What is this property of functions called: $f(a,b,c,d)=g(x(a,b),y(c,d))$ In general (I didn't want to put this long formula in the title): $f(a_1,a_2,\dots,a_k)=g(x_1(a_{i_{1,1}},\dots,a_{i_{1,n_1}}),x_2(a_{i_{2,1}},\dots,a_{i_{2,n_2}}),\dots,x_l(a_{i_{l,1}},\dots,a_{i_{l,n_l}}))$, where $l<k$ and $a_{i_{1,1}}$ through $a_{i_{l,n_l}}$ are a permutation of $a_1$ through $a_l$.
Basically, $f$ can be "separated" into a function $g$ applied to the result of $l$ functions which take disjoint subsequences that form a partition of $f$'s arguments.
A concrete example: $f(a,b,c,d)=ab+cd$; here $g(x,y)=x+y$, $x(a,b)=ab$ and $y(c,d)=cd$. A function that wouldn't be separable like that would be $ab/c+cd$.
 A: You need to be more specific about what properties you want to require of the functions $g, x_i$.  If they can be totally generic (so we are working at the level of set theory), then every function has this property.  For instance, if we are working on $\mathbb{R}$, you could take
$x_1(a,b) = (a,b) \in \mathbb{R}^2$, $x_2$ similarly, and then think of $g$ as a function taking two inputs which are elements of $\mathbb{R}^2$ (which it can then pick apart and apply $f$ to).  In fact this is nothing but the associativity of the Cartesian product.
Even if you want all the functions to map back into $\mathbb{R}$, no generality is lost.  There exists a bijection $\phi$ from $\mathbb{R}^2$ to $\mathbb{R}$.  Then just take $x_i = \phi$, and $g(p,q) = f(\phi^{-1}(p), \phi^{-1}(q))$.
If you want to require all the functions in sight to have some sort of regularity property, such as continuity, then maybe we are getting somewhere... 
A: This operation on functions is composition.  I wouldn't call it a "property of" functions.
