Negation of logical statement in graph theory I am given the following statement i am going to attempt to translate it into something alitle easier for me to understand then negate it.
EDIT:
Statement:
For every set $ X \subset V(G) \smallsetminus \{ u,v \} $ either every vertex of $X$ is adjacent to $u$ and nonadjacent to $v$, or every vertex is adjacent to $v$ and non adjacent to $u$.
I believe what this is saying is:
For every set $ X \subset V(G) \smallsetminus \{ u,v \} $   then $\forall x \in X$,  $x$ is adjacent to $u$ or $v$ but not both.
Negation:
There exists a set $ X \subset V(G) \smallsetminus \{ u,v \} $ and $\exists x \in X $ s.t. $x$ is nonadjacent to $u$ and $v$ or $x$ is adjacent to $u$ and $v$.
Attempt to translate statement back:
There exists a set  $ X \subset V(G) \smallsetminus \{ u,v \} $ where there exists a vertex of $X$ that is adjacent to $u$ and $v$, or a vertex that is non adjacent to $u$ and $v$.
Does this all make sense?
 A: I don't mean to be punctilious, but I think a clarification is needed.
The statement "For every set $X \subseteq V(G) \smallsetminus \{u,v\}$ either every vertex of $X$ is adjacent to $u$ and nonadjacent to $v$, or every vertex [of $X$] is adjacent to $v$ and non adjacent to $u$" can be formalized by the formula $A$ as follows:
\begin{align}\label{first}
A \ = \ \forall \, X \subseteq V(G)\smallsetminus &\{u,v\} : \\
\forall x \big(x \in X &\to (Adj(x,u) \land \lnot Adj(x,v))\big) \lor \forall x \big(x \in X \to (\lnot Adj(x,u) \land Adj(x,v)) \big)
\end{align}
The statement "For every set $X \subseteq V(G) \smallsetminus \{u,v\}$ then $\forall x \in X$ $x$ is adjacent to $u$ or $v$ but not both" can be formalized by the formula $B$ as follows:
\begin{align}\label{second}
B \ = \ \forall \, X \subseteq V(G)\smallsetminus &\{u,v\} : \\
\forall x \Big(x \in X &\to \big((Adj(x,u) \land \lnot Adj(x,v)) \lor (\lnot Adj(x,u) \land Adj(x,v)) \big) \Big)
\end{align}
But the two formulas are not equivalent! More precisely, $A$ implies $B$ but $B$ does not imply $A$. This means that you could have a graph $G$ such that $B$ is true but $A$ is false.
For instance, think of a graph $G$ whose set of vertexes is $V(G) = \{u,v,x_1,x_2\}$ (with $x_1 \neq x_2$) and whose only edges are $\{x_1, u\}$ and $\{x_2, v\}$: then, $B$ is true but $A$ is false (take $X = \{x_1, x_2\}$).
The formula $A$ requires that all the vertexes different from $u$ and $v$ have to be adjacent to one and only one vertex among $u$ and $v$, and it has to be the same for all of them.
The formula $B$ just requires that every vertex different from $u$ and $v$ has to be adjacent to $u$ or $v$, not necessarily all to $u$ or all to $v$.
The negations of the formulas $A$ and $B$ (up to logical equivalences) are
\begin{align}\label{neg}
\lnot A \ = \ \exists \, X \subseteq V(G)\smallsetminus &\{u,v\} : \\
\exists x \big(x \in X &\land (\lnot Adj(x,u) \lor Adj(x,v))\big) \land \exists x \big(x \in X \land (Adj(x,u) \lor \lnot Adj(x,v)) \big)
\\
\\
\lnot B \ = \ \exists \, X \subseteq V(G)\smallsetminus &\{u,v\} : \\
\exists x \Big(x \in X &\land \big((\lnot Adj(x,u) \lor Adj(x,v)) \land (Adj(x,u) \lor \lnot Adj(x,v))\big) \Big)
\end{align}
The formula $\lnot A$ means that there exist a set $X \subseteq V(G)\smallsetminus \{u,v\}$ and vertexes $x,y \in X$ such that $x$ is either adjacent to $v$ or non-adjacent to $u$, and $y$ is either adjacent to $u$ or non-adjacent to $v$.
The formula $\lnot B$ means that there exist a set $X \subseteq V(G)\smallsetminus \{u,v\}$ and a vertex $x \in X$ such that:


*

*$x$ is either adjacent to $v$ or non-adjacent to $u$, and

*$x$ is either adjacent to $u$ or non-adjacent to $v$.


Clearly, $\lnot B$ implies $\lnot A$ but the converse does not hold, see the example of graph $G$ above. 
