1. Lattice and Boolean Algebra

2. Lattice and Boolean 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

3. Lattice and Boolean Algebra
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

4. 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

5. 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

6. Lattice and Boolean Algebra
Examples
distributive
non-distributive
Lattice and Boolean Algebra

7. 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

8. Lattice and 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

9. Lattice and 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

10. 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

11. Lattice and Boolean Algebra
Example
prove the idempotent laws given Huntington’s postulates:
a = a  0
= a  aa
= (a  a)  (a  a)
= (a  a)  1
= a  a
Lattice and Boolean Algebra

12. 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

13. Lattice and Boolean Algebra
Examples B3
P({a,b,c})
{n  | n|30}
Lattice and Boolean Algebra

14. Isomorphic Boolean Algebra
Two Boolean algebras A, ,  , , 0A,1A and B, ,  , , 0B,1B are isomorphic iff there is a mapping f:AB, such that
for arbitrary a,b  A, f(ab) = 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

15. Lattice and Boolean Algebra
De Morgan’s Theorem
De Morgan’s Laws hold
These equations can be generalized
Lattice and Boolean Algebra

16. 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

17. 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

18. 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

19. 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

20. 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 x
f(x) a 1 a
Lattice and Boolean Algebra

21. 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) x y f g
fg
fg f g 1
Example
Lattice and Boolean Algebra

22. 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

23. 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

24. 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