Theorem: Let (L, ≤) be a lattice. If a, b, c ∈ L, then
(1) a ∨ (b ∧ c) ≤ (a ∨ b) ∧ (a ∨ c)
(2) a ∧ (b ∨ c) ≥ (a ∧ b) ∨ (a ∧ c)
These inequalities are called “Distributive Inequalities”.
Proof: We have
a ≤ a ∨ b and a ≤ a ∨ c............... (i)
Also, by the above theorem, if x ≤ y and x ≤ z in a lattice, then x ≤ y ∧ z. Therefore (i) yields
a ≤ (a ∨ b ) ∧ (a ∨ c)................... (ii)
Also
b ∧ c ≤ b ≤ a ∨ b
and
b ∧ c ≤ c ≤ a ∨ c ,
that is, b ∧ c ≤ a ∨ b and b ∧ c ≤ a ∨ c and so, by the above argument, we have
b ∧ c ≤ (a ∨ b) ∧ (a ∨ c).................. (iii)
Also, again by the above theorem if x ≤ z and y ≤ z in a lattice, then
x ∨ y ≤ z
Hence, (ii) and (iii) yield
a c (b ∨ c) ≤ ( a ∨ b) ∧ (a ∨ c)
This proves......... (1).
The second distributive inequality follows by using the principle of duality.
Theorem: (Modular Inequality) : Let (L, ≤) be a lattice. If a, b, c ∈ L, then a ≤ c if and only if a ∨ (b ∧ c) ≤ (a ∨ b) ∧ c
Proof: We know that
a ≤ c <-> a ∨ c = c................ (1)
Also, by distributive inequality,
a ∨ (b ∧ c) ≤ (a ∨ b) ∧ (a ∨ c)
Therefore using (1) a ≤ c if and only if
a ∨ (b ∧ c) ≤ (a ∨ c) ∧Ù c,
which proves the result.
The modular inequalities can be expressed in the following way also:
(a ∧ b) ∨ (a ∧ c) ≤ a ∧ [b ∨ (a ∧ c)]
(a ∨ b) ∧ (a ∨ c) ≥ a ∨ [b ∧ (a ∨ c)]
Example: Let (L, ≤) be a lattice and a, b, c ∈ L. If a ≤ b ≤ c, then
(i) a ∨ b = b ∧ c, (ii) (a ∧ b ) ∨ (b ∧ c) = (a ∨ b) ∧ ( a ∨ c).
Solution: (i) We know that
a ≤ b <-> a ∨ b = b
and
b ≤ c <-> b ∧ c = b
Hence a ≤ b ≤ c implies
a ∨ b = b ∧ c.
(ii) Since a ≤ b and b ≤ c, we have
a ∨ b = a and b ∨ c = b
Thus
(a ∨ b) ∧ (b ∨ c) = a ∧ b = b, since a ≤ b <-> a ∨ b = b.
Also, a ≤ b ≤ c a ≤ c by transitivity. Then a ≤ b and a ≤ c a ∨ b = b , a ∨ c = c
and so
(a ∨ b ) ∧ (a ∨ c) = b ∧ c = b since b ≤ c <-> b ∧ c = b.
Hence
(a ∧ b) ∨ (b ∧ c) = b = (a ∨ b) ∧ (a ∨ c),
which proves (ii).
