A symmetric function While working on a research problem on fuzzy metric spaces, I came across a special symmetric function $F_n:X^n\times (0,\infty)\to [0,1]$  i.e.
\begin{equation*}
F_n(x_1,x_2,\dots,x_n,t)=F_n(x_{\pi(1)},x_{\pi(2)},...,x_{\pi(n)},t) 
\end{equation*}
for every permutation $\pi$ of $\{1,2,...,n\}$ such that
\begin{equation*}
[F_n(x_1,x_2,...,x_n,t)]^{n-2}=\prod_{1\le i_1<i_2<\dots<i_{n-1}\le n} F_{n-1}(x_{i_1},x_{i_2},\dots x_{i_{n-1}},t)
\end{equation*}
Where $[F_n]^m=F_n\ast F_n\ast\dots(m \quad\text{times})$, $\ast$ being continuous $t$-norm.
I am interested in finding a relation between  $F_n(x_1,x_2,\dots,x_n,t)$ and $F_2(x_i,x_j,t), (1\le i<j\le n$). Any suggestions on how to approch the problem?
Note: Here is a somewhat similar situation in graph theory. if we take $F_n$ as the sum of all distances $d(x_i,x_j),1\le i<j\le n$ between different pairs of vertices $x_i$ and $x_j$ of a complete graph $K_n$  with vertex set $X=\{x_1,x_2,\dots,x_n\}$. Then $F_n:X^n\to \mathbb{R}$ represents a symmetric function in variables $x_1,x_2,\dots,x_n\in X$ such that
\begin{equation} 
(n-2)F_n(x_1,x_2,...,x_n)=\sum_{1\le i_1<i_2<\dots<i_{n-1}\le n} F_{n-1}(x_{i_1},x_{i_2},\dots x_{i_{n-1}})
\end{equation}
And We have
\begin{equation} 
F_n(x_1,x_2,...,x_n)=\sum_{1\le i<j\le n} F_2(x_i,x_j)
\end{equation}
 A: We prove that for each $t$
$\mbox{(1) } F_n(x_1,x_2,...,x_n,t)]^{(n-2)!}=\left[\prod_{1\le i<j\le n} F_2(x_i,x_j,t)\right]^{(n-2)!}$
even without the symmetry assumption by straight induction by $n$, starting from $n=3$. For $n=3$ Claim (1) is the same as the condition connecting functions $F_{n}$ and $F_{n-1}$ (I understand the products with respect to the binary operation $*$). Assume that we already have proved the Claim (1) for a number $n-1$. Then 
$$[F_n(x_1,x_2,...,x_n,t)]^{(n-2)!}=\left[\prod_{1\le i_1<i_2<\dots<i_{n-1}\le n} F_{n-1}(x_{i_1},x_{i_2},\dots x_{i_{n-1}},t)\right]^{(n-3)!}=
\prod_{1\le i<j\le n} [F_2(x_i,x_j,t)]^{ (n-2)!}$$ (this follows from the commutativity and associativity of the operation $*$) and a remark that each pair $(i,j)$ ($i<j$) is in exactly $n-2$ sequences $(i_1,\dots,i_{n-1})$ where $1\le i_1<\dots<i_{n-1}\le n$, that is in all such sequences except $(1,\dots,\hat{i},\dots,j,\dots,n-1,n)$ and $(1,\dots, i,\dots, \hat{j},\dots,n-1,n)$, where $\hat{k}$ means that the number $k$ is skipped from the sequence. 
If for each $n>0$ a function $a\mapsto [a]^n$ is strictly monotone then we can drop the powers and obtain that
$$F_n(x_1,x_2,...,x_n,t)=\prod_{1\le i<j\le n} F_2(x_i,x_j,t).$$
