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I very often have a problem with choice between $\to$ and $\wedge$ - when I solve some problem about writing formula in second order, but also in first order.

To be more precisely, lets consider formula in second order which express that each node has even degree, our graphs are finite and has no loops, are undirected.
Let $\phi(X)$ express that cardinality of set $X$ is even. It maybe expressed in second order logic with using only existentional quantifiers.

Lets look at:
(1)$\forall_{v\in V}\exists_{X} [(x\in X \to E(v,x)) \wedge \phi(X)]$
(2)$\forall_{v\in V}\exists_{X} [(x\in X \to E(v,x)) \to \phi(X)]$
After edition I consider also (3)
(3)$\forall_{v\in V}[[\exists_{X} (x\in X \to E(v,x))] \to \phi(X)]$

Believe me, to common sense I cant see a difference. However, in case of: $(x\in X \to E(v,x))$ I can see a sense - right side of implication is neccessary condition for left side of implication. And it seems to be ok, because $X$ should be set of adjacent nodes for $v$.
Of course, my formulas are not in second order, but they can be easily adapted. Can you help me with my doubts ?

Edition
$\phi'(X)$ should express that $X$ has even cardinality.
$\phi'(X) = \exists_{X_1\subseteq X}\exists_{X_2\subseteq X}\exists_f[(x\in X_1\to x\notin X_2)\wedge (x\in X_2\to x\notin X_1) \wedge \text{$f$ is bijection between $X_1$ and $X_2$}]$
$\phi'(X)$ is true iff $X$ is of even cardinality or is infinite. We should also force finitness of this set:
$\phi(X) = \phi'(X) \wedge (\forall_{x\in X}\neg\phi'(X\setminus\{x\})$

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    The difference is that if for every set $X$ the "condition" $(x∈X→E(v,x))$ is *false*, then the first formula is *false* while the second is *true* (irrespective of $\phi(X)$). – 2017-02-10
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    I try to read the first formula : for each vertex $v$ of graph $V$, there is the set $X$ such that : (the elements of $X$ are "connected" to $v$ **and** the cardinality of $X$ is even). – 2017-02-10
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    Typically, $\exists$ needs $\land$. – 2017-02-10
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    Ok, so correct sentence expressing that each node has even degree is (1), at least using your interpretation in second message – 2017-02-10
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    What about interpretation of (2) ? **for each vertex $v$ of graph $V$, if there is the set1 $X$ such that : *(if the elements of $X$ are "connected" then $X$ cardinality of $X$ is even*)**. – 2017-02-10
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    And look at (3), please: for each node $v$: if node $v$ has set of neighbours then it should be of even cardinality (in other words even cardinality is necessary condition for existence of set of neighbours) – 2017-02-10
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    The issue with 3) is that now $\phi(X)$ is no more in the scope of the existential quantifier. – 2017-02-10
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    Allright, so (1) is only possible formula. Can you show me how to adapt (1) for second order ? It is more difficult than I expected. – 2017-02-10
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    Second order ? Instead of $x \in X$ write $X(x)$ and the quantifier $\exists X$ will do the job. – 2017-02-10
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    @MauroALLEGRANZA I don't undertand. At the front of formula cant be quatinying over elements in first order. How to move $\forall_v$ behind $\exists_X$ ? – 2017-02-10
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    And what about correcntess of $\phi(X)$ ? I edited my question (at the bottom of message) – 2017-02-10
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53433/discussion-between-haskell-fun-and-mauro-allegranza). – 2017-02-10
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    About edit: $\phi(X)$ must express a property of set $X$; specifically, $X$ has an even number of elements (correct ?) If so, the formula defining it must have $X$ as free var. Compare with : $\text {Even}(n) := \exists z (n=2 \times z)$. – 2017-02-11
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    Corrected. yes, it is about even cardinality (as you remeber - the general aim is even degree of each node in graph) – 2017-02-11
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    @MauroALLEGRANZA what do you think now ? – 2017-02-11

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