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Boolean Algebra of Discrete Mathematics

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Boolean Algebra of Discrete Mathematics

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  1. Discrete structureslecture notesModule-IIIBoolean Algebra and Applications B.Tech (CSE) V Semester Faculty: Dr. NeetuFaujdar

  2. Module III : Boolean Algebra and Its Applications • Definition of Boolean Algebra, Laws of Boolean Algebra. • Basic Theorems, Boolean Functions – Disjunctive Normal Form, Conjunctive Normal Form, Duality Principle. • Boolean Expression – Sum of Products, Product of Sum, Minterm and Maxterm, Karnaugh map. • Applications of Boolean Algebra.

  3. Books and References Text: • C.L. Liu, Elements of Discrete Mathematics, Tata McGraw Hill, Second Edition • Kenneth H. Rosen, Discrete Mathematics and its Applications, Mc.Graw Hill References: • J.P.Tremblay & R. Manohar, “Discrete Mathematical Structure with Applications to Computer Science” Mc.GrawHill • Colmun, Busby and Ross, Discrete Mathematical Structure, PHI, 6th Edition,

  4. Definition of Boolean Algebra

  5. Definition of Boolean Algebra • Boolean Algebra is used to analyze and simplify the digital (logic) circuits. It uses only the binary numbers i.e. 0 and 1. • It is also called as Binary Algebra or logical Algebra. Boolean algebra was invented by George Boole in 1854. • Let B be a nonempty set with two binary operations + and ∗, a unary operation , and two distinct elements 0 and 1. Then B is called a Boolean algebra if the following axioms hold where a, b, c are any elements in B.

  6. (B, +, *,’,0,1) 0 is the the zero element and 1 is the unit element and a’ is the complement of a.

  7. Logic Gate Examples of Boolean Algebra

  8. Truth Table Examples of Boolean Algebra

  9. Rules in Boolean Algebra

  10. Following are the important rules used in Boolean algebra. • Variable used can have only two values. Binary 1 for HIGH and Binary 0 for LOW. • Complement of a variable is represented by an overbar (-). Thus, complement of variable B is represented as . Thus if B = 0 then = 1 and B = 1 then = 0. • ORing of the variables is represented by a plus (+) sign between them. For example ORing of A, B, C is represented as A + B + C.

  11. Continue- • Logical ANDing of the two or more variable is represented by writing a dot between them such as A.B.C. Sometime the dot may be omitted like ABC.

  12. Laws of Boolean Algebra

  13. There are many types of Boolean Laws. 1. Commutative law • Any binary operation which satisfies the following expression is referred to as commutative operation. • A . B = B . A    The order in which two variables are AND’ed makes no difference • A + B = B + A    The order in which two variables are OR’ed makes no difference • Commutative law states that changing the sequence of the variables does not have any effect on the output of a logic circuit.

  14. Continue- 2. Associative law • This law states that the order in which the logic operations are performed is irrelevant as their effect is the same. • A + (B + C) = (A + B) + C = A + B + C    (OR Associate Law) • A(B.C) = (A.B)C = A . B . C    (AND Associate Law)

  15. Continue- 3. Distributive law • Distributive law states the following condition. • A(B + C) = A.B + A.C    (OR Distributive Law) • A + (B.C) = (A + B).(A + C)    (AND Distributive Law) 4. Annulment Law • A term AND´ed with a “0” equals 0 or OR´ed with a “1” will equal 1   (i) A . 0 = 0    A variable AND’ed with 0 is always equal to 0 (ii) A + 1 = 1    A variable OR’ed with 1 is always equal to 1

  16. Continue- 5. Identity Law • A term OR´ed with a “0” or AND´ed with a “1” will always equal that term   (i) A + 0 = A   A variable OR’ed with 0 is always equal to the variable (ii) A . 1 = A    A variable AND’ed with 1 is always equal to the variable

  17. Continue- 6. Idempotent Law • An input that is AND´ed or OR´ed with itself is equal to that input   (i) A + A = A    A variable OR’ed with itself is always equal to the variable (ii) A . A = A    A variable AND’ed with itself is always equal to the variable

  18. Continue- 7. Complement Law • A term AND´ed with its complement equals “0” and a term OR´ed with its complement equals “1”   (i) A . A = 0    A variable AND’ed with its complement is always equal to 0 (ii) A + A = 1    A variable OR’ed with its complement is always equal to 1

  19. Continue- 8. Double Negation Law • A term that is inverted twice is equal to the original term (i)   = A     a double complement of a variable is always equal to the variable. 9. Absorptive Law • This law enables a reduction in a complicated expression to a simpler one by absorbing like terms.   (i) A + (A.B) = A    (OR Absorption Law) (ii) A(A + B) = A    (AND Absorption Law)

  20. Examples of Boolean Expression Simplification

  21. Continue-

  22. Continue-

  23. Continue-

  24. Proofs of Famous Laws of Boolean Algebra

  25. 1. Associative Law (Associate Law of Addition) • Statement: Associative law of addition states that OR ing more than two variables i.e. mathematical addition operation performed on variables will return the same value irrespective of the grouping of variables in an equation. It involves in swapping of variables in groups. • The Associative law using OR operator can be written as • A+(B+C) = (A+B)+C Proof: If A, B and C are three variables, then the grouping of 3 variables with 2 variables in each set will be of 3 types, such as (A + B), (B + C) and(C + A).

