Equivalence Between Sentences in $\mathcal{L}^{w}_{II}$ and in $\mathcal{L}_{\omega_1 \omega}$ I am trying to solve a logic question from the Ebbinghaus book "Mathematical Logic". 
Let  $\mathcal{L}^{w}_{II}$ be the system corresponding to Weak Second-Order Logic, where quantification is only allowed over finite sets (and relations). $\mathcal{L}_{\omega_1 \omega}$ is first-order logic equipped with infinite disjunction. 
Question 2.7 of Chapter IX - Extension of First-Order Logic says: 

Show that for every $\mathcal{L}^{w}_{II}$-sentence $\phi$ there is an 
  $\mathcal{L}_{\omega_1 \omega}$-sentence $\psi$ with the same models (that is, for all $S$-structures $\mathcal{A}$, $\mathcal{A} \vDash_{w} \phi$ iff $\mathcal{A} \vDash \psi$. Conclude from this that the Löwenheim-Skolem Theorem holds for $\mathcal{L}^{w}_{II}$. 

Initially, I thought about doing a proof by induction on the structure of the sentence $\phi$ and remembered that this may be problematic: for instance, if $\phi := \exists x \phi'$, then $\phi$ is a sentence, but $\phi'$ may not be a sentence, thus blocking us from using the inductive hypothesis. My next attempt was to try to generalize the hypothesis for arbitrary formulas, but I don't know how to do that in this specific case: it appears to me that an assignment in $\mathcal{L}^{w}_{II}$ assigns standard variables AND relation variables, while an assignment in $\mathcal{L}_{\omega_1 \omega}$ just assigns standard variables. 
How do I use induction in this case? 
For the inductive step, I believe the difficulty of the problem is in the case where we want to express: 

$\mathcal{I} \vDash_{w} \exists X^n \phi$ $\quad$ :iff there is a finite $C \subset A^n$ such that $\mathcal{I} \frac{C}{X^n} \vDash \phi$ 

using an infinite disjunction. I believe this can be done as:

C is finite $\iff$ C has one element $\lor$ C has two elements $\lor$ $\ldots$ $\lor$ C has $n$ elements $\lor \ldots$

Is this the correct path to take?
Thanks in advance. 
 A: You're right that you want to prove this by induction over arbitrary formulas, and you're right that free second order variables cause a problem. 
So instead let's prove by induction on the number of second-order quantifiers that for every $\mathcal{L}^w_{II}$-formula $\varphi(x_1,\dots,x_n)$ with no free second-order variables, there is an equivalent $\mathcal{L}_{\omega_1,\omega}$-formula $\varphi'(x_1,\dots,x_n)$. Here equivalent means that $M\models \varphi(a_1,\dots,a_n)$ if and only if $M\models \varphi'(a_1,\dots,a_n)$ for every $S$-structure $M$ and every tuple $(a_1,\dots,a_n)\in M^n$. 
The first-order formula building operations can be handled trivially, since $\mathcal{L}_{\omega_1,\omega}$ is also closed under these operations. So it suffices to consider the case of $\exists X\, \varphi(X,y_1,\dots,y_m)$, where $X$ is an $n$-ary relation variable. We can't apply induction to $\varphi(X,y_1,\dots,y_m)$ directly, since it has a free relation variable, so we have to rewrite our formula first. Here's were we can use your idea about cardinality. 
We want to take a disjunction over all the possible sizes of $X$. For each size $k$ of $X$, we can replace $\exists X$ by $\exists \overline{x}_1\dots\exists \overline{x}_k$, where each $\overline{x}_i = (x_{i,1},\dots,x_{i,n})$ is an $n$-tuple of variables. Then we can go into the formula and look for atomic instances of $X$. Everywhere we see  $X(t_1,\dots,t_n)$, we can replace this atomic formula by $\bigvee_{i=1}^k \overline{t} = \overline{x}_i$, by which I mean $$\bigvee_{i=1}^k \bigwedge_{j = 1}^n (t_j = x_{i,j}).$$ Altogether, we have $$\exists X\, \varphi(X,y_1,\dots,y_m) \iff \bigvee_{k\in \omega} \exists \overline{x_1}\dots\exists \overline{x_k}\, \varphi\left(\bigvee_{i=1}^k \overline{z} = \overline{x}_i,y_1,\dots,y_m\right)$$
Now each disjunct in this infinite disjunction is a $\mathcal{L}^w_{II}$-formula with no free second order variables and one less second-order quantifier, so we can apply the inductive hypothesis to each disjunct, and we're done. 
