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}. (???)