DISTRIBUTIVE LATTICES FACT 1: For any lattice <A,≤>: 1 and 2 and 3 and 4 hold in <A,≤>: The distributive inequalities: 1. for every a,b,c ∈ A: (a ∧ b) ∨ (a ∧ c) ≤ a ∧ (b ∨ c) 2. for every a,b,c ∈ A: a ∨ (b ∧ c) ≤ (a ∨ b) ∧ (a ∨ c) 3. for every a,b,c ∈ A: (a ∧ b) ∨ (b ∧ c) ∨ (a ∧ c) ≤ (a ∨ b) ∧ (b ∨ c) ∧ (a ∨ c) The modular inequality: 4. for every a,b,c ∈ A: (a ∧ b) ∨ (a ∧ c) ≤ a ∧ (b ∨ (a ∧ c)) FACT 2: For any lattice <A,≤>: 5 holds in <A,≤> iff 6 holds in <A,≤> iff 7 holds in <A,≤>: 5. for every a,b,c ∈ A: a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) 6. for every a,b,c ∈ A: a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) 7. for every a,b,c ∈ A: a ∨ (b ∧ c) ≤ b ∧ (a ∨ c). A lattice <A,≤> is distributive iff 5. holds. FACT 3: For any lattice <A,≤>: 8 holds in <A,≤> iff 9 holds in <A,≤>: 8. for every a,b,c ∈ A:(a ∧ b) ∨ (a ∧ c) = a ∧ (b ∨ (a ∧ c)) 9. for every a,b,c ∈ A: if a ≤ b then a ∨ (b ∧ c) = b ∧ (a ∨ c) A lattice <A,≤> is modular iff 8. holds. FACT 4: Every distributive lattice is modular. Namely, let <A,≤> be distributive and let a,b,c ∈ A and let a ≤ b. a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) [by distributivity] = b ∧ (a ∨ c) [since a ∨ b =b]. The pentagon: The diamond: o 1 o 1 o z o y x o o y o z o x o 0 o 0 THEOREM 5: A lattice is modular iff the pentagon cannot be embedded in it. PROOF: 1. The pentagon is not modular. x ≤ y x ∨ ( y ∧ z) = x y ∧ (x ∨ z) = y The class of modular lattices is defined by identity 8, hence it is closed under sublattices: every sublattice of a modular lattice is itself a modular lattice. If the pentagon can be embedded in a lattice, then that lattice has a non-modular sublattice, hence it is not modular. 2. Let <A,≤> be a non-modular lattice. Let a,b,c ∈ A, and a ≤ b and let a ∨ (b ∧ c) ≠ b ∧ (a ∨ c). We first show: a < b, ¬(c ≤ a), ¬(a ≤ c), ¬(c ≤ b), ¬(b ≤ c) a < b Namely: a ≤ b. If a = b, then a ∨ (b ∧ c) = a ∨ (a ∧ c) = a and b ∧ (a ∨ c) = a ∧ (a ∨ c) = a. Hence a ∨ (b ∧ c) = b ∧ (a ∨ c), contradicting the assumption. Hence a ≠ b. Hence a < b. ¬(a ≤ c) Namely: If a ≤ c, then a ∨ c = c, hence b ∧ (a ∨ c) = b ∧ c. Since a ≤ b and a ≤ c, a ≤ b ∧ c. Then a ∨ (b ∧ c) = b ∧ c. Hence b ∧(a ∨ c) = a ∨ (b ∧ c), contradicting the assumption. Hence ¬(a ≤ c). ¬(c ≤ a) Namely: If c ≤ a, then c ≤ b, hence b ∧ c = c, hence a ∨ (b ∧ c) = a ∨ c = a. Similarly, b ∧ (a ∨ c) = b ∧ a = a. Then b ∧(a ∨ c) = a ∨ (b ∧ c), contradicting the assumption. Hence ¬(c ≤ a). ¬(b ≤ c) Namely: If b ≤ c, then b ∧ c = b. Then a ∨ (b ∧ c) = b. Also a ∨ c = c, hence b ∧ (a ∨ c) = b. Then b ∧(a ∨ c) = a ∨ (b ∧ c), contradicting the assumption. Hence ¬(b ≤ a). ¬(c ≤ b) Namely: If c ≤ b, then b ∧ c = c, hence a ∨ (b ∧ c) = a ∨ c. Since a ≤ b and c ≤ b, a ∨ c ≤ b. Hence b ∧ (a ∨ c) = (a ∨ c). Then b ∧(a ∨ c) = a ∨ (b ∧ c), contradicting the assumption. Hence ¬(c ≤ b). Now let: 1' = a ∨ c z = c y = b ∧ (a ∨ c) x = a ∨ (b ∧ c) 0' = b ∧ c What we are going to prove is: 0' < x < y < 1' 0' < z < 1' ¬(x ≤ z), ¬(z ≤ x), ¬(y ≤ z), ¬(z ≤ y) x ∧ z = y ∧ z = 0' x ∨ z = y ∨ z = 1' 0' < x Namely: b ∧ c ≤ a ∨ (b ∧ c) If b ∧ c = a ∨ (b ∧ c), then a ≤ b ∧ c, and hence a ≤ c. But ¬(a ≤ c). Hence b ∧ c ≠ a ∨ (b ∧ c). Hence 0' < x. x < y Namely: (2.) under FACT 1, tells us that a ∨ (b ∧ c) ≤ (a ∨ b) ∧ (a ∨ c). Since a < b, a ∨ b = b, Hence we know that: a ∨ (b ∧ c) ≤ b ∧ (a ∨ c). By the assumption, it follows that: a ∨ (b ∧ c) < b ∧ (a ∨ c). Hence x < y. y < 1' Namely: b ∧ (a ∨ c) ≤ a ∨ c If b ∧ (a ∨ c) = a ∨ c, then a ∨ c ≤ b, and hence c ≤ b. But ¬(c ≤ b). Hence b ∧ (a ∨ c) ≠ a ∨ c. Hence y < 1'. 0 < z Namely: b ∧ c ≤ c If b ∧ c = c, c ≤ b. But ¬(c ≤ b). Hence b ∧ c ≠ c. Hence 0' < z. z < 1' Namely: c ≤ a ∨ c If c = a ∨ c, then a ≤ c. But ¬(a ≤ c). Hence c ≠ a ∨ c. Hence z < 1'. z ∧ y = 0' Namely: z ∧ y = c ∧ (b ∧ (a ∨ c)) = (c ∧ (a ∨ c)) ∧ b = b ∧ c = 0' ¬(z ≤ y), ¬(y ≤ z) Namely: If z ≤ y, then z ∧ y = z. Since z ∧ y = 0', then z = 0'. But 0' < z. Hence ¬(z ≤ y). Similarly ¬(y ≤ z). z ∨ x = 1' Namely: z ∨ x = c ∨ (a ∨ (b ∧ c)) = (c ∨ (b ∧ c)) ∨ a = a ∨ c = 1' ¬(z ≤ x), ¬(x ≤ z) Namely: If z ≤ x, then z ∨ x = x. Since z ∨ x = 1', then z = 1'. But z < 1' Hence ¬(z ≤ x). Similarly, ¬(x ≤ z). z ∧ x = 0' Namely: z ∧ y = 0' and 0' ≤ x, hence z ∧ y ≤ x. z ∧ y ≤ z, hence z ∧ y ≤ z ∧ x. x ≤ y, hence z ∧ x ≤ y. z ∧ x ≤ z, hence z ∧ x ≤ z ∧ y. Hence z ∧ x = z ∧ y, hence z ∧ x = 0'. z ∨ y = 1' Namely: z ∨ x = 1' and y ≤ 1', hence y ≤ z ∨ x. z ≤ z ∨ x, hence z ∨ y ≤ z ∨ x. x ≤ y, hence x ≤ z ∨ y. z ≤ z ∨ y, hence z ∨ x ≤ z ∨ y. Hence z ∨ y = z ∨ x , hence z ∨ y = 1'. What we have shown now is that {0',x,y,z,1'} ⊆ A is closed under join and meet. Hence <{0',x,y,z,1'},≤⌠{0',x,y,z,1'}> is a sublattice of <A,≤>. But, of course, this sublattice of <A,≤> is isomorphic to the pentagon, hence the pentagon can be embedded in <A,≤>. ◄ THE STORY IN PICTURES For the non-modular lattice, the story is simple. Let <A,≤> be a lattice, let a ≤ b, and let a ∨ (b ∧ c) < b ∧ (a ∨ c). As we have seen, we easily esablish that then a < b, and c is independent of a and of b, and, for that matter, of a ∨ (b ∧ c) and of b ∧ (a ∨ c). We get structure AF b ∨ c b a ∨ c b ∧ (a ∨ c) c a ∨ (b ∧ c) a b ∧ c a ∧ c