Proving deduction theorem without a system The semantic version of the deduction theorem says
$\Sigma ,A\vDash B$ iff $\Sigma \vDash(A\to B)$
I know that there exists proof for this in formal systems. But how can i prove this without using a system.
 A: Since we're working with semantics, we'll need to talk about structures. First remember what $\models$ means: 


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*"$\Sigma, A\models B$" means that every model of $\Sigma\cup\{A\}$ satisfies $B$.

*"$\Sigma\models A\implies B$" means that every model of $\Sigma$ satisfies $A\implies B$. 
Now do you see a way to prove one from the other? HINT: to prove the latter from the former, what can you say about a counterexample to "$\Sigma\models A\implies B$"? Similarly, think about a counterexample to the former - why does the existence of such a thing mean that the latter is false?
A: $\Sigma , A \vDash B$ iff (definition $\Sigma \vDash \varphi$)
for all interpretations $I$: If $I\vDash \Sigma \cup \{ A \}$ then $I\vDash B$ iff (definition $I \vDash \Sigma$)
for all interpretations $I$: If $I\vDash \varphi$ for all $\varphi \in \Sigma \cup \{ A \}$ then $I\vDash B$ iff (pure logic)
for all interpretations $I$: If $I\vDash \varphi$ for all $\varphi \in \Sigma$ then if $I \vDash A$ then $I\vDash B$ iff (semantics $\rightarrow$)
for all interpretations $I$: If $I\vDash \varphi$ for all $\varphi \in \Sigma$ then $I \vDash A \rightarrow B$ iff (definition $I \vDash \Sigma$)
for all interpretations $I$: If $I\vDash \Sigma$ then if $I \vDash A \rightarrow B$ iff (definition $\Sigma \vDash \varphi$)
$\Sigma \vDash A \rightarrow B$
