Is it true that if not $\alpha \vDash D$ implies $\alpha \vDash \neg D$? If it is not true that $\alpha \vDash D$, with $D$ arbitrary formula, is it true that $\alpha \vDash \neg D$?
I think that this assertion is false, but I cannot find counterexamples.
Thanks in advance for the help!
 A: If $\alpha$ is a formula, we have that :

$\alpha \vDash D$ iff for every valuation $v$, if $v(\alpha)=T$, then $v(D)=T$.

Thus to say that : "it is not true that $\alpha \vDash D$" means that, negating the above condition :

not $\alpha \vDash D$ iff there is a valuation $v_0$ such that $v_0(\alpha)=T$ and $v_0(D)=F$.

Clearly, from $v_0(D)=F$, follows that $v_0(\lnot D)=T$, but this is not enough to conclude with : $\alpha \vDash \lnot D$, because in order to assert it, we must have that $v(\lnot D)=T$, for all $v$ such that $v(\alpha)=T$, and not only for a single $v_0$.

The counter example is with a propositional letter $p_i$ that is not in $\alpha$:
Thus, with $D := p_i$, for every valuation $v$ such that $v(\alpha)=T$, we can always extend it to $v_1$ such that $v_1(\alpha)=v(\alpha)$ and $v_1(p_i)=F$, and thus we have not $\alpha \vDash D$, or extend it to $v_2$ as above such that $v_2(p_i)=T$, showing that not $\alpha \vDash \lnot D$.


If (as per Henning's answer) we "read" $\alpha$ not as a formula but as a structure $\mathfrak A$, then for any formula $\varphi$ :

$\mathfrak A \vDash \varphi$ or $\mathfrak A \vDash \lnot \varphi$.

A: That depends on what kind of thing $\alpha$ is.
If $\alpha$ is a set of assumptions, then Mauro's answer applies.
On the other hand, if $\alpha$ is a particular world (variously called a valuation, an interpretation, a structure, a model ...), then exactly one of $\alpha\vDash D$ and $\alpha\vDash\neg D$ will hold for any wff $D$.
The $\vDash$ that appears in the two cases are "different symbols" each with a separate definition (or, more honestly, there are two different concepts that happen to be notated with the same symbol for convenience).
