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Lecture 2
Chapter 2.1 – 2.5
2.2 Basic Definitions
•
Natural Numbers:
–
None negative integers
•
•
Debate whether to include 0
Set:
–
A collection of objects, usually having a common property
•
If S is a set, and x and y are certain elements then:
x ∈ S means x is a element of the set S
y ∉ S means y is not an element of set S
•
A set with denumerable number of elements is specified by braces:
A = {1, 2, 3, 4}
•
Element
–
•
Field
–
•
Any of the distinct objects that make up a set
A set of elements, together with two binary operators (addition and multiplication with the latter excluding
zero) , each having the properties of Closure, Associativity, Commutativity, Identity, Inverse, and both
operator combining to give Distributivity.
A binary operator defined on a set S of elements is a rule that assigns, to each pair of S elements, a
unique element from S
–
Consider:
a*b=c
* is a binary operator if it specifies a rule for finding c from the pair (a, b) and also if a, b, c ∈ S
•
•
Postulate/Axiom
–
•
is not a binary operator if a, b ∈ S and if c ∉ S
A premise or starting point of reasoning
Theorem
–
A statement that has been proven on the basis of previously established statements, such as other
theorems—and generally accepted statements, such as axioms.
2.2 Basic Definitions
• Closure Postulate:
– A set S is closed with respect to a binary operator if, for
every pair of elements of S, the binary operator
specifies a rule for obtaining a unique element of S.
• The set of natural numbers S = {1, 2, 3, 4, …} is closed w.r.t.
the binary operator + by the rules of arithmetic addition since
for any a, b ∈ N, there is a unique c ∈ N such that a + b = c.
• The set of natural numbers N = {1, 2, 3, 4, …} is not closed
w.r.t. the binary operator - by the rules of arithmetic
subtraction since for any a, b ∈ N, there is a c ∉ N
2 – 3 = -1 and 2, 3 ∈ N but (-1) ∉ N
– Boolean Algebra satisfies:
• Closure w.r.t. + operator
• Closure w.r.t. · operator
2.2 Basic Definitions
• Associative Law:
– A binary operator * on a set S is said to be associative
if:
(x * y) * z = x * (y * z) for all x, y, z, ∈ S
Satisfied by:
(x · y) · z = x · (y · z)
(x + y) + z = x + (y + z)
• Commutative Law:
– A binary operator * on a set S is said to be
commutative if:
x * y = y * x for all x, y, ∈ S
Satisfies by:
x· y = y· x
x+y = x+y
2.2 Basic Definitions
• Identity Element:
– A set S is said to have an identity element w.r.t. a binary operation * on S if
there exists an element e ∈ S with the property that:
e * x = x * e = x for all x ∈ S
additive identity element:
denoted by 0
x + 0 = 0 + x = x for any x ∈ S
multiplicative identity element:
denoted by 1
x · 1 = 1 · x = 1 for any x ∈ S
– Note: The set of natural number N = {1, 2, 3, …} has no identity element since 0
does not belong to the set
• Inverse
– A set S is having the identity element e w.r.t. a binary operator * is said to have
an inverse whenever, for every x ∈ S, there exists an element y ∈ S such that
x*y = e
Additive Inverse:
For every a in S, there exists an element −a in S, such that a + (−a) = 0.
Multiplication Inverse:
For any a in S other than 0, there exists an element a−1 in S, such that a · a−1 = 1.
2.2 Basic Definitions
• Distributive Law:
– If * and · are two binary operators on a set S, * is
said to be distributive over · whenever
x * (y · z) = (x * y) · (x *z)
– Distributivity of multiplication over addition:
For all a, b and c in S, the following equality holds:
a · (b + c) = (a · b) + (a · c).
2.2 Basic Definitions
• A field is an example of an algebraic structure
– An algebraic structure consists of one or more sets
closed under one or more operations, satisfying some
axioms.
– An axiom is a statement or proposition on which an
abstractly defined structure is based.
• A field is a set of elements, together with two
binary operators, each having properties 1-5 and
both combining to give property 6.
– The set of real numbers, together with the binary
operators + and ·, forms the field of real numbers.
2.2 Basic Definitions
• The operator and postulates have the following
meanings for ordinary algebra:
–
–
–
–
–
–
The binary operator + defines addition
The additive identity is 0
The additive inverse defines subtraction
The binary operator · defines multiplication
The multiplication identity is 1
For a ≠ 0, the multiplicative inverse of a = 1/a defines
division (a · 1/a = 1)
– The only distributive law applicable is that of · over +
a · (b + c ) = (a · b) + (a · c)
2.2 Basic Definitions
• Is the set of natural numbers a field? Why?
No, as there is no negative number(s) to satisfy
the Additive Inverse (a + (-a) = 0)
• Is the set of integers a field? Why?
No, as there is no negative number(s) to satisfy
the Additive Inverse (a + (-a) = 0)
Is the set of real numbers a field? Why?
