Second order logic - what does it mean following formula $\phi$? $E$ is edge relation, however $inc(u, e)$ guarantees that $u$ is end of edge $e$. $V$ is set of verticles.
$$\phi = \exists_{C\subseteq E} span(C)$$
$$span(C) = \forall_{X\subseteq V} [\exists_{u\in V}\exists_{v\in V} (u\in X \wedge v\notin X) \to(\exists_{e\in C}\ \ inc(u,e) \wedge\ inc(v, e))]$$  
For me, it guarantees that each node is end at least of one edge in subset of edges $C$.  But, then I don't undersatnd why it is so complicated, it was sufficient to say:  $span(C) = \forall_{v\in V} \exists_{e\in C}(inc(v,e))$ 
 A: I think that formula is even simpler. In the inner existential quantifier we can set $v = u$ thus:
$$\mathtt{span}(C) = \forall_{X\subseteq V} [\exists_{u\in V} (u\in X \land u\notin X) \to(\text{whatever})]$$
that is, the precondition of the implication is false, and the whole is true as long as $V \neq \varnothing$. Thus $\mathtt{span}(C) \iff V \neq \varnothing$ (or just say it's always true because $V \neq \varnothing$ is a sane assumption).
What I think the autor meant is:
\begin{align}
\mathtt{span'}(C) = &\forall_{X\subseteq V} \Bigg(\\
  &\Big(\exists_{u\in V}\exists_{v\in V} (u\in X \land v\notin X)\Big) \to\\
  &\Big(\exists_{u\in V}\exists_{v\in V}\exists_{e \in C} \big(u\in X \land v\notin X\land \mathtt{inc}(u,e) \land \mathtt{inc}(v, e)\big)\Big)\Bigg)
\end{align}
That formula says a subset of edges $C$ satisfies property $\mathtt{span'}$ iff:


*

*$\forall X \subseteq V$ — for all partitions (one part is $X$ the other one is $V \setminus X$),

*$\exists u,v \in V.\ (u \in X \land v \in V \setminus X) \to \ldots$ — if that partition is not trivial, then...

*$\exists_{u\in V}\exists_{v\in V}\exists_{e \in C} \big(u\in X \land v\in V\setminus X\land \mathtt{inc}(u,e) \land \mathtt{inc}(v, e)\big)$ — there is an edge in $C$ that crosses that partition.


In other words, $\mathtt{span'}(C)$ iff for any non-trivial partition $C$ has an edge that (somehow) connects these parts. If a subset has this particular property, people usually say that it spans the graph.
That property is much more that what you suggest. Take graph $K_4$ and let $C = \{(1,2), (3,4)\}$, that is:
\begin{align}
&V = \{1,2,3,4\}, \\
&E = \big\{\{u,v\} \mid u,v \in V, u\neq v\big\}, \\
&\mathtt{inc}(v,e) \iff v \in e,\\
&C = \big\{\{1,2\}, \{3,4\}\big\}.
\end{align}
It satisfies your property because 
\begin{align}
&\mathtt{inc}\big(1,\{1,2\}\big) \\
&\mathtt{inc}\big(2,\{1,2\}\big) \\
&\mathtt{inc}\big(3,\{3,4\}\big) \\
&\mathtt{inc}\big(4,\{3,4\}\big) \\
\end{align}
but it does not satisfy $\mathtt{span'}$. Consider $X = \{1,2\}$. Then we can satisfy $\exists_{u\in V}\exists_{v\in V} (u\in X \land v\notin X)$ with $u = 1$ and $v = 3$, but there are no $u$ and $v$ that would make $$\exists_{u\in V}\exists_{v\in V}\exists_{e \in C} \big(u\in X \land v\notin X\land \mathtt{inc}(u,e) \land \mathtt{inc}(v, e)\big)$$ true.
I hope this helps $\ddot\smile$
