Equivalence of Semantic Consequence and Logical Implication

Theorem
Let $U = \left\{{\phi_1, \phi_2, \ldots, \phi_m, \ldots}\right\}$ be a countable set of logical formulas.

Let $\psi$ be a logical formula.

Then $U \models \psi$ iff $U \vdash \psi$.

That is, logical consequence is equivalent to logical implication.

Necessary Condition
This is a statement of the Extended Soundness Theorem of Propositional Calculus:

Let $\mathbf H$ be a countable set of logical formulas.

Let $\mathbf A$ be a logical formula.

If $\mathbf H \vdash \mathbf A$, then $\mathbf H \models \mathbf A$.

Sufficient Condition
Let $\mathbf H$ be a countable set of logical formulas.

Let $\mathbf H \models \mathbf A$.

Then $\mathbf H \cup \left\{{\neg \mathbf A}\right\}$ has no models.

By the Compactness Theorem of Propositional Calculus‎, there is a finite subset $\mathbf H_0 \subseteq \mathbf H$ such that $\mathbf H_0 \cup \left\{{\neg \mathbf A}\right\}$ has no models.

Then $\mathbf H_0 \models \mathbf A$.

By the statement of the Extended Completeness Theorem of Propositional Calculus $\mathbf H_0 \vdash \mathbf A$.

Hence $\mathbf H \vdash \mathbf A$.

Comment
There are two things being proved here:


 * $(1): \quad$ Suppose we have a sequent $\phi_1, \phi_2, \ldots, \phi_m, \ldots \vdash \psi$, the validity of which has been established, for example, by a tableau proof.

The result:
 * if $\phi_1, \phi_2, \ldots, \phi_m, \ldots \vdash \psi$ then $\left\{{\phi_1, \phi_2, \ldots, \phi_m, \ldots}\right\} \models \psi$

establishes that if all the propositions $\phi_1, \phi_2, \ldots, \phi_m, \ldots$ evaluate to true, then so does $\psi$.

This establishes that propositional logic is sound.


 * $(2): \quad$ Suppose we have determined that $\left\{{\phi_1, \phi_2, \ldots, \phi_m, \ldots}\right\} \models \psi$.

The result:
 * if $\left\{{\phi_1, \phi_2, \ldots, \phi_m, \ldots}\right\} \models \psi$ then $\phi_1, \phi_2, \ldots, \phi_m, \ldots \vdash \psi$

establishes that if we can show that there is a model for a proposition, then we will be able to find a tableau proof for it.

This establishes that propositional logic is complete.