MELJUN CORTES HANDOUTS Boolean Algebra
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BooleanAlgebra LogicDesignand Switching * Property of STI Page1of 33 an algebra that is applied only to binary numbers has a list of operations that it can use to simplify its equations 1 _________________ ________ ________ ________ ________ ________ ________ ________ ________ ________ ________ ________ BooleanAlgebra LogicDesignand Switching * Property of STI Page3of 33 There exists a set of K objects or elements, subject to an equivalence relation, denoted “=”, which satisfies the principle of substitution. defines that variables can be defined using a (Boolean) algebraic expression by using the symbol “=” 3 __________________ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ BooleanAlgebra LogicDesignand Switching * Property of STI Page2of 33 a set of rules defined on how a set of numbers can be manipulated also called Huntington’s Postulates used to derive certain theorems applicable assuming certain conditions Huntington’s Postulates I. There exists a set of K objects or elements, subject to an equivalence relation, denoted “= ”, which satisfies the principle of substitution. IIa. A rule of combination “+ ” is defined s uch that A + Bis in K whenever both A and Bare in K . IIb. A rule of combination “· ” is defined s uch that A · B(abbreviated AB) is in K whenever both A and B are in K . IIIa. There exists an element 0 in Ksuch that, for every A in K , A + 0 = A. IIIb. There exists an element 1 in K such that, for every A in K, A · 1 = A. IV. Commutative Property a. Add it io n: A + B = B + A b. Mu lt i pl i c at ion: A · B = B · A V. DistributiveLaw a. A + ( B· C ) = ( A + B) · ( A + C ) b. A · (B + C ) = ( A · B) + ( A · C ) VI. For every element A in K , there exists an element A’ such that A · A’ = 0 and A + A’ = 1. VII. There are at least two elements X and Y in K such that X = Y Sourc e:Hill,F.andPeterson ,G.,“IntroductiontoSwitchi ngTheoryandLogica lDesign”,pp.4 2 -43 2 _________________ ________ ________ ________ ________ ________ ________ ________ ________ ________ ________ ________ BooleanAlgebra LogicDesignand Switching * Property of STI Page4of 33 A rule of combin ation “+” is de fined such tha t A + B is in K whenever both A and B are in K. defines the addition of binary numbers whenever two binary numbers are added, their sum is also a binary number 4 __________________ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______
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Transcript of MELJUN CORTES HANDOUTS Boolean Algebra
8/8/2019 MELJUN CORTES HANDOUTS Boolean Algebra
http://slidepdf.com/reader/full/meljun-cortes-handouts-boolean-algebra 1/9
BooleanAlgebra
LogicDesignand Switching
* Property of STI
Page1of 33
an algebra that is applied only to binary
numbers
has a list of operations that it can use to simplifyits equations
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LogicDesignand Switching
* Property of STI
Page3of 33
There exists a set of K objects or elements, subject
to an equivalence relation, denoted “=”, whichsatisfies the principle of substitution.
defines that variables can be defined using a(Boolean) algebraic expression by using the
symbol “=”
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BooleanAlgebra
LogicDesignand Switching
* Property of STI
Page2of 33
a set of rules defined on how a set of numbers can bemanipulated
also called Huntington’s Postulates
used to derive certain theorems applicable assuming certainconditions
Huntington’s Postulates
I. There exists a set of K objects or elements, subject to anequivalence relation, denoted “= ”, which satisfies the principle ofsubstitution.
IIa. A rule of combination “+” is defined such that A + B is in K whenever both A and B are in K .
IIb. A rule of combination “·” is defined such that A · B (abbreviated AB) is in K whenever both A and B are in K .
IIIa. There exists an element 0 in K such that, for every A in K , A + 0 =
A.IIIb. There exists an element 1 in K such that, for every A in K, A · 1 =
A.
IV. Commutative Property
a. Ad di ti on: A + B = B + A
b . M ul tipl i ca tion : A · B = B · A
V. Distributive Law
a. A + ( B · C ) = ( A + B) · ( A + C )
b. A · (B + C ) = ( A · B) + ( A · C )
VI. For every element A in K , there exists an element A’ such that
A · A’ = 0
and
A + A’ = 1.