We need to check now which of the lines in this picture represent 'smaller than' and which represent 'smaller or equal than'. This is done in the following picture: thick and double lines represent 'smaller than', thin lines represent 'smaller or equal than': b 1' y z = c x 0' a Homomorphisms can contract thin lines, but not thick or double lines. This means that the picture really stands for two structures: A1, where a < x, A1 = AF, and A2 where a = x. A2 is the pentagon. We know then that A contains the sublattice in the picture, [{a,b,c}], which is either AF, which contains the pentagon as a sublattice, or the pentagon itself. THEOREM 6: A modular lattice is distributive iff the diamond cannot be embedded in it. PROOF: 1. The diamond is modular, but not distributive. That the diamond is modular follows from theorem 5. Obviously the pentagon cannot be embedded in it. The diamond is not distributive: y ∨ (x ∧ z) = y (y ∨ x) ∧ (y ∨ z) = 1 The class of distributive lattices is defined by identity 5, hence it is closed under sublattices: every sublattice of a distributive lattice is itself a distributive lattice. If the diamond can be embedded in a lattice, then that lattice has a non-distributive sublattice, hence it is not distributive. 2. Let <A,≤> be a modular, non-distributive lattice. Let a,b,c ∈ A and let a ∧ (b ∨ c) ≠ (a ∧ b) ∨ (a ∧ c) . By (1.) of FACT 1, this means that: (a ∧ b) ∨ (a ∧ c) < a ∧ (b ∨ c). This implies that: ¬(a ≤ b), ¬ (b ≤ a), ¬ (a ≤ c), ¬(c ≤ a), ¬(b ≤ c), ¬(c ≤ b). Namely: If a ≤ b, then a ∧ b = a. Hence (a ∧ b) ∨ (a ∧ c) = a ∨ (a ∧ c) = a Then a < a ∧ (b ∨ c), which is impossible. Hence ¬(a ≤ b). If b ≤ a, then a ∧ b = b. Then (a ∧ b) ∨ (a ∧ c) = b ∨ (a ∧ c). Hence b ∨ (a ∧ c) < a ∧ (b ∨ c). But, since <A,≤> is modular and b ≤ a, b ∨ (a ∧ c) = a ∧ (b ∨ c). Contradiction. Hence ¬(b ≤ a). Obviously, the same argument shows that ¬(a ≤ c) and ¬(c ≤ a). If b ≤ c, then a ∧ (b ∨ c) = a ∧ c. Hence (a ∧ b) ∨ (a ∧ c) < (a ∧ c). This is obviously impossible. Hence ¬(b ≤ c). Similarly, ¬(c ≤ b). Now let: 1' = (a ∨ b) ∧ (a ∨ c) ∧ (b ∨ c) x = (a ∧ 1') ∨ 0' y = (b ∧ 1') ∨ 0' z = (c ∧ 1') ∨ 0' 0' = (a ∧ b) ∨ (a ∧ c) ∨ (b ∧ c) 0 ≤ 1' . This is (3.) of FACT 1. 0' ≤ x, 0' ≤ y, 0' ≤ z. Obviously. x ≤ 1', y ≤ 1', z ≤ 1'. Namely: a ∧ 1' ≤ 1' and 0 ≤ 1', hence ((a ∧ 1') ∨ 0' ≤ 1', hence x ≤ 1'. Similarly, y ≤ 1' and z ≤ 1'. What we are going to prove is: 0' < x < 1, 0' < y < 1', 0 < z < 1' ¬(x ≤ y), ¬(y ≤ x), ¬(x ≤ z), ¬(z ≤ x), ¬(y ≤ z), ¬(z ≤ y) x ∧ y = x ∧ z = y ∧ z = 0' x ∨ y = x ∨ z = y ∨ z = 1' We will start by deriving some useful facts.