Yes, all the field axioms are satisfied
2.3 Axiomatic Definition of Boolean Algebra
• In 1937, Claude Shannon founded both digital computer and digital
circuit design theory when, as a 21-year-old master's degree student at
the Massachusetts Institute of Technology (MIT), he wrote
his thesis demonstrating that electrical applications of boolean
algebra could construct and resolve any logical, numerical relationship.
• His thesis was published in the 1938 issue of the Transactions of the
American Institute of Electrical Engineers , and proved that a two-valued
Boolean algebra (whose members are most commonly denoted 0 and 1,
or false and true) can describe the operation of two-valued electrical
switching circuits.
• In this work, Shannon proved that boolean algebra and binary
arithmetic could be used to simplify the arrangement of
the electromechanical relays that were used then in telephone call routing
switches. He next expanded this concept, and he also proved that it would
be possible to use arrangements of relays to solve problems in Boolean
algebra.
2.3 Axiomatic Definition of Boolean Algebra
• Would certainly want
to minimize the
number of mechanical
relays used!
2.3 Axiomatic Definition of Boolean Algebra
•
Boolean algebra is an algebraic structure defined by a set of elements, B, together
with two binary operators + and ·, provided that the following postulates are
satisfied:
1.
2.
(a) The structure is closed with respect to +
i.e. for all a, b in S, both a + b are in S
(b) The structure is closed with respect to ·
i.e. for all a, b in S, both a · b are in S
(a) The element 0 is an identity element with respect to +
x+0 = 0+x = x
(b) The element 1 is an identity element with respect to ·
x·1 = 1·x = x
3.
(a) The structure is commutative with respect to +
x+y=y+x
(b) The structure is commutative with respect to ·
x·y=y·x
4.
(a) The operator · is distributive over +
x · (y + z) = (x · y) + (x · z)
(b) The operator + is distributive over ·
x + (y · z) = (x + y) · (x + z)
5.
For every element x ∈ B, there exists an element x’ ∈ B called the complement of x such
that
(a) x + x’ = 1
(b) x · x’ = 0
6. There exist at least two elements x, y ∈ B such that x ≠ y
2.3 Axiomatic Definition of Boolean Algebra
• Compare Boolean Algebra with arithmetic and ordinary
algebra
– The Distributive Law of + over · is valid for Boolean algebra but
not for ordinary algebra
x + (y · z) = (x + y) · (x + z)
– Boolean algebra does not have additive or multiplicative
inverses; therefore there are no subtraction or division
operations
• Subtraction Inverse: a + (−a) = 0
• Multiplicative Inverse: a · a−1 = 1
– Postulate 5 defines the complement operator which is not
available in ordinary algebra
x + x’ = 1
x · x’ = 0
– Ordinary algebra operates on real numbers, which constitutes
an infinite set of elements. Boolean algebra deals with a set B
comprised of only two elements, 0 and 1.
2.3 Axiomatic Definition of Boolean Algebra
• Two –Valued Boolean Algebra
– Defined on a set of two elements, B = {0, 1} with
rules for the two binary operators + and⋅as shown
in the following operator tables:
x
y
x∙y
x
y
x+y
x
x’
0
0
0
0
0
1
0
0
0
1
0
0
1
1
0
1
1
0
0
1
0
1
1
0
1
1
1
1
1
0
1
1
AND
OR
NOT
2.3 Axiomatic Definition of Boolean Algebra
• Show that the Huntington postulates are valid
for the set B = {0, 1} and the two binary
operators + and ⋅
– Closed: From binary operator tables, the result of
AND, OR, and NOT result in 0 or 1 and 0 ∈ B
– Identity Elements: From tables, we see that:
0+0=0
1⋅1=1
0+1=1+0=1
1⋅0=0⋅1=0
which establishes the two identity elements, 0
for + and 1 for ⋅ per postulate 2
– Identity Commutative Law: obvious from the
binary operator tables
2.3 Axiomatic Definition of Boolean Algebra
– Distributive Law: x⋅(y+z) = (x⋅y)+(x⋅z)
x
y
z
y+z
x⋅(y+z)
x⋅y
x⋅z
(x⋅y)+(x⋅z)
0
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
1
0
1
0
0
0
0
0
1
1
1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
1
1
1
0
1
1
1
1
0
1
1
1
0
1
1
1
1
1
1
1
1
1
values are the same
2.3 Axiomatic Definition of Boolean Algebra
– Distributive Law: x+(y ⋅ z) = (x+y) ⋅ (x+z)
x
y
z
y⋅z
x+(y⋅z)
x+y
x+z
(x+y) ⋅(x+z)
0
0
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
1
0
0
0
1
0
0
0
1
1
1
1
1
1
1
1
0
0
0
1
1
1
1
1
0
1
0
1
1
1
1
1
1
0
0
1
1
1
1
1
1
1
1
1
1
1
1
values are the same
2.3 Axiomatic Definition of Boolean Algebra
– Postulate 1:
• From the complement table
x + x’ = 1 since 0 + 0’ = 0 + 1 =1 and 1 + 1’ = 1 + 0 = 1
x ⋅ x’ = 0 since 0 ⋅ 0’ = 0 ⋅ 1 = 0 and 1 ⋅ 1’ = 1 ⋅ 0 = 0
– Postulate 6:
• As the two-values Boolean algrebra has two elements, 1 and
0, with 1 ≠ 0
• Thus a we have defined a two-valued Boolean
algebra having a set of two elements, 1and 0, two
binary operators equivalent to AND and OR, and
a complement operator equivalent to the NOT
operator.
