Can we prove this proposition without thinking semantics? 
Let $A$ be a set of propositional symbols, $\alpha$ ba a WFF on $A$ and $M$ be a subset of $A$. And let $M^+: = M \cup \{(\neg a): a\in (A-M)\}$.   Then, only one of $M^+ \vdash \alpha$ or $M^+ \vdash \neg \alpha$ be true.

I think this proposition is true. Because the propositional logic system is complet. But can we prove this proposition only in syntactic ways without thinking semantics? And please let me know if this proposition has a name.
 A: Sure, this can be proven purely syntactically. It's probably easiest just to do a straight induction on formulas. To minimize the amount of work, assume $\neg$ and $\land$ are the only connectives (all others being defined in terms of these).
The base case is a proposition variable $p\in A$. By definition of $M^{+}$, either $p\in M^{+}$ or $\neg p\in M^{+}$ so we can prove $M^{+}\vdash p$ or $M^{+}\vdash\neg p$ immediately by assumption.
For the $\land$ case, we have two cases: either $M^{+}\vdash\alpha_i$ for $i\in\{1,2\}$ in which case we can immediately prove $M^{+}\vdash\alpha_1\land\alpha_2$, or $M^{+}\vdash\neg\alpha_i$ for some $i\in\{1,2\}$ from which it is easy to prove $M^{+}\vdash\neg(\alpha_1\land\alpha_2)$.
Finally, for the $\neg$ case, we have either $M^{+}\vdash\alpha$ or $M^{+}\vdash\neg\alpha$ by induction hypothesis, and we want to show $M^{+}\vdash\neg\alpha$ or $M^{+}\vdash\neg\neg\alpha$. Clearly, if we have $M^{+}\vdash\neg\alpha$, we are done immediately. Otherwise, we have $M^{+}\vdash\alpha$ from which we can prove $M^{+}\vdash\neg\neg\alpha$ with a standard proof.
Normally the problem is the base case. We usually can't say anything one way or the other about proposition variables.
Now, to be precise, this doesn't quite prove the claim. This only proves that $M^{+}\vdash\alpha$ or $M^{+}\vdash\neg\alpha$. It doesn't show that only one of these is true. If you can show $M^{+}\nvdash\bot$, that would establish it though. Alternatively, you could modify the above proof fairly easily to cover the exact claim given an assumption (or proof) of consistency, i.e. $\nvdash\bot$.
