Equivalence of Semantic Consequence and Logical Implication

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

Let $$\psi$$ be a logical formula.

Then $$U \models \psi$$ iff $$\phi_1, \phi_2, \ldots, \phi_m \vdash \psi$$.

That is, logical consequence is equivalent to logical implication.

Necessary Condition
First we show that:
 * If $$\phi_1, \phi_2, \ldots, \phi_m \vdash \psi$$, then $$\left\{{\phi_1, \phi_2, \ldots, \phi_m}\right\} \models \psi$$.

We use a proof by induction on the length of tableau proofs.

Because $$\phi_1, \phi_2, \ldots, \phi_m \vdash \psi$$, there is a tableau proof of $$\psi$$ from the premises $$\phi_1, \phi_2, \ldots, \phi_m$$.

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition:


 * "For all sequents $$\phi_1, \phi_2, \ldots, \phi_m \vdash \psi$$ which have a tableau proof of length $$n \in \N^*$$, it is true that $$\left\{{\phi_1, \phi_2, \ldots, \phi_m}\right\} \models \psi$$."

Basis for the Induction
If $$n = 1$$, then the tableau proof must be of the form:

since all other proofs need more than one line.

This is the sequent that proves the Law of Identity.

(Note that here it can only be the case that $$m = 1$$, and we are showing that if $$\phi \vdash \phi$$ then $$\phi \models \phi$$.)

Clearly any model $$\mathcal{M}$$ such that $$\mathcal{M} \left({\phi}\right) = T$$ will imply that $$\mathcal{M} \left({\phi}\right) = T$$, and hence $$\phi \models \phi$$ (trivially).

Hence if $$\phi \vdash \phi$$ then $$\phi \models \phi$$ and so $$P(1)$$ is true.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true for all $$k \ge 1$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

That is, from our induction hypothesis:
 * "For all sequents $$\phi_1, \phi_2, \ldots, \phi_m \vdash \psi$$ which have a tableau proof of length $$k \in \N^*$$, it is true that $$\left\{{\phi_1, \phi_2, \ldots, \phi_m}\right\} \models \psi$$."

... we need to show:
 * "For all sequents $$\phi_1, \phi_2, \ldots, \phi_m \vdash \psi$$ which have a tableau proof of length $$k+1 \in \N^*$$, it is true that $$\left\{{\phi_1, \phi_2, \ldots, \phi_m}\right\} \models \psi$$."

Induction Step
This is our induction step:

Let us assume that the proof of
 * $$\phi_1, \phi_2, \ldots, \phi_m \vdash \psi$$

has a tableau proof of length $$k+1$$.

The proof will be of the following structure:

What we need to ask now is: what was the justification for the last line?

Let us consider all possible cases.


 * Suppose the last rule is the rule of conjunction: $\and \mathcal{I}$.

Then we know that $$\psi$$ is of the form $$\psi_1 \and \psi 2$$ and the justification on line $$k+1$$ refers to two lines further up on which we have $$\psi_1$$ and $$\psi_2$$ as the conclusions.

Using the induction hypothesis, we conclude that both $$\psi_1$$ and $$\psi_2$$ evaluate to true.

Inspecting the truth table for conjunction we see that $$\psi_1 \and \psi 2$$ evaluates to $$T \and T = T$$ as well.


 * Suppose the last rule is the rule of or-elimination: $\or \mathcal{E}$.

Then we must have proved, or assumed, some formula $$\eta_1 \or \eta_2$$, referred to via $$\or \mathcal{E}$$ on line $$k+1$$.

So we have a shorter proof:
 * $$\phi_1, \phi_2, \ldots, \phi_m \vdash \eta_1 \or \eta_2$$

and so $$\eta_1 \or \eta_2$$ evaluates to $$T$$ by the induction hypothesis.

By the truth table for disjunction we see that either $$\eta_1$$ or $$\eta_2$$ evaluates to $$T$$.

WLOG assume that $$\eta_1$$ evaluates to $$T$$.

Using $$\or \mathcal{E}$$ we had to assume $$\eta_1$$ and $$\eta_2$$ and show $$\psi$$ for each of these cases.

So, consider the first case.

So we have a shorter proof:
 * $$\phi_1, \phi_2, \ldots, \phi_m, \eta_1 \vdash \psi$$

and all the formulas on the LHS evaluate to $$T$$.

By the induction hypothesis, this implies that $$\psi$$ evaluates to $$T$$ as well.

The case in which $$\eta_2$$ evaluates to $$T$$ is worked in exactly the same way.


 * The rest of the possible proof rules can be checked similarly, and can all be verified that they behave semantically in the same way that their truth tables do.

So $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * "For all sequents $$\phi_1, \phi_2, \ldots, \phi_m \vdash \psi$$ which have a tableau proof of length $$n \in \N^*$$, it is true that $$\left\{{\phi_1, \phi_2, \ldots, \phi_m}\right\} \models \psi$$."

Sufficient Condition
Next we show that:
 * If $$\left\{{\phi_1, \phi_2, \ldots, \phi_m}\right\} \models \psi$$, then $$\phi_1, \phi_2, \ldots, \phi_m \vdash \psi$$.

Suppose we have $$\left\{{\phi_1, \phi_2, \ldots, \phi_m}\right\} \models \psi$$.

First we show that $$\phi_1 \implies \left({\phi_2 \implies \left({\phi_3 \implies \left({\ldots \left({\phi_m \implies \psi}\right) \ldots}\right)}\right)}\right)$$ is a tautology.

We have from the Extended Rule of Implication that:
 * $$\phi_1 \implies \left({\phi_2 \implies \left({\phi_3 \implies \left({\ldots \left({\phi_m \implies \psi}\right) \ldots}\right)}\right)}\right)$$

means the same thing as
 * $$\phi_1, \phi_2, \ldots, \phi_m \vdash \psi$$.

From the same source, we see this is equivalent to $$\left({\phi_1 \and \phi_2 \and \ldots \and \phi_m}\right) \implies \psi$$.

This can evaluate to $$F$$ only if all $$\phi_1, \phi_2, \ldots, \phi_m$$ evaluate to $$T$$ and $$\psi$$ evaluates to $$F$$.

But this contradicts the fact that $$\left\{{\phi_1, \phi_2, \ldots, \phi_m}\right\} \models \psi$$.

So we have that $$\models \phi_1 \implies \left({\phi_2 \implies \left({\phi_3 \implies \left({\ldots \left({\phi_m \implies \psi}\right) \ldots}\right)}\right)}\right)$$.

Comment
There are two things being proved here:


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

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

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

This establishes that propositional logic is sound.


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

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

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.