VII. There are at least two elements X and Y in K such that X = Y
Source:Hill,F.andPeterson,G.,“ IntroductiontoSwitchingTheoryandLogicalDesign”,pp.42-43
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* Property of STI
Page4of 33
A rule of combination “+” is defined such that A + B
is in K whenever both A and B are in K.
defines the addition of binary numbers
whenever two binary numbers are added, their
sum is also a binary number
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BooleanAlgebra
LogicDesignand Switching
* Property of STI
Page5of 33
A rule of combination “ ·” is defined such that A · B
(abbreviated AB) is in K whenever both A and Bare in K.
defines the multiplication of binary numbers
whenever two binary numbers are multiplied,
their product is also a binary number
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LogicDesignand Switching
* Property of STI
Page7of 33
There exists an element 1 in K such that, for every
a in K, A · 1 = A.
defines the one element in the binary
numbering system
whenever one is multiplied to any binary
number, the value of the number to which one
was multiplied will not change
one is defined such that when it is multiplied to
a binary number, the binary number will retain
its value
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BooleanAlgebra
LogicDesignand Switching
* Property of STI
Page6of 33
There exists an element 0 in K such that, for every
A in K, A + 0 = A.
defines the zero element in the binary
numbering system
whenever zero is added to any binary number,the value of the number to which zero was
added will not change
zero is defined such that when it is added to abinary number, the binary number will retain its
value
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LogicDesignand Switching
* Property of STI
Page8of 33
Cummutative Property of Addition: A + B = B + A.
reversing the order of two binary numbers inperforming addition is shown as not to effect the
resulting sum
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BooleanAlgebra
LogicDesignand Switching
* Property of STI
Page9of 33
Cummutative Property of Multiplication: A · B = B ·
A.
reversing the order of two binary numbers in
performing multiplication is shown as not to
effect the resulting product
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Page11of 33
Distributive Property of Multiplication: A · (B + C) =
(A · B) + (A · C).
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BooleanAlgebra
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* Property of STI
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Distributive Property of Addition: A + (B · C) = (A +
B) · (A + C).
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* Property of STI
Page12of 33
For every element A in K, there exists an element
A’ such that A · A’ = 0 and A + A’ = 1.
defines the complement ( A’ ) of any variable A
such that when A’ is multiplied to A, the
resulting product is zero
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BooleanAlgebra
LogicDesignand Switching
* Property of STI
Page13of 33
There are at least two elements X and Y in K such
that X ¹ Y.
defines at least two elements in K that are not
equal
Assign the following expressions to variables X
and Y :
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* Property of STI
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Theorem 1 (a) X + X = X (b) X · X = X
Theorem 2 (a) X + 1 = 1 (b) X · 0 = 0
Theorem 3 (X’)’ = X
Theorem 4 (a) X+ (Y+Z ) = (X+Y) +Z (b) X(YZ) = (XY)Z
Theorem 5 (a) (X +Y)’ = X’Y’ (b) (XY)’ = X’+Y’
Theorem 6 (a) X + XY = X (b) X(X+Y) = X
Theorem 3 is sometimes called INVOLUTION and Theorems
6a&b are forms of ABSORPTION. Other forms of absorption
will be presented in a few examples later.
the number of combinations is directly related to
the number of variables
2# of variables = # of combinations
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BooleanAlgebra
LogicDesignand Switching
* Property of STI
Page14of 33
Boolean Algebra
1. valid only for a set havingtwo elements (0 and 1)
2. although this propertycould be derived orproven to hold true theHuntington’s postulatesdo not include
Associative Property
3. the distributive propertyof + over · is valid sinceits set have only twoelements (0 and 1)
4. Additive andMultiplicative Inversesare not defined
