Satisfiability Proof Question Exercise: Prove that $\Gamma\models A$
  iff $\Gamma\cup\{\neg A\}$
  is not satisfiable.
Proof: We must prove two clauses:


*

*$\Gamma\models A\Rightarrow
  \Gamma\cup\{\neg A\}$
is not satisfiable

*$\Gamma\cup\{\neg A\}$
is not satisfiable $\Rightarrow\Gamma\models A$
Suppose (1) is not true. In such a case, we could make the following claim (by distributing the negation): $\Gamma\models A\text{ and }\Gamma\cup\{\neg A\}\text{ is satisfiable}$
This would require the existence of a valuation v
  where $v^{*}(A)=\text{T}$
  and $v^{*}(\neg A)=\text{T}$
 , or -- simplified -- $v^{*}(A)=\text{F}$
 . But a sentence cannot be both T and F under the same valuation v
  of $\mathcal{L}$
 , so (1) must be true.
Suppose now that (2) is not true. In such a case, we could make the following claim (by distributing the negation): $\Gamma\cup\{\neg A\}\text{ is not satisfiable}\text{ and }\Gamma\nvDash A$
Since $\Gamma\cup\{\neg A\}$
  is not satisfiable, there must be some S
  where $S\in\Gamma$
 , such that $\{S\}\cup\{\neg A\}$
  is not satisfiable. For $\{S,\neg A\}$
  to be unsatisfiable, S
  must be A
 . Trivially, then, $A\in\Gamma$
  and so, $\Gamma\models A$. Therefore, (2) must be true. $\Diamond$
I only got 4/10 points for this proof and before I go to office hours I just wanted to ask if I did something obviously wrong/stupid (which might very well be the case). For what it's worth, here are my professor's notes: http://i.imgur.com/aNNNnsd.jpg?1
 A: I'll prove (2) not by contradiction but by contraposition. Assume $\Gamma$ doesn't entail $A$. Then there is at least one valuation that makes every sentence in $\Gamma$ true but $A$ false. But this same valuation makes every sentence in $\Gamma\cup\left\{¬A\right\}$ true. So $\Gamma\cup\left\{¬A\right\}$ is satisfiable.
A: For statement (1), where your professor noted that you need to be more explicit with your arguments, I guess he meant that you should go by the definitions:
By definition, $\Gamma\models A$ means that there is a valuation $v$ such that if for all formulas $B\in \Gamma$, $v^*(B)=\text{T}$, then $v^*(A)=\text{T}$, too.
On the other hand, $\Gamma\cup\{\neg A\}$ being satisfiable means by definition that there is a valuation $u$ such that $u^*(\neg A)=\text{T}\Rightarrow u^*(A)=\text{F}$. But $u$ is a valuation for which $\forall B\in\Gamma, u^*(B)=\text{T}$, so it should also valuate A as true, but it doesn't. Hence the contradiction.
A: Your proof isn't clear at all.
On one hand, $\Gamma \models A$ means by definition that every every models of $\Gamma$ satisfies $A$. On the other hand, $\Gamma \cup \{\neg A\}$ not satisfiable means that there is no model of $\Gamma \cup \{\neg A\}$. So the exercise is completely trivial once you know that a model can not satisfies a formula and its negation.
If you want to write it, it goes as : $\Gamma \models A$ iff for every structure $\mathcal M$, $\mathcal M \models \Gamma$ implies $\mathcal M \models A$ iff there is no structure $\mathcal M$ such that $\mathcal M \models \Gamma$ and $\mathcal M \models \neg A$ iff $\Gamma \cup \{\neg A\}$ is not satisfiable.
