What is the definition of a congruence relation on an order-theoretic lattice? A congruence relation on a
(as an algebraic structure defined)
lattice $(X,\wedge,\vee)$
is an equivalence relation $\equiv$ on $X$
that preserves the lattice structure, i.e,
for all $x_1$, $x_2$ , $y_1$ and $y_2$ in $X$ it satisfies

*

*$x_1\equiv x_2$ and $y_1\equiv y_2$ implies $x_1\wedge y_1\equiv x_2\wedge y_2$,
and

*$x_1\equiv x_2$ and $y_1\equiv y_2$ implies $x_1\vee y_1\equiv x_2\vee y_2$.

See e.g. Grätzer's book "Lattice Theory: Foundation", page 36.
A lattice $(X,\wedge,\vee)$, defined as an algebraic structure,
gives rise to a lattice $(X,\leq)$
defined as a poset,
and vice versa,
by letting $x\leq y$
if and only if $x\wedge y = x$ and if and only if $x\vee y = y$,
and conversely
by letting $x\wedge y$ and $x\vee y$ be the greatest lower bound and least upper bound of $x$ and $y$, respectively
(see e.g. Wikipedia).
Due to this equivalence of a lattice as an algebraic structure and as a poset,
I would expect that there is a definition of a congruence relation on a lattice as a poset that is equivalent to the one above for lattices as an algebraic structure.
Is this indeed the case?
And if so, what does it look like?
My guess is that it is as follows:
for all $x_1$, $x_2$, $y_1$ and $y_2$ in $X$,

*

*$y_1\equiv y_2$ and $x_1\leq y_1$ and $x_2\leq y_2$ implies $x_1\equiv x_2$,
and

*$x_1\equiv x_2$ and $x_1\leq y_1$ and $x_2\leq y_2$ implies $y_1\equiv y_2$.

But is this correct?
I do not see how to derive these two conditions from the initial algebraic ones.
Suppose that the algebraic ones hold,
and that $x_1$, $x_2$, $y_1$ and $y_2$ are elements in $X$
satisfying $x_1\leq y_1$ and $x_2\leq y_2$.
Then $x_i\wedge y_i = x_i$ and $x_i\vee y_i = y_i$ for $i=1,2$.
Supposing $y_1\equiv y_2$,
then the desired $x_1\equiv x_2$ would follow if $x_1\wedge y_1\equiv x_2\wedge y_2$,
but the latter I only obtain if I take $x_1\equiv x_2$ to begin with.
Also, in the opposite direction I seem to have more than I need, which makes me think my guess is wrong:

*

*supposing $y_1\equiv y_2$ (but not necessarily $x_1\equiv x_2$),
then since $x_i\wedge y_i\leq y_i$ for $i=1,2$ it follows that
$x_1\wedge y_1\equiv x_2\wedge y_2$,
and

*supposing $x_1\equiv x_2$ (but not necessarily $y_1\equiv y_2$),
then since $x_i\leq x_i\vee y_i$ for $i=1,2$ it follows that
$x_1\vee y_1\equiv x_2\vee y_2$.

 A: That is not correct.
Consider a chain with $3$ (or more) elements, say
$$0 \prec a \prec 1.$$
There, you have the congruence $\theta = \Theta(a,1) = \{(0,0),(a,a),(a,1),(1,a),(1,1)\}$, and $0$ is not related either with $a$ or $1$, but $0 \leq a$ and $a \leq 1$, so your conjecture would yield $(0,a) \in \theta$.
Regarding your main question, you seem to seek a concept of congruence on a poset that would restrict to a lattice congruence if that poset happens to be a lattice.
Again, that doesn't exist.
Take for example, as a lattice $\mathbf L$ the one whose Hasse diagram is depicted next:

Its congruence lattice is isomorphic to the lattice itself (easy to check).
Now, drop one of the operations (either $\wedge$ or $\vee$, it doesn't matter) and you get a semi-lattice with the same Hasse diagram (so, the same poset), but its congruence lattice is the following one:

Hence the congruence lattice really depends on the operations.
More generally, if $\mathbf L$ is any lattice, then the congruence lattice of $\mathbf L$ is distributive.
But if $\mathbf L'$ is the semi-lattice that arises from dropping one of the lattice operations, then the congruence lattice of the semi-lattice $\mathbf L'$ is distributive iff $\mathbf L$ is a chain.
(This follows from a well-known result which states that a semi-lattice has a distributive congruence lattice iff that semi-lattice doesn't contain the four-element lattice above as a sub-semi-lattice.)
