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I have an intuitive idea that, given some set of formulas $Γ$, and two formulas $A, B \not\in Γ$, $((Γ\cup{A}) \models B)↔(Γ \models (A→B))$. I can rationalize this as, if the left side of the biconditional is true, then whenever $A$ is true, $B$ is true, and if $A$ is false, nothing about $B$'s truth value is said. The same thing is descriptive of the RHS, and so they evaluate to true if both do and not true (false) if the other does.

How would I formalize a proof of this property, or express it in a rigorous fashion (in the language of logic)?

I'm aware that the logical connectives have valuation functions, such that, given valuations for the propositional letters/variables, any formula can be evaluated. But to my understanding, logical entailment ($\models$) is a metalogical concept and is not modeled the same way, but I cannot see any reason why it should not be.

Should the property be proven by creating an evaluation function for $\models$ as a connective, and prove the property for all valuations of $Γ, A,$ and $B$?

I feel that any answer to this will allow extrapolation to understanding other such metalogical properties as well, so feel free to be more general with your answers, but any insight into this particular problem will be greatly appreciated as well!

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Daniel Fischer Mar 5 '15 at 19:37
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Being $\Gamma$ a set of formulae, it is more correct to write :

$\Gamma \cup \{ A \} \vDash B$ iff $\Gamma \vDash A \to B$.

Having said that, the relation of tautological implication (or tautological consequence) is a relation between a set of formulae ($\Gamma$) called premises and a formula ($\tau$) called the conclusion defined as follows :

$\Gamma$ tautologically implies $\tau$ (written : $\Gamma \vDash \tau$) iff every truth assignment for the sentence symbols in $\Gamma$ and $τ$ that satisfies every member of $\Gamma$ also satisfies $τ$.

If $\Gamma$ is a finite set of formulae, i.e. $\Gamma = \{ \varphi_1, \varphi_2,\ldots,\varphi_n \}$, we have that :

$\varphi_1, \varphi_2,\ldots,\varphi_n \vDash \psi$ iff $\vDash \varphi_1 \land \varphi_2 \land \ldots \land \varphi_n \to \psi$.

Thus, in order to show that $\Gamma \cup \{ A \} \vDash B$ iff $\Gamma \vDash A \to B$ we have to consider the two cases :

$(\Rightarrow)$ : assume that $\Gamma \cup \{ A \} \vDash B$ and consider a truth assignment that satisfy all formulae in $\Gamma$. We have two possibilities: $A$ is satisfied by this truth assignment. But then, being $B$ tautologically implied by $\Gamma \cup \{ A \}$, we have that also $B$ is satisfied by this truth assignment.

Thus, $A \to B$ is true, i.e. it is satisfied by the truth assignment that satisfy all formulae in $\Gamma$.

Second possibility : $A$ is not satisfied by the above truth assignment. Then, by truth conditions for $\to$, $A \to B$ is true; again it is satisfied by the truth assignment that satisfy all formulae in $\Gamma$.

In both ways, $A \to B$ is satisfied by the truth assignment that satisfy all formulae in $\Gamma$, and thus we conclude that :

$\Gamma \vDash A \to B$.

The same for $(\Leftarrow)$ : assume that $\Gamma \vDash A \to B$ and consider a truth assignment that satisfy all formulae in $\Gamma$ and in addition satisfy also $A$.

The assumption implies that $A \to B$ is also satisfied by this truth assignment and thus we have that both $A$ and $A \to B$ are satisfied by it, and this (again by properties of $\to$) implies that also $B$ is satisfied.

This means that :

$\Gamma \cup \{ A \} \vDash B$.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – user642796 Mar 4 '15 at 2:46

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