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Lattice and Boolean Algebra. Algebra. An algebraic system is defined by the tuple A,o 1 , …, o k ; R 1 , …, R m ; c 1 , … c k , where, A is a non-empty set, o i is a function A p i A, p i is a positive integer, R j is a relation on A, and c i is an element of A. Lattice.
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Algebra • An algebraic system is defined by the tuple A,o1, …, ok; R1, …, Rm; c1, … ck, where, A is a non-empty set, oi is a function Api A, pi is a positive integer, Rj is a relation on A, and ci is an element of A. Lattice and Boolean Algebra
Lattice • The lattice is an algebraic system A, , , given a,b,c in A, the following axioms are satisfied: • Idempotent laws: a a = a, a a = a; • Commutative laws: a b = b a, a b = b a • Associative laws: a (b c) = (a b) c, a (b c) = (a b) c • Absorption laws: a (a b) = a, a (a b) = a Lattice and Boolean Algebra
Lattice - Example • Let A={1,2,3,6}. • Let a b be the least common multiple • Let a b be the greatest common divisor • Then, the algebraic system A, , satisfies the axioms of the lattice. Lattice and Boolean Algebra
Distributive Lattice • The lattice A, , satisfying the following axiom is a distributive lattice 5. Distributive laws: a (b c) = (a b) (a c), a (b c) = (a b) (a c) Lattice and Boolean Algebra
Examples distributive non-distributive Lattice and Boolean Algebra
Complemented Lattice • Let a lattice A, , have a maximum element 1 and a minimum element 0. For any element a in A, if there exists an element xa such that a xa = 1 and a xa = 0, then the lattice is a complemented lattice. • Find complements in the previous example Lattice and Boolean Algebra
Boolean Algebra • Let B be a set with at least two elements 0 and 1. Let two binary operations and , and a unary operation are defined on B. The algebraic system B, , , , 0,1 is a Boolean algebra, if the following postulates are satisfied: • Idempotent laws: a a = a, a a = a; • Commutative laws: a b = b a, a b = b a • Associative laws: a (b c) = (a b) c, a (b c) = (a b) c • Absorption laws: a (a b) = a, a (a b) = a • Distributive laws: a (b c) = (a b) (a c), a (b c) = (a b) (a c) Lattice and Boolean Algebra
Boolean Algebra • Involution: • Complements: a a = 1, a a = 0; • Identities: a 0 = a, a 1 = a; • a 1 = 1, a 0 = 0; • De Morgan’s laws: Lattice and Boolean Algebra
Huntington’s Postulates • To verify whether a given algebra is a Boolean algebra we only need to check 4 postulates: • Identities • Commutative laws • Distributive laws • Complements Lattice and Boolean Algebra
Example • prove the idempotent laws given Huntington’s postulates: a = a 0 = a aa = (a a) (a a) = (a a) 1 = a a Lattice and Boolean Algebra
Models of Boolean Algebra • Boolean Algebra over {0,1} B={0,1}. B, , , , 0,1 • Boolean Algebra over Boolean Vectors Bn = {(a1, a2, … , an) | ai {0,1}} Let a=(a1, a2, … , an) and b = (b1, b2, … , bn) Bn define a b = (a1 b1, a2 b2, … , an bn) a b = (a1 b1, a2 b2, … , an bn) a=(a1, a2, … , an) then Bn, , , , 0,1 is a Boolean algebra, where, 0 = (0,0, …, 0) and 1 = (1,1, …, 1) • Boolean Algebra over Power Set Lattice and Boolean Algebra
Examples P({a,b,c}) {n | n|30} B3 Lattice and Boolean Algebra
Isomorphic Boolean Algebra • Two Boolean algebras A, , , , 0A,1A and B, , , , 0B,1B are isomorphic iff there is a mapping f:AB, such that • for arbitrary a,b A, f(ab) = f(a)f(b), f(a b) = f(a) f(b), and f(a) = f(a) • f(0A ) = 0B and f(1A ) = 1B An arbitrary finite Boolean algebra is isomorphic to the Boolean algebra Bn, , , , 0,1 Question: define the mappings for the previous slide. Lattice and Boolean Algebra
De Morgan’s Theorem • De Morgan’s Laws hold • These equations can be generalized Lattice and Boolean Algebra
Definition • Let Bn, , , , 0,1 be a Boolean algebra. The variable that takes arbitrary values in the set B is a Boolean variable. The expression that is obtained from the Boolean variables and constants by combining with the operators , , and parenthesis is a Boolean expression. If a mapping f:Bn B is represented by a Boolean expression, then f is a Boolean function. However, not all mappings f:Bn B are Boolean functions. Lattice and Boolean Algebra
Theorem • Let F(x1, x2, …, xn) be a Boolean expression. Then the complement of the complement of the Boolean expression F(x1, x2, …, xn) is obtained from F as follows • Add parenthesis according to the order of operations • Interchange with • Interchange xi with xi • Interchange 0 with 1 Example Lattice and Boolean Algebra
Principle of Duality • In the axioms of Boolean algebra, in an equation that contains , , 0, or 1, if we interchange with , and/or 0 with 1, then the other equation holds. Lattice and Boolean Algebra
Dual Boolean Expressions • Let A be a Boolean expression. The dual AD is defined recursively as follows: • 0D = 1 • 1D = 0 • if xi is a variable, then xiD = xi • if A, B, and C are Boolean expressions, and A = B C, then AD= BD CD • if A, B, and C are Boolean expressions, and A = B C, then AD= BD CD • if A and B are Boolean expressions, and A = B, then Lattice and Boolean Algebra
Examples • Given xy yz = xy yz xz the dual (x y)(y z) = (x y)(y z)(x z) • Consider the Boolean algebra B={0,1,a,a} check if f is a Boolean function. f(x) = xf(0) xf(1) f(x) = x a x 1 f(a) = a a a 1 = a Lattice and Boolean Algebra
Logic Functions • Let B = {0,1}. A mapping Bn B is always represented by a Boolean expression–a two-valued logic function. f g = h f(x1,x2,…,xn) g(x1,x2,…,xn) = h(x1,x2,…,xn) f = g f(x1,x2,…,xn) = g(x1,x2,…,xn) Example Lattice and Boolean Algebra
Logical Expressions • Constants 0 and 1 are logical expressions • Variables x1,x2,…,xn are logical expressions • If E is a logical expression, then E is one • If E1 and E2 are logical expressions, then (E1 E2) and (E1 E2) are also logical expressions • The logical expressions are obtained by finite application of 1 - 4 Lattice and Boolean Algebra
Evaluation of logical Expressions • An assignment mapping :{xi} {0,1} (i = 1, … , n) • The valuation mapping |F| of a logical expression is obtained: • |0| = 0 and |1| = 1 • If xi is a variable, then | xi | = (xi) • If F is a logical expression, then |F| = 1 |F| = 0 • If F and G are logical expressions, then |F G| = 1 (|F| = 1 or |G| = 1) • If F and G are logical expressions, then |F G| = 1 (|F| = 1 and |G| = 1) Example: F:x y z (x) = 0, (y) = 0, (z) = 1 Lattice and Boolean Algebra
Equivalence of Logic Expressions • Let F and G be logical expressions. If |F| = |G| hold for every assignment , then F and G are equivalent ==> F G • Logical expressions can be classified into 22n equivalence classes by the equivalence relation () Lattice and Boolean Algebra