Let's say that it is true that: a conclusion A is a logical consequence of a set of premises T iff negation of A is in T and T is not finitely performable. Formally written: (T ⊨ A) ⇔ T ⋃ {¬A} is not satisfiable.
My question is the following: What Is the logical consequence of saying that It is NOT TRUE that the conclusion A is a logical consequence of a set of premises T? Is the logical consequence that T ⋃ {¬A} is always satisfiable?
I'll start from the end and answer your question - no, that's not the logical consequence. I will now explain why.
Your theorem concerning the relation between entailmaent is correct. Let's write it again.
$ T \models A \Leftrightarrow T\cup \{ \neg A\} $ is not satisfiable
This is logically equivalent to:
$ \neg ( T \models A ) \Leftrightarrow \neg ( T\cup \{ \neg A\} $ is not satisfiable )
Meaning:
$ \neg ( T \models A ) \Leftrightarrow ( T\cup \{ \neg A\} $ is satisfiable )
Now the thing is to bear in mind what satisfinability is: A set of formulas is satisfable $ \Leftrightarrow $ there exists a model in which all the formulas receive as truth-value of true in that model. In our case - there exists a model where T receives a True truth-value while $ \neg A $ received a truth value of True also. Which means that $A$ receives a truth-value of False. Which is clearly our definition when the premises do not entail the consequence in an argument.
Hope this helps.