Let $(X, \land, \lor)$ be a Distributive and Limited lattice.

Given $x \in X$ , we seek to prove that, to any $y, y^{'} \in X$ such that $x ⊥ y$ and $x ⊥ y^{'}$ implies that $y = y^{'}$ .

We have that

⎩ ⎨ ⎧ x \land y x \lor y x \land y^{'} x \lor y^{'} = 0 = 1 = 0 = 1

1) $y^{'} \leq y$

x \lor y = 1 ⟹ (x \lor y) \land y^{'} = y^{'} = (x \land y^{'}) \lor (y \land y^{'}) = y \land y^{'}

Therefore, $y^{'} = y \land y^{'} ⟹ y^{'} \leq y$ , by the Partial Ordering induces by the lattice.

2) $y \leq y^{'}$

x \lor y^{'} = 1 ⟹ (x \lor y^{'}) \land y = y = (x \land y) \lor (y^{'} \land y) = y^{'} \land y

Therefore, $y = y \land y^{'} ⟹ y \leq y^{'}$ .

Thus, $y = y^{'}$ , and every point in $X$ has a unique complement.