How to prove a $\mathcal{L}$-structure holds satisfaction for a $\mathcal{L}$-formula? I'm just trying to come up with following problem:
Assume language $\mathcal{L}$ be $\{S,<\}$, where $S$ is a unary function and $<$ is a binary relation. Let $\phi$ be the formula $(\forall x)(\exists y)(Sx<y)$.
A) Find an $\mathcal{L}$-structure $\mathcal{A}$ such that $\mathcal{A} \models \phi$.
B) Prove it.

Here is my solution:
A:
I consider my candidate $\mathcal{L}$-structure as following: I assume function $S$ works like a successor function on $\mathbb{N}$, mathematically $S^{\mathcal{A}}: n \rightarrow n+1, n \in \mathbb{N}$. Furthermore, $<$ is just like a comparing operator on $\mathbb{N}$: $<^{\mathcal{A}}$.
I define a variable function $s$ as following $s: v_i \rightarrow 2i$ and replace $x$ and $y$ with $v_1$ and $v_2$, respectively. Then, applying $s$ onto the $\mathcal{L}$-formula leads to:
$s(Sv_1) = S^{\mathcal{A}}(v_1^{\mathcal{A}}) = S^{\mathcal{A}}(2) = 3$
$s(v_2) = v_2^{\mathcal{A}}= 4$
as $3<4$ then by my candidate $\mathcal{L}$-structure, $\mathcal{A} \models \phi $
B:
Now I must prove the results of prior section:
$\mathcal{A} \models \phi[s]$ iff $\forall a \in \mathbb{N}, (\exists v_2) (Sv_1<v_2) s[v_1|a]$
$\mathcal{A} \models \phi[s]$ iff $\forall a \in \mathbb{N}, \exists b \in \mathbb{N}, (Sv_1<v_2) s[v_1|a] s[v_2|b]$
Now, I have no idea about the next step. Any help?
 A: This doesn't directly answer how to finish the proof you've started, but I think it's worth pointing out as it does answer the given question overall:
There is a much simpler solution to this problem.  Choose as your structure $\mathcal{A}$ with universe $A$ the 3 element set $\{a,b,c\}$ with both $S_\mathcal{A}$ and $<_\mathcal{A}$ equal to $\{(a,b) (b,c) (c,a)\}$.
The proof can just be explicitly showing in a table that for each $x \in A$, there's a $y \in A$ such that $(S(x),y) \in <_\mathcal{A}$:
$$
\begin{array}{cc|c}
x & y & S(x) & (S(x),y) & (S(x),y) \in <_\mathcal{A} \\
\hline
a & c & b & (b,c) & T\\
b & a & c & (c,a) & T\\
c & b & a & (a,b) & T\\
\end{array}
$$
Thus $\mathcal{A} \models \phi$
In fact, even a structure with one element $a$, with $a < a$ and $S(a) = a$ satisfies $\phi$ (technically even a totally empty structure satisfies $\phi$ but that might be a little too trivial).  Your structure would be more fitting if it was required to be infinite, but at least as you've written it there's no such requirement.
