How to prove 'if $A\models B$, then $\models A\Rightarrow B$ ' by contradiction? I need to find a book or an article with a proof (by contradiction) of the following theorem:
'if $A\models B$, then $\models A\Rightarrow B$'
I need it because I need to check if my proof is ok.
It has to be proof of semantic version, not syntactic.
EDIT: PLEASE CHECK MY PROOF :)
Let's (1) assume that $\alpha \models \beta$. Let's assume (2) that $\not\models \alpha\Rightarrow \beta$. From (2) we have: 'exists interpretation $v$ such that $v(\alpha)=1$ and $v(\beta)=0$'. From (1) we have: 'for any interpretation $v$: if $v(\alpha)=1$, then $v(\beta)=1$'. We remember that from (2) we have 'there exists interpretation $v$ such that $v(\alpha)=1$', so adding this information to (1) we get that for any $v$: $v(\beta)=1$. So we have contradiction: 'there exists $v$ such that $v(\beta)=0$' contradicts 'for any $v$: $v(\beta)=1$'.
 A: I can't offer you a book or an article that offers a proof of that specific result by contradiction. Here is a proof by contradiction:
If $\not\models A \Rightarrow B$, then there is a structure $M$ such that $M \models A$ but $M \not\models B$. So $A \models B$ (which means that $B$ holds in any model in which $A$ holds) must be false. Hence if $A \models B$, then $\models A \Rightarrow B$.
A direct proof is (to my mind) easier and more natural: if $A \models B$, then $B$ holds in any model in which $A$ holds, but then in any model, either $A$ does not hold or $B$ does, i.e., $\models A \Rightarrow B$.
A: Let's (1) assume that $\alpha \models \beta$. Let's assume (2) that $\not\models \alpha\Rightarrow \beta$. From (2) we have: 'exists interpretation $v$ such that $v(\alpha)=1$ and $v(\beta)=0$'. From (1) we have: 'for any interpretation $v$: if $v(\alpha)=1$, then $v(\beta)=1$'. We remember that from (2) we have that there exists interpretation $v$ such that $v(\alpha)=1$, so adding this information to (1) we get that for any $v$: $v(\beta)=1$. So we have contradiction: 'there exists $v$ such that $v(\beta)=0$' contradicts 'for any $v$: $v(\beta)=1$'.
