Proof. Given x∧z = y∧z and x∨z = y∨z, thenand similarly, y ≤ x, so that x = y. Conversely, neither M 3 nor N 5 can be subalgebras of a cancellative slew lattice. £ A lattice (L; ∨, ∧) is complete if every subset X of L has a supremum (an element u ≥ x for all x in X, with u being the least such element in L) denoted by sup(X) and an infimum (an element v ≤ x for all x in X, with v being the greatest such element in L) denoted by inf(X). In particular, a complete lattice has a greatest element 1 and a least element 0. Conversely, a lattice with both least and greatest elements 0 and 1 is complete if all subsets have suprema, or equivalently, if all subsets have infima. Finally, in any complete lattice, we let 0 = sup(∅) and 1 = inf(∅).
Lattices and universal algebraAn algebra is any system, A = (A:occurs, subalgebras may be indicated by their underlying sets.) Under inclusion, ⊆, the subalgebras of an algebra A form a complete lattice Sub(A) with greatest element A, least element the smallest subalgebra containing ∅ and meets given by intersection. If none of the operations are nullary, then the least subalgebra is the empty subalgebra, ∅. If there are no operations, then Sub(A) is the lattice 2 A .Recall that a congruence on A = (A:Under inclusion, ⊆, the congruences on A form a complete lattice Con(A). Its greatest element is the universal relation ∇ = A×A relating all elements in A. Its least element is the identity relation Δ. Suprema and infima in Con(A) are calculated as in the lattice Equ(A) of all equivalences on A. In particular, infima in Con(A) are given by intersection. £Recall that an element c in a lattice (L; ∨, ∧) is compact if for any subset X of L, c ≤ supX implies that c ≤ supY for some finite subset Y of X. (Every cover can be reduced to a finite cover.) An algebraic lattice is a complete lattice for which every element is a supremum of compact elements. The proof of the following result is easily accessible in the literature Theorem 1.1.4. Given an algebra A = (A: f 1 , f 2 , …, f r ), both Sub(A) and Con(A) are algebraic lattices.
I: PreliminariesProof. That χ is a homomorphism follows easily from the associative, commutative and distributive laws. By cancellation, χ is one-to-one. Upon composing with either coordinate projections, it clearly it mapped onto each factor. £ Corollary 1.1.10. Every nontrivial distributive lattice is a subdirect product of C 1 . £We return to the variety of all lattices. On any lattice, consider the polynomial M(x, y, z) = (x∨y) ∧ (x∨z) ∧ (y∨z) that was implicit in the proof of Theorem 1.5. M satisfies the identitiesGiven an algebra A = (A; f 1 , …, f r ) on which a ternary operation M(x, y, z) satisfying these identities is polynomial-defined using the operations of A, then Con(A) is distributive. In general, if a ternary function M can be defined from the functions symbols of a variety V such that M satisfied these identities on all algebras in V, then the congruence lattices of all algebras in that variety are distributive and V is said to be congruence distri...