Relation Isomorphism preserves Lattice Structure

Theorem
Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.

Let $\struct {A, \RR}$ be a lattice.

Then $\struct {B, \SS}$ is also a lattice.

Proof
Let $\struct {A, \RR}$ be a lattice.

Recall the definition:

From Relation Isomorphism preserves Ordering:
 * $\SS$ is an ordering on $B$.

Let $\phi: \struct {A, \RR} \to \struct {B, \SS}$ be a relation isomorphism.

We need to show that for all $x, y \in A$, the ordered set $\struct {\set {\map \phi x, \map \phi y}, \SS}$ admits both a supremum and an infimum.

Let $x, y \in A$.

Then $\struct {\set {x, y}, \RR}$ admits both a supremum and an infimum.

Let $c = \map \sup {\set {x, y}, \RR}$.

Then by definition of supremum:
 * $\forall a \in \set {x, y}: a \mathrel \RR c$
 * $\forall d \in A: c \mathrel \RR d$

where $d$ is an upper bound of $\struct {\set {x, y}, \RR} \subseteq A$.

Now consider the image of $\set {x, y}$ under $\phi$.

By definition of relation isomorphism:
 * $\forall \map \phi s \in \set {\map \phi x, \map \phi y}: \map \phi c \mathrel \SS \map \phi s$
 * $\forall \map \phi d \in B: \map \phi c \mathrel \SS \map \phi d$

where $\map \phi d$ is an upper bound of $\struct {\set {\map \phi x, \map \phi y}, \SS} \subseteq B$.

So by definition of supremum:
 * $\map \phi c = \map \sup {\set {\map \phi x, \map \phi y}, \SS}$

That is, $\struct {\set {\map \phi x, \map \phi y}, \SS}$ admits a supremum.

Using a similar technique it can be shown that:
 * If $c = \map \inf {\set {x, y}, \SS}$, then:
 * $\map \phi c = \map \inf {\set {\map \phi x, \map \phi y}, \SS}$

Hence $\struct {\set {\map \phi x, \map \phi y}, \SS}$ admits both a supremum and an infimum.

That is, $\SS$ is a lattice ordering and so $\struct {B, \SS}$ is a lattice.