What property of a function describes $f\left(f\left(f\left(\phi, x\right), y\right), z\right) == f\left(\phi, z\right)$?

Question

How would you describe a function that has the relationship $f\left(f\left(f\left(\phi, x\right), y\right), z\right) == f\left(\phi, z\right)$? given that $f\left(\phi, x\right)$ is idempotent (i.e., $f\left(f\left(\phi, x\right), x\right) == f\left(\phi, x\right)$?

Is this simply a property of idempotent functions?

An application of this function can be seen in a programming language where a method object.SetProperty(x) changes the state of the object. However, subsequent calls to SetProperty() overwrites the existing state. Thus the following are considered equivalent:

object = new Object;
object.SetProperty(x);
object.SetProperty(y);
object.SetProperty(z);

and

object = new Object;
object.SetProperty(z);

Answer 1

I think the property you are looking for is the following: $$ f(f(\phi, x),y) = f(\phi,y) \text{ for all } x, y. \tag{1} $$ If $f$ satisfies (1), then this both implies that $f$ is idempotent (by your definition) and also implies your condition, $f\left(f\left(f\left(\phi, x\right), y\right), z\right) = f\left(\phi, z\right)$.

So, I think (1) is the condition you want. I am not aware of a name for this property, except that it is similar to something called a "right zero" or "right-absorbing-element". So, I would suggest that if $f$ satisfies (1), then you could call $f$ right-absorbing. This is meant to refer to the fact that the right-most applied argument $y$ "absorbs" the other applied argument $x$.

Is this simply a property of idempotent functions?

No, the property (1) is not true of all idempotent functions. For example, if $\phi$ is a set and $f(\phi, x)$ represents adding the element $x$ to $\phi$ (phi.add(x)), this is idempotent (because adding the same element twice to a set is the same as adding it once), but not right-absorbing (because adding $x$ and then adding $y$ isn't the same as just adding $y$).

Answer 2

I have never heard of a particular name for that property in that context, however in probability there is the adjective memoryless which corresponds to this idea.

See Poisson / Exponential variables, as well as Markov processes.