$) assuming weak completeness ( if $q$ is a valid in the Boolean logic $(A,F)$ then $q\in F$ )
Theorem: For any proposition $q$ and any set $S$ of propositions in $(A,F)$, If $S\models q$ ,then $S\vdash q$.
His proof (bolding is mine ) :
Suppose $S\models q$. Let $\hat S$ be the set of (provable) consequences of $S$. Then $\hat S$ is a filter that includes $F$. We must show that $q$ is in $\hat S$. The assumption $S\models q$ implies that in the Boolean logic $(A,\hat S)$ the proposition $q$ is valid. therefore $q$ is provable (in symbols, $\vdash q$) in $(A,\hat S)$ by the (weak) completeness theorem for $(A,s)$. In other words, $q\in \hat S$,as was to be shown
My question is: I don't see why this bolded sentence is true. Why is assuming that $S\models q$ implies that $q$ is valid in $(B,\hat S)$?
What I understand so far is that we have the Boolean logic $(A,F)$ and we have the filter $\hat S$ of all consequences from $(A,F)$. Now we define a new congruence relation induced by the filter $\hat S$ on $A$ to get the Boolean logic $(A,\hat S)$ but I can not see how does this entail the implication mentioned above.
Any help?