how to prove or disprove "If a formula is not logical consequence of a set of formulas, its negation is."? The definition of Logical Consequence that my Logical Analysis teacher gave us is the following:
"Let C = {$F_1, F_2, ..., F_n$}  be a set of formulas and Q a formula.
You say Q is logical consequence of set C if every interpretation that is a model on C is also a model in Q."
Now I want to prove or disprove that if a formula is not logical consequence of a set of formulas, its negation is.
I know that $V_I(F) = 1$   , is a model for formula F.
But how do I even begin the demonstration?
How do I prove or disprove it?
 A: Let $C$ be the axioms for a group and let $Q$ be $\forall x\;\forall  y\;(xy=yx).$ There are models of $C+Q$ (e.g. a one-element group) and models of $C+\neg Q$ (e.g. the permutation group $S_3$) so neither $Q$ nor $\neg Q$ is a consequence of $C.$
A: Consider the definition of logical consequence and apply it to "the formula $Q$ is not a logical consequence of the set $\mathcal C$".
This means that there is a truth assignment $v$ such that $v(F_i)=$ t for every $F_i \in \mathcal C$ and $v(Q)=$ f. (Please note "there is a $v$".)
To say that $\lnot Q$ is a logical consequence of $\mathcal C$ means that for every truth assignment $v$ we have that: if $v(F_i)=$ t, for every $F_i$, then $v(\lnot Q)=$ t, i.e. $v(Q)=$ f.
This is not as what we hav found above.
Thus, we suspect that the result does not hold.
To disprove it, we have to manufacture a suitable counterexample.
A simple one is:

$\mathcal C = \{ p \lor q \}$ and $Q= p$.

We have that: $p \lor q \nvDash p$ but also $p \lor q \nvDash \lnot p$.
For the first result, consider a truth assignment $v$ such that $v(p)=$ f and $v(q)=$ t.
For the second one, consider a truth assignment $v$ such that $v(p)=$ t.
