A system is said to be dualised if there is an operator * that determines for each feature f a dual feature f*, such that f** = f, and such that any statement S about a subset of these features, say S = s(a, b, c, ...) remains true if each feature is replaced by its dual. That is S* = s(a*, b*, c*, ...) is also true.
A lattice is a set L equipped with two binary operations n and u which are each commutative [x n y = y n x, x u y = y u x] and associative [x n (y n z) = (x n y) n z, x u (y u z) = (x u y) u z] and each is absorptive with respect to the other [x n (x u y) = x, x u (x n y) = x]. From the symmetry of these laws the system is evidently dualised.
Theorem: The lattice operations n and u are idempotent. Proof: In the first absorptive law replace y by x n z, then x = x n (x u (x n z)) = x n x by the second absorptive law. Dually x = x u x.
Theorem: The relation x R y defined by x n y = y that is ‘x is an n-left-identifier of y’ or ‘y is an n-right-absorber of x’ is an ordering. Proof: (1) By idempotency x = x n x, that is x R x, so R is reflexive. (2) If x R y and y R x this means x n y = y and y n x = x so by commutation x = y. So R is antisymmetric. (3) If x R y and y R z this means x n y = y and y n z = z, so x n z = x n (y n z) = (x n y) n z = y n z = z, using associativity. So R is transitive.
The operations u and n are those of finding the greatest lower bound and least upper bound with respect to R. Dually the relation x S y defined by x u y = y is an order relation.
A lattice in which x ≥ z implies x n (y u z) = (x n y) u z for all x, y, z; or dually x ≤ z implies x u (y n z) = (x u y) n z for all x, y, z, is a modular lattice. This postulate is self-dual so modular lattices remain dual systems.
Theorem: A lattice is modular if and only if it contains no sublattice isomorphic to the pentagonal lattice {m < a < w; m < b < c < w}. Proof: (1) The pentagonal lattice is non-modular since c n (a u b) = c and (c n a) u b = b and b ≠ c, and any lattice containing this will be nonmodular. (2) If a lattice is nonmodular then we can find x, y, z with x > z and x n (y u z) ≠ (x n y) u z, whence by the modular inequality x n (y u z) > (x n y) u z. Thus the elements {y n x < y < y u z; y n x < (x n y) u z < x n (y u z) < y u z} form a pentagonal sublattice.
A distributive lattice is one in which each of the operations is distributive over the other, i.e. x n (y u z) = (x n y) u (x n z) and x u (y n z) = (x u y) n (x u z). It follows that duality is preserved.
Theorem: A distributive lattice is modular. Proof: If x ≥ z then x n z = z and substituting in the first distributive law we get x n (y u z) = (x n y) u z. Dually if x ≤ z we have x u z = z and substitution in the second distribitive law also gives the dual modular equality x u (y n z) = (x u y) n z.
Observation: For a lattice to be distributive it is only necessary to postulate one of the distributive laws, because the other can then be deduced. In fact either distributive law is equivalent to the single self-dual distributive law: (x n y) u (y n z) u (z n x) = (x u y) n (y u z) n (z u x). (Proof?)
Theorem: A modular lattice is distributive if and only if it has no sublattice isomorphic to the 5-element tetrahedral lattice {m < a < w, m < b < w, m < c < w}. Proof: (1) In this lattice we find (a n b) u (b n c) u (c n a) = m and (a u b) n (b u c) n (c u a) = w and m ≠ w, so it is not distributive and no lattice containing it is distributive. (2) Conversely if the lattice is not distributive there must be some elements x, y, z for which the self-dual axiom is false, in which case the sublattice {m = (x n y) u (y n z) u (z n x), a = (m n x) u w, b = (m n y) u w, c = (m n z) u w, w = (x u y) n (y u z) n (z u x)} is of the above type.
Theorem: A lattice is distributive if and only if ‘a n b = a n c and a u b = a u c implies b = c’. Proof: (1) If the lattice were non-distributive it would contain a five-element sublattice isomorphic to the pentagonal or tetrahedral lattice, in neither of which is b = c, although a n b = a n c and a u b = a u c. (2) Conversely a n b = a n c and a u b = a u c implies b = b u (a n b) = b u (a n c) = (b u a) n (b u c) = (a u b) n (b u c) = (a u c) n (b u c) = (a n b) u c = (a n c) u c = c.
In a terminate lattice there exist a first element o and a last element i such that o ≤ x ≤ i for all elements x of the lattice. If such elements exist they are unique. The first and last elements have the properties o n x = o, o u x = x, i n x = x, i u x = i.
Theorem: If x u y = o then x = y = o and if x n y = i then x = y = i. Proof: x = x n (x u y) = x n o = o and similarly for y and dually for u.
An element y is said to be independent of the set X if y n (uX) = o, where uX = u{x, x', x'', ..} = x u x' u x'' u ... Otherwise y is dependent on the set. And the set X forms an independent set if each element of X is independent of the others, i.e. x n u(X − {x}) = o.
Theorem: In a modular lattice, if y is independent of an independent set X then {y} ∪ X is an independent set. Proof: Since y is independent of X it is only necessary to show that x is independent of the set X − {x} ∪ {y}. (???)