  26. Continue- According to associative law (A + B + C) = (A + B) +C = A + (B + C) = B + (C + A) We know that, A + AB = A (according to Absorption law) Now let’s assume that, x = A + (B + C) and y = (A + B) + C According to associative law, we need to prove that x = y. Now, find Ax = A [ A + (B + C) ] = AA +A (B + C) = A + AB + AC → since AA = A = (A+ AB) + AC = A + AC → since A + AB = A = A → since A + AC = A

  27. Continue- Therefore Ax = A Similarly, for Bx = B [ A + (B + C) ] = AB +B (B + C) = AB + BB + BC = AB + B + BC → since BB = B = (B+ BC) + AB = B + AB → since B + BC = B = B → since B + AB = B

  28. Continue- • Using these above equations, we can say that the relation between A, B, C and + operator doesn’t change when multiplied by other variable like x, such as xy = yx = x = y. yx = ((A + B) + C) x = (A + B) x + Cx = (Ax + Bx) + Cx = (A + B) + C = y xy = (A + (B + C)) y = Ay + (B + C) y = Ay + (By + Cy) = A + (B + C) = x So x = y, which means A + (B + C) = (A + B) + C = B + (A +C)

  29. Example of Associative Law • Take three variables 0, 1 and 0, then • According to associative law, • (0 + 1) + 0 = 0 + (1 + 0) • 1 + 0 = 0 + 1 • 1 = 1 • Hence associative law is verified. • Hence the Associative law is proved, (A + B + C) = (A + B) +C = A + (B + C) = B + (C + A)

  30. 2. Proof of Distributive Law • This is the most used and most important law in Boolean algebra, which involves in 2 operators: AND, OR. • Statement1: • The multiplication of two variables and adding the result with a variable will result in same value as multiplication of addition of the variable with individual variables. • In other words, ANDing two variables and ORing the result with another variable is equal to AND of ORing of the variable with the two individual variables. • Distributive law can be written as • A + BC = (A + B)(A + C) • This is called OR distributes over AND.

  31. Continue- • Proof: • If A, B and C are three variables then • A + BC = A * 1 + BC → since A*1 = A • = A (1 + B)+ BC → since 1 + B = 1 • = A * 1 + AB + BC • = A *(1 + C) + AB + BC → since A*A = A*1 = A • = A *(A + C) + B (A + C) • = (A + C) (A + B) • A + BC = (A + B) (A + C) • Hence, distributive law is proved.

  32. Continue- • Statement 2: • The addition of two variables and multiplying the result with a variable will result in same value as addition of multiplication of the variable with individual variables. • In other words, ORing two variables and ANDing the result with another variable is equal to OR of ANDing of the variable with the two individual variables. • Distributive law can be written as • A (B+C) = (A B) + (A C) • This is called AND distributes over OR.

  33. Continue- • Proof: • A (B + C) = A (B*1) + A (C*1) → since 1 * B = B, 1 * C = C • = [(AB)*(A*1)] + [(AC) *(A*1)] • =[(AB) * A] + [(AC) *A] • = (A +1) (AB + AC) • = (AB +AC) → since 1 + A = 1 • Hence, distributive law is proved.

  34. Continue- • Example: • Take three variables 0, 1 and 0, then • According to distributive law, • 0 (1 + 0) = (0*1) + (0*0) • 0 (1) = (0) + (0) • 0 = 0 • Hence, distributive law is verified.

  35. 3. Proof of Commutative Law • Statement: • Commutative law states that the inter-changing of the order of operands in a Boolean equation does not change its result. • Using OR operator → A + B = B + A (A∪B=B ∪A) • Using AND operator → A * B = B * A (A ∩B=B ∩A) • This law is also has more priority in Boolean algebra.

