I'm trying to prove Proposition 4.12 from Blackburn, de Rijke and Venema's "Modal Logic" (page 196). The proposition is as follows:
A logic $\Lambda$ is strongly complete with respect to a class of structures S iff every $\Lambda$-consistent set of formulas is satisfiable in some $\mathfrak{S} \in S$. A logic $\Lambda$ is weakly complete with respect to a class of structures S iff every $\Lambda$-consistent formula is satisfiable in some $\mathfrak{S} \in S$
As weak completeness follows from strong completeness, only the proof for strong completeness matters. The authors present a proof from the right to left case of strong completeness by contraposition, which is as follows:
Suppose $\Lambda$ is not strongly complete with respect to S. Thus, there is a set of formulas $\Gamma \cup \{\varphi\}$ such that $\Gamma \Vdash_S \varphi$ but $\Gamma \nvdash_\Lambda \varphi$. Then $\Gamma \cup \{\neg \varphi\}$ is $\Lambda$-consistent, but is not satisfiable on any structure in S.
Which I found to not be particularly hard, however, the left to right direction has left me stumped. I tried a direct proof, but could not find any way to relate completeness and satisfiability. Similarly for a proof by contraposition.
I feel like I'm missing some rather obvious detail, but after nearly an entire day of trying to prove this, I don't see what I could be missing. Thanks in advance.
Again by contraposition.
Assume that there is $\Gamma \cup \{ \phi \}$ that is $\Lambda \text {-cons}$ but not satisfiable.
Being not sat we have: $\Gamma \cup \{ \phi \} \vDash \bot$; but $\Lambda \text {-cons}$ means: $\Gamma \cup \{ \phi \} \nvdash \bot$, and thus the logic in not strongly complete.