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

By the Compactness Theorem of Propositional Calculus‎, there is a finite subset $$\mathbf{H}_0 \subseteq \mathbf{H}$$ such that $$\mathbf{H}_0 \cup \left\{{\mathbf{A}}\right\}$$ 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 for the Direct Proof
There are two things being proved here:


 * 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.


 * 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.

Necessary Condition
Let $$\mathbf{H}$$ be a countable set of logical formulas.

Let $$\mathbf{H} \not \vdash \mathbf{A}$$.

Not ready....

??????????????

Then $$\mathbf{H} \cup \left\{{\mathbf{A}}\right\}$$ has models     (1).

By the Compactness Theorem of Propositional Calculus‎, there is a finite subset $$\mathbf{H}_0 \subseteq \mathbf{H}$$ such that $$\mathbf{H}_0 \cup \left\{{\mathbf{A}}\right\}$$ has no models      (2).

??????????????

Contradiction

It occurs because of (1) and (2). So $$\mathbf{H} \vdash \mathbf{A}$$

By the ???statement of the Extended Completeness Theorem of Propositional Calculus $$\mathbf{H}_0 \models \mathbf{A}$$.

Hence $$\mathbf{H} \models \mathbf{A}$$.

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

Let $$\mathbf{H} \not \models \mathbf{A}$$.

Then $$\mathbf{H} \cup \left\{{\mathbf{A}}\right\}$$ has models     (1).

By the Compactness Theorem of Propositional Calculus‎, there is a finite subset $$\mathbf{H}_0 \subseteq \mathbf{H}$$ such that $$\mathbf{H}_0 \cup \left\{{\mathbf{A}}\right\}$$ has no models      (2).

Because of (1) and (2), a contradiction. So $$\mathbf{H} \models \mathbf{A}$$.

???? What about $$\mathbf{H}_0 \models \mathbf{A}$$ ????

By the statement of the Extended Completeness Theorem of Propositional Calculus $$\mathbf{H}_0 \vdash \mathbf{A}$$.

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

Comment for the Proof by Contradiction
Not ready... initially hide it from reader:

The step (1) uses ...

The step (2) uses ...