I'm stuck solving Exercise 3.23 on page 108 of in 'Propositional and Predicate Calculus: A Model of Argument' by Derek Goldrei:
Show that the following general statements are equivalent. (Γ and Σ are sets of formulas and φ is a formula.)
(E) For all Γ, if every finite subset of Γ is satisfiable, then so is Γ.
(F) For all Σ and φ, if Σ ⊨ φ then Δ ⊨ φ, for some finite subset Δ of Σ.
The author suggests solving this purely in terms of satisfiability, logical consequence, etc., without using any results about derivations such as the completeness theorem.
Here is what I have tried:
In one direction, I suppose that (E) holds and use this to prove the contrapositive of (F). So I assume that for every finite subset Δ of Σ it is not true that φ is a logical consequence of Δ. This implies that there is some truth assignment that satisfies Δ and makes φ false. In other words, every finite subset Δ of Σ is satisfiable. By (E) this implies that Σ is satisfiable. But how do I show that φ is not a logical consequence of Σ?
In the other direction, I suppose that (F) holds and attempt to prove the contrapositive of (E). I assume that Γ is not satisfiable. This means that any formula φ is a logical consequence of Γ, including a contradiction. By (F) this implies that ⊥ is a logical consequence of some finite subset Δ of Γ. This in turn implies that Δ is not satisfiable. In other words, there is some finite subset of Γ that is not satisfiable, which is the consequent of (E). Is this correct?
To prove E$\implies$F, given $\Sigma$ and $\varphi$ as in F, apply E to $\Gamma = \Sigma\cup\{\neg\varphi\}$.
For the converse, given $\Gamma$ as in E, apply F to $\Sigma=\Gamma$ and $\varphi=\bot$ (where $\bot$ is always false).