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$E$ 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: $span(C) = \forall_{v\in V} \exists_{e\in C}(inc(v,e))$

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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 $\ddot\smile$

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    Thanks you very much for such clear answer! Maybe, you would like help here? http://math.stackexchange.com/questions/2128595/eulerian-path-in-second-order-write-formula/2129289#2129289 Nobody helped in successful way. – 2017-02-10