A subtlety with $\vDash$? The symbol $\vDash$ seems to have two different meanings: to show logical consequence and to show truth in a model. These seem to be two different things referred to by the same symbol as the following shows.
As a symbol of logical consequence the following subtlety arises (A is a set of wffs, B is a wff):

*

*A $\vDash$ B (If A is true then always B is true)

*A $\nvDash$ B (If A is true then not always is B true or ~B is always true)

*A $\vDash$ ~B (If A is true then always ~B is true)

All three cases mean different situations, they are not equivalent.
Now, as a symbol to indicate truth it behaves differently from that of logical consequence (from now on A is a model and B a wff within the model):

*

*A $\vDash$ B (B is true in A)

*A $\nvDash$ (B is not true in A)

*A $\vDash$ ~B (~B is true in A)

So here 2. and 3. mean the same thing, they are equivalent.
Is that (my) insight correct? If not, please point out where and why.
 A: Yes, this is correct. It might help to observe that there is yet one more usage, which appropriately subsumes both those you ask about: for $\mathbb{K}$ a class of structures and $\varphi$ a sentence, we write $\mathbb{K}\models\varphi$ iff each $\mathcal{A}\in\mathbb{K}$ satisfies $\varphi$. This generalizes the (original!) "truth-in-a-structure" definition by taking $\mathbb{K}=\{\mathcal{A}\}$, and generalizes "semantic consequence" definition by taking $\mathbb{K}=Mod(T)$ where $Mod(T)=\{\mathcal{A}: \mathcal{A}\models T\}$.
Note in this context that we have "bivalence" (= the coincision of $\not\models$ and $\models\neg$) iff $\mathbb{K}$ contains exactly one element up to elementary equivalence; one way this can happen is if $\mathbb{K}$ is in fact a singleton, and another way this can happen is if $\mathbb{K}=Mod(T)$ for a complete theory $T$ (bigger theories have smaller model classes).
Incidentally, in my opinion it is extremely helpful in this context to use different fonts/alphabets to denote different types of object. For example, I would only write "$A\models B$" if $A$ and $B$ were objects of the same type.
