There is an implication that I can't quite understand:
Given a set $\Sigma$ of well-formed-formulas and a well-formed-formula $\tau$, if $\Sigma_0\cup \{\tau\}$ is satisfiable for every finite $\Sigma_0\subseteq\Sigma$, then $\Sigma\cup\{\tau\}$ is finitely satisfiable
But I don't understand. Isn't the converse of the statement true, because finite satisfiability of $\Sigma\cup\{\tau\}$ implies the finite satisfiability of $\Sigma$?:
Given a set $\Sigma$ of well-formed-formulas and a well-formed-formula $\tau$, if $\Sigma\cup\{\tau\}$ is finitely satisfiable, then $\Sigma_0\cup \{\tau\}$ is satisfiable for every finite $\Sigma_0\subseteq\Sigma$