  36. Continue- First Law : First law states that the union of two sets is the same no matter what the order is in the equation. Proof : A ∪ B = B ∪ A Consider the first law, A ∪ B = B ∪ A Let x ∈ A ∪ B. If x ∈ A ∪ B then x ∈ A or x ∈ B x ∈ A or x ∈ B x ∈ B or x ∈ A [according to definition of union] x ∈ B ∪ A x ∈ A ∪ B => x ∈ B ∪ A Therefore, A ∪ B ⊂ B ∪ A ---------------------(1)

  37. Continue- Consider the first law in reverse, B ∪ A = A ∪ B Let x ∈ B ∪ A. If x ∈ B ∪ A then x ∈ B or x ∈ A x ∈ B or x ∈ A x ∈ A or x ∈ B [according to definition of union] x ∈ A ∪ B x ∈ B ∪ A => x ∈ A ∪ B Therefore, B ∪ A ⊂ A ∪ B --------------------------(2) From equation 1 and 2 we can prove A ∪ B = B ∪ A

  38. Continue- Second Law : Second law states that the intersection of two sets is the same no matter what the order is in the equation. Proof : A ∩ B = B ∩ A Consider the second law, A ∩ B = B ∩ A Let x ∈ A ∩ B. If x ∈ A ∩ B then x ∈ A and x ∈ B x ∈ A and x ∈ B x ∈ B and x ∈ A [according to definition of intersection] x ∈ B ∩ A x ∈ A ∩ B => x ∈ B ∩ A Therefore, A ∩ B ⊂ B ∩ A ---------------------(3)

  39. Continue- Consider the second law in reverse, B ∩ A = A ∩ B Let x ∈ B ∩ A. If x ∈ B ∩ A then x ∈ B and x ∈ A x ∈ B and x ∈ A x ∈ A and x ∈ B [according to definition of intersection] x ∈ A ∩ B x ∈ B ∩ A => x ∈ A ∩ B Therefore, B ∩ A ⊂ A ∩ B -----------------------(4) From equation 3 and 4 we can prove the Commutative Property A ∩ B = B ∩ A

  40. Continue- • Example: • Take 2 variables 1 and 0, then • 1 + 0 = 0 + 1 • 1 = 1 • Similarly, • 1 * 0 = 0 * 1 • 0 = 0

  41. 3. Proof of Absorption Law • Absorption law involves in linking of a pair of binary operations. • i. A+AB = A • ii. A(A+B) = A • iii. A+ĀB = A+B • iv. A.(Ā+B) = AB • 3rd and 4th laws are also called as Redundancy laws. • Statement 1: A + AB = A

  42. Continue- • Proof: • A + AB = A.1 + AB → since A.1 = A • =A(1+B) → since 1 + B = 1 • = A.1 • = A • Statement 2:A (A + B) = A • Proof: • A (A + B) = A.A + A.B • = A+AB → since A . A = A • = A (1 + B) • = A.1 • = A

  43. Continue- • Statement 3:A + ĀB = A + B • Proof: • A + ĀB = (A + Ā) (A + B) → since A+BC = (A+B)(A+C) using distributive law • = 1 * (A + B) → since A + Ā = 1 • =A + B • Statement 4: A * (Ā+B) = AB • Proof: A * (Ā + B) = A. Ā + AB • = AB → since A Ā = 0

  44. Basic theorems of Boolean algebra

  45. 1. Principle of Duality • Suppose E is an equation of set algebra. The dual of denoted as E*is the equation obtained by replacing each occurrence of • ∪ by ∩, • ∩ by ∪ • U by ∅ • ∅ by U in E. • Example, the dual of C= (U ∩ A) ∪ (B ∩ A) is C*= (∅ ∪ A) ∩ (B ∪ A) • The principle of duality, states that if any equation E is an identity then its dual E *is also an identity.

  46. Continue- • Statement: • Duality principle states that “The Dual of the expression can be achieved by replacing the AND operator with OR operator, along with replacing the binary variables, such as replacing 1 with 0 and replacing 0 with 1”.This law explains that, replacing the variables doesn’t change the value of the Boolean function. • But while interchanging the names of the variables, we must change the binary operators also. “If the operators and variables of an equation or function that produce no change in the output of the equation, though they are interchanged is called “Duals”.

  47. Continue- • The Duality principle is also known as “De Morgan Duality”, which states that ‘Interchanging of Duals pairs in Boolean algebra will result in same output of the equation’.

  48. Continue- • There is one special type of operation in duality that is ‘Self-dual’. A self-dual operation processes the input to the output, without making any changes to it. So this is also called “Do nothing operation”. • Example: • If we have the Boolean equation like A + B = 0, then the equation formed by replacing the variable 0 with 1 and replacing the OR operator with AND operator is A * B = 1. This means both the Boolean functions are represents the operation of logic circuit. • As per Duality principle, if A, B are two variables then both the equations A + B = 0 and A * B = 1 are true in case of same logic circuit.

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