Name of the monotone-like property "$f(x)\geq f(x\wedge y)$ implies $f(x\vee y)\geq f(x)$" on a lattice

Question

Let $(X,\leq)$ be a lattice. There is a function $f$ that has the following property.

$f(x)\geq f(x\wedge y)$ implies $f(x\vee y)\geq f(x)$

where strict inequality on the left-hand side implies strict inequality on the right-hand.

At a first glance, the property looks like some type of monotonicity as $f$ should be increasing in its arguments.. but it is more general than the monotonicity.

Anyone happens to know the name of this property?

Answer 1

Let me rewrite the question here:

Let $(X,\leq)$ be a lattice. There is a function $f$ that has the following properties:

1. $f(x)\geq f(x\wedge y)$ implies $f(x\vee y)\geq f(x)$,
2. strict inequality on the left-hand side implies strict inequality on the right-hand side.

Claim I. A function $f$ satisfies Condition 1. if and only if $f$ is a monotone function.
Claim II. A function $f$ satisfies both 1. and 2. if and only if $f$ is a constant function.

Reasoning for Claim 1.
Assume that $f$ satisfies the implication in Condition 1. We have $$\begin{array}{rll} y\geq x&\to x=x\wedge y&\\ &\to f(x)=f(x\wedge y)&\textrm{($f$ is a function)}\\ &\to f(x\vee y)\geq f(x)&\textrm{(Condition 1.)}\\ &\to f(y)\geq f(x)&\textrm{($x\vee y =y$)}. \end{array} $$ This shows that $y\geq x$ implies $f(y)\geq f(x)$. Conversely, if $f$ is any monotone function with domain $X$, then since $x\vee y\geq x$ we have $f(x\vee y)\geq f(x)$. This shows that the conclusion of the implication in Condition 1. always holds for monotone functions.

Reasoning for Claim 2.
To prove that a function $f$ satisfying Conditions 1. and 2. is constant, it suffices to prove that $f(x)=f(y)$ whenever $x,y\in X$ are comparable. (Justification: Suppose that we have shown that $f(x)=f(y)$ when $x$ and $y$ are comparable. Given arbitrary $u, v\in X$, both $u$ and $v$ are comparable with $u\wedge v$, so $f(u) = f(u\wedge v) = f(v)$, implying that $f$ is constant.)
We argue by contradiction. Assume that $f$ satisfies Conditions 1. and 2. but that $X$ contains $x\geq y$ for which $f(x)\neq f(y)$. Necessarily $x>y$. By Condition 1., $f$ is monotone, so we must have $f(x)>f(y)\;\;(=f(x\wedge y))$. Applying Conditions 1.+2. to $f(x)>f(x\wedge y)$, we must have $f(x\vee y)>f(x)$. But $x=x\vee y$, so this reduces to $f(x)>f(x)$, which is false. This contradiction proves Claim 2.