Equivalence of two biconditionals of propositional metalogic

Question

In application to propositional metalogic, I am told that the following two biconditionals are equivalent:

(i) Γ is satisfiable iff every finite subset of Γ is satisfiable.

(ii) Γ ⊨ α iff some finite subset Δ is such that Δ ⊨ α.

where, Γ is a set of well-formed formulas, α is some arbitrary well-formed formula, and '⊨' represents "the antecedent logically implies the consequent."

but how is this possible if one biconditional is making a claim of satisfaction and the other of logical implication? I simply don't see how the two could be equivalent.

Answer 1

Hint

We have to prove it using the definitions of satisfiability, logical consequence, etc.

Consider one case : (i) $\implies$ (ii), and consoder the subcase : $\implies$ of (ii).

We have :

(i) $ \ \Gamma$ is satisfiable iff every finite subset of $\Gamma$ is satisfiable

and we assume that :

$\Gamma \vDash \alpha$.

We want to prove that :

for some finite $\Delta \subseteq \Gamma : \Delta \vDash \alpha$.

Assume not, i.e. for all finite $\Delta \subseteq \Gamma : \Delta \nvDash \alpha$.

This means that : for any finite $\Delta \subseteq \Gamma$, there is a valuation $v_{\Delta}$ such that $v_{\Delta}(\delta)=$t, for any $\delta \in \Delta$, and $v_{\Delta}(\alpha)=$f.

But if $v_{\Delta}(\alpha)=$f, then $v_{\Delta}(\lnot \alpha)=$t, i.e. $\Delta \cup \{ \lnot \alpha \}$ is satisfiable.

This hold for any finite $\Delta \subseteq \Gamma$, and thus, by (i) : $\Gamma \cup \{ \lnot \alpha \}$ is satisfiable, contradicting : $\Gamma \vDash \alpha$.