Intuition of the function defined in first order logic According to first-order logic, $\models_{\mu} \forall x \phi [s]$ if and only if for every $d \in |\mu|$, we have $\models_{\mu} \phi [s(x|d)]$, where 
$$ s(x|d) = \left\{
     \begin{array}{lr}
       s(y) & : y \neq x\\
       d & : y=x
     \end{array}
   \right.$$
Question: What is the intuition behind the function $s(x|d)$? 
From what I know, $\models_{\mu} \forall x \phi [s]$ means that for any $x$ given, the well-formed formula $\phi$ must be true with regard to $s$. 
But from the definition, I don't see the connection here. 
 A: The "device" represented by the function $s$ and its "variation" $s(x|d)$ is needed in order to translate into the terms of the semantical specification the condition for the assertability of : $\forall x \varphi(x)$.
The function $s : V \to |\mathfrak A|$ maps variables of the language to elements of the domain of the interpretation, i.e. into objects; we can say that the function $s$ assign a "temporary" reference to the variables.
Thus $s(x) \in |\mathfrak A|$.
If, e.g., we consider as domain $\mathbb N$, then $s(x)=n$ means that the fucntion $s$ maps the variable $x$ to the number $n$.
Here are some examples of $s : V \to \mathbb N$ :

$s(v_i)=0$, i.e. all the variables are mapped to the number zero
$s(v_i)=i$, i.e. $s(v_1)=1, s(v_2)=2, \ldots$
$s(v_i)=2^i$, i.e. $s(v_1)=2, s(v_2)=4, \ldots$.

This "device" is used into the semantical clause of Enderton's book [page 83] :

$\mathfrak A \vDash Pt_1[s]$ iff $\overline s(t_1) \in P^{\mathfrak A}$,

i.e. we say that the structure $\mathfrak A$ satisfy the atomic formula $Pt_1$ with $s$ when the "temporary" reference $\overline s(t_1)$ "induced" by $s$ to the term $t_1$ [see the inductive definition : page 83] is an objcet of the domain of $\mathfrak A$ such that $P^{\mathfrak A}$ holds of it.
Thus $s(x|d)$, with $d$ element of the domain, is the fucntion $s'$ that coincides with $s$ except eventually for the objcet corresponding to $x$, that has been "changed" to $d$.
When we say :

$\mathfrak A \vDash ∀x \varphi[s]$ if and only if for every $d \in |\mathfrak A|$, we have $\mathfrak A \vDash \varphi[s(x|d)]$,

we express the fact that, whatever will be the object $d \in |\mathfrak A|$ assigned as reference to $x$, $d$ will satisfy $\varphi$ in $\mathfrak A$, i.e.

$d \in \varphi^{\mathfrak A}$.

Thus, if whatever will be the object $d \in |\mathfrak A|$, $d$ will satisfy $\varphi$ in $\mathfrak A$, this means that $\forall x \varphi(x)$ is true in $\mathfrak A$.
