Usually an idempotent function is $f: S \rightarrow S$ such that $f \circ f = f$. But I need something slightly different: $f: S \times S \rightarrow S$, such that:
For $x, y, z \in S$:
- $f(x, y) = f(y, x)$ --- this is just commutativity
- $f(x, x) = c(x)$ --- $c \,$ can be any function
- $f(x, f(x,x)) = f(x,c(x)) = c(x)$ --- to guarantee idempotence
- $f(x, f(y, z)) = f(f(x, y), z)$ --- associativity
In "infix" notation the requirements amounts to:
- $x + y = y + x$
- $x + x = c(x)$
- $x + x + x = x + c(x) = c(x)$
- $x + (y + z) = (x + y) + z$
Under my definition, $f$ is not an idempotent function as it doesn't make sense to write $f \circ f$.
So I try to visualize the graph of $f(x,y)$. Axiom 1 requires that the graph is symmetric along $x = y$. Axiom 2 is automatically satisfied if $f$ is a single-valued function along $x = y$.
My problem is to understand axiom 3 (including its geometric interpretation). What kind of constraint does it impose on $f$?
Thanks!