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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

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

$\mathcal{A} \models \phi[s]$ iff $\forall a \in \mathbb{N}, \exists b \in \mathbb{N}, (Sv_1

Now, I have no idea about the next step. Any help?

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    In order to prove $\forall$ you have to consider **all** variable fucntions $s$; thus, you have not to "impose" some rule on $s$. The argument needs only that the - with domain $\mathbb N$ - the unary function $S$ is interpreted as the "successor" function : $n+1 : \mathbb N \to \mathbb N$. – 2017-01-13
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    Thus, whatever the variable function $s$ is, if $s(v_1)=n$, then : $\mathcal A \vDash ϕ[s]$ iff $∀n \in \mathbb N,(∃v_2)(Sv_1 < v_2)s[v_1 \mid n]$. In order to satisfy it, we have to find a $v_2$-variant of $s$ (call it $s'$) such that $s'(v_2)=m$ and $n+1 < m$, and this we can for sure do. – 2017-01-13
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    @MauroALLEGRANZA: Thanks for the points. – 2017-01-13

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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.