A logic exercise (finite satisfiability) I am new here and this is my first post.
Had a logic exercise, thought of a solution but I am not sure.

If the wff $(\phi_1 \wedge  \phi_2 \wedge... \wedge \phi_n)\rightarrow \phi_0$ is valid and S is a set of wffs such that any finite subset of S is satisfiable then prove that for some of the sets:
  $$
S\cup\{\phi_{0}\},S\cup\{\neg\phi_{1}\},\ldots,S\cup\{\neg\phi_{n}\}
$$
  has every finite subset satisfiable  without using the compactness or the completeness-soundness theorems.

My suggestion for a solution is that the wff is equivalent to $\neg(\phi_{1}\wedge\phi_{2}\wedge...\wedge\phi_{n})\vee\phi_{0} $ that means that, for every interpretation $M$, either i)  $\phi_0$ must be true in $M$ or ii)$(\phi_{1}\wedge\phi_{2}\wedge...\wedge\phi_{n})$ must be false in $M$ and then treat either case separately. In the first case (i) $S\cup{\phi_0}$ is finite satisfiable in the second (ii) some the other sets is since at least one $\phi_i$ must false. I believe i can work the details but is my initial thought right? Is in the right direction for the proof? 
 A: Let $X' \subseteq S$. We want to prove that $X' \subseteq S \cup \{\phi_0\}$ or $X' \subseteq S \cup \{\neg \phi_1\}$ or ... or $X' \subseteq S \cup \{\neg \phi_n\}$ such that $X'$ is finite and satisfiable ('finite satisfiable, for short). Now we state our assumptions:

Suppose $\vDash (\phi_1 \wedge \phi_2 \wedge ... \wedge \phi_n) \rightarrow \phi_0$ and that $X \subseteq S$ is finite satisfiable.

Now we pick up your suggestion of treating the above formula in terms of disjuntion, since in classic logic $\vDash_v (\phi \vee \psi)$ means $max(v(\phi),v(\psi))=1$, which is to say that

$\vDash_v \phi$ or $\vDash_v \psi$.

The statement we want to prove is a disjuntion. Therefore, we divide the proof in two cases:
Case 1: $X' \subseteq S \cup \{\phi_0\}$.

We recall that since $\vDash (\phi_1 \wedge \phi_2\wedge ... \phi_n) \rightarrow \phi_0$, then the antecedent is false or the consequent true. Let $M'$ be an interpretation such that $M'(\phi_0)=1$. Then we got our result since by our assumption $X \subseteq S$ is finite satisfiable. Now let $M''(\phi_0)=0$. We have to guarantee that the other disjunct holds whenever $M''(\phi_1 \wedge \phi_2 \wedge ... \wedge \phi_n)=0$ (see the following case).

Case 2: $X' \subseteq S \cup \{\neg \phi_1\}$ or $X' \subseteq S \cup \{\neg \phi_2\}$ or ... or $X' \subseteq \cup \{\neg \phi_n\}$.

We just need to take in consideration the interpretations $M''$ above described. Since $M''(\phi_1 \wedge \phi_2 \wedge ... \wedge \phi_n)=0$, it means there exists at least one $\phi_k$ such that $M''(\phi_k)=0$. Then since our assumption that $X \subseteq S$ is finitely satisfiable, it follows that $X' \subseteq S \cup \{\neg \phi_k\}$ is finitely satisfiable.

