- Residuated Boolean algebra
In
mathematics , a residuated Boolean algebra is aresiduated lattice whose lattice structure is that of a Boolean algebra. Examples include Boolean algebras with the monoid taken to be conjunction, the set of all formal languages over a given alphabet Σ under concatenation, the set of all binary relations on a given set "X" under relational composition, and more generally the power set of any equivalence relation, again under relational composition. The original application was torelation algebra s as a finitely axiomatized generalization of the binary relation example, but there exist interesting examples of residuated Boolean algebras that are not relation algebras, such as the language example.Definition
A residuated Boolean algebra is an algebraic structure ("L", ∧, ∨, ¬, 0, 1, •, I, , /) such that: (i) ("L", ∧, ∨, •, I, , /) is a
residuated lattice , and:(ii) ("L", ∧, ∨, ¬, 0, 1) is a Boolean algebra.An equivalent signature better suited to the
relation algebra application is ("L", ∧, ∨, ¬, 0, 1, •, I, , ) where the unary operations "x" and "x" are intertranslatable in the manner ofDe Morgan's laws via:"x""y" = ¬("x"¬"y"), "x""y" = ¬("x"¬"y"), and dually /"y" and "y" as: "x"/"y" = ¬(¬"x""y"), "x""y" = ¬(¬"x"/"y"),with the residuation axioms in the
residuated lattice article reorganized accordingly (replacing "z" by ¬"z") to read:("x""z")∧"y" = 0 ⇔ ("x"•"y")∧"z" = 0 ⇔ ("z""y")∧"x" = 0This De Morgan dual reformulation is motivated and discussed in more detail in the section below on conjugacy.
Since residuated lattices and Boolean algebras are each definable with finitely many equations, so are residuated Boolean algebras, whence they form a finitely axiomatizable variety.
Examples
1. Any Boolean algebra, with the monoid multiplication • taken to be conjunction and both residuals taken to be
material implication "x"→"y". Of the remaining 15 binary Boolean operations that might be considered in place of conjunction for the monoid multiplication, only five meet the monotonicity requirement, namely 0, 1, "x", "y", and "x"∨"y". Setting "y" = "z" = 0 in the residuation axiom "y" ≤ "x""z" ⇔ "x"•"y" ≤ "z", we have 0 ≤ "x"
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