2.4 Basic Theorems and Properties of Boolean Algebra
• Duality:
– Every algebraic expression deducible from the
postulates of Boolean algebra remain valid if the
operators and identify elements are interchanged
– For the two-valued Boolean algebra, the identity
elements and the elements of the set B are the
same : 1 and 0
– For the dual of an algebraic expression, simply
interchange OR and AND operators and replace 1’s
with 0’s and 0’s with 1’s
2.4 Basic Theorems and Properties of Boolean Algebra
• Basic Theorems
2.4 Basic Theorems and Properties of Boolean Algebra
• Operator Precedence
1.
2.
3.
4.
Parentheses
NOT
AND
OR
2.5 Boolean Functions
• A Boolean function is described by an algebraic expression and consists of
binary variables, the constants 1 and 0, and the logic operation symbols
– Consider:
F1 = x + y’z
F2 = x’y’z + x’yz + xy’
x
y
z
y’z
x+y’z
x’y’z
x’yz
xy’
x’y’z + x’yz + xy’
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
1
0
1
0
0
0
0
0
0
0
0
1
1
0
0
0
1
0
1
1
0
0
0
1
0
0
1
1
1
0
1
1
1
0
0
1
1
1
1
0
0
1
0
0
0
0
1
1
1
0
1
0
0
0
0
2.5 Boolean Functions
F1 = x + y’z
y’
y’z
2.5 Boolean Functions
F2 = x’y’z + x’yz + xy’ = x’z(y’ + y) = x’z + xy’
x
y
z
x’z
xy’
x’z + xy’
x’y’z
x’yz
xy’
x’y’z + x’yz + xy’
0
0
0
0
0
0
0
0
0
0
0
0
1
1
0
1
1
0
0
1
0
1
0
0
0
0
0
0
0
0
0
1
1
1
0
1
0
1
0
1
1
0
0
0
1
1
0
0
1
1
1
0
1
0
1
1
0
0
1
1
1
1
0
0
0
0
0
0
0
0
1
1
1
0
0
0
0
0
0
0
2.5 Boolean Functions
2.5 Boolean Functions
• Algebraic Manipulation
– Term: A gate is required for each term and each
variable within the term is an input to the gate
– Literal: A single variable within a term, in
complemented or uncomplemented form
– Reducing the number of terms, number of literals, or
both may result in a simpler circuit
F2 = x’y’z + x’yz + xy’
has three terms and eight literals
F2 = x’z + xy’
has two terms and four literal
2.5 Boolean Functions
• Complement of a Function
– The complement of function F is F’ and is obtained by the
interchange of 0’s with 1’s and 1’s with 0’s in the value of F
– The complement of a function may be derived
algebraically using DeMorgan’s theorem
(A + B)’ = A’B’
theorem 5a (DeMorgan’s)
– Extended to three variables
(A + B + C)’ = (A + x)’
where x = B + C
= A’x’ theorem 5a (DeMorgan’s)
= A’(B + C)’ substitute B + C = x
= A’(B’C’)
theorem 5a (DeMorgan’s)
= A’B’C’
theorem 4b (associative)
– In General
(A + B + C + D + … + F)’ = A’B’C’D’ … F’
(ABCD … F)’ = A’ + B’ + C’ +D’ + … + F’
2.5 Boolean Functions
• Example 2.3
– Find the complement of the previous functions F1 and F2
• F1 = x + y’z
F1’ = (x + y’z)’
= x’ ⋅ (y’z)’
= x’ ⋅ y + z’
• F2 = x’y’z + x’yz + xy’
F2’ = (x’y’z + x’yz + xy’)’
= ((x’y’z)’ ⋅ (x’yz)’ ⋅ (xy’))’
= (x + y + z’) ⋅ (x + y’ + z’) ⋅ (x’ + y)
= (a + y)(a + y’) ⋅ (x’ +y)
= (aa + a(y’ + y) + yy’) ⋅ (x’ + y)
= (a + a + 0) ⋅ (x’ + y)
= a + (x’ + y)
= (x + z’) ⋅ (x’ + y)
• F2 = x’z + xy’
F2’ = (x’z + xy’)’
= (x’z)’ ⋅ (xy’)’
= (x + z’) ⋅ (x’ + y)
let x + z’ = a
recall xx = x, xx’ = 0,
recall x + x = x
x + x’ = 1
Homework
• 2.2 a – c (10 points ea.)
• 2.5 (for 2.2 a-c) (10 points ea.)
• 2.11a, b (10 points ea.)
• 2.14 a-c (10 points ea.)
Total of 110 points