5. Subtraction and Divisionoperations are alsoundefined
6. the definition of acomplement is defined
7. especially defined tohave the property ofduality
(Decimal) Algebra
ordinary (decimal)
algebra is applicable to aset having infinite number
of elements
not applicable
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* Property of STI
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Solution:
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BooleanAlgebra
LogicDesignand Switching
* Property of STI
Page17of 33
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Theorem 2(a): X + 1 = 1
Truth table:
Theorem 2(b): X · 0 = 0
Truth table:
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BooleanAlgebra
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* Property of STI
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Theorem 1(a): X + X = X
Truth table:
Theorem 1(b): X · X = X
Truth table:
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* Property of STI
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Theorem 3: ( X’ )’ = X Truth table:
Y = X’
Y’ = (X’)’
Y’ = X = (X’)’
Theorem 4(a): X + (Y + Z ) = (X + Y ) + Z
Truth table:
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BooleanAlgebra
LogicDesignand Switching
* Property of STI
Page21of 33
Theorem 4(b): X · (Y · Z ) = (X · Y ) · Z
X · (Y · Z) = X · (Y · Z + 0) Postulate 3a
= X · (Y + 0) · (Z + 0) Postulate 5a
= ((X · Y) + (X · 0)) · Z Postulate 5b
Postulate 3a
= ((X · Y) + 0) · Z Theorem 2b
= (X · Y) · Z Postulate 3a
Truth table of X · (Y · Z ):
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LogicDesignand Switching
* Property of STI
Page23of 33
(X + Y) + X’Y’ = 0
((X + Y) + X’) · ((X + Y)+Y’) Postulate 5a
((X + X’)+ Y) · (X +(Y + Y’)) Postulate 4
(1 + Y) · (X+ 1) Postulate 6
1 · 1 Theorem 2a
1 Postulate 3b
Truth table:
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BooleanAlgebra
LogicDesignand Switching
* Property of STI
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Theorem 5(a): (X + Y )’ = X’Y’
(DeMorgan’s Theorem)
Postulate 6
A · A’ = 0
and
A + A’ = 1
A = ( X + Y )
A’ = ( X + Y )’
X ’Y ’ = ( X + Y )’
(X + Y) · (X + Y)’ = 0 (X + Y) · X’Y’ = 0
and
(X + Y) + (X + Y)’ = 1 (X + Y) + X’Y’ = 1
(X + Y) · X’Y’ = 1
X’Y’ · (X + Y) Postulate 4b
((X’Y’) · X)+((X’Y’) · Y) Postulate 5b
((X’X) · Y’)+((YY’) · X’) Postulate 4b
(0 · Y’) + (X’ · 0) Theorem 6
0 + 0 Theorem 2b
0 Postulate 3a
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Theorem 5(b): (XY )’ = X’ + Y’
(DeMorgan’s Theorem)
Postulate 6
(XY) · (XY)’ = 0 (XY) · X’ + Y’ = 0
and
(XY) + (XY)’ = 1 (XY) + X’ + Y’ = 1
(XY)·(X’ + Y’) = 0
((XY) · X’)+((XY) · Y’) Postulate 5b
((X’X) · Y)+((YY’) · X) Postulate 4b
(0 · Y) + (X · 0) Theorem 6
0 + 0 Theorem 2b
0 Postulate 3a
(XY) + (X’ + Y’) = 1
((X’+ Y’)+ X) · ((X’+ Y’)+Y) Postulate 5a
((X+X’)+ Y’) · (X’+(Y + Y’)) Postulate 4a
(1+ Y’) · (X’ + 1) Postulate 6
1 · 1 Theorem 2a
1 Postulate 3b
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BooleanAlgebra
LogicDesignand Switching
* Property of STI
Page25of 33
Truth table:
Theorem 6(a): X + XY = X
X + XY = (X · 1) + (X · Y) Postulate 3b
= X · (1 + Y) Postulate 5b
= X · 1 Theorem 2a
= X Postulate 3b
Truth table:
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* Property of STI
Page27of 33
Using the Postulates and Theorems presented
previously, prove the following Theorems andwrite down their respective truth tables:
1. X +X ’Y =X +Y Theorem6c
2. X (X ’+Y )=XY Theorem6d
3. X ’+XY =X ’+Y Theorem6e
4. X ’(X +Y )=X ’Y ’ Theorem6f
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BooleanAlgebra
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* Property of STI
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Theorem 6(b): X (X + Y ) = X
X(X+Y) = (X · X) + (X · Y) Postulate 5b
= X + XY Theorem 1b
= X Theorem 6a
Truth table:
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* Property of STI
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Solution:
1. X + X ’Y =X + Y
X + X ’Y = (X+XY)+X’Y Theorem 6a
= X+Y(X+X’) Distributive Property
(Postulate 5b)
= X+Y ·1 Postulate 6
= X+Y Postulate 3
Truth table:
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BooleanAlgebra
LogicDesignand Switching
* Property of STI
Page29of 33
2. X (X ’+Y )=XY
X ( X ’+Y ) = XX’+XY Distributive Property
(Postulate 5b)
= 0+XY Postulate 6
= XY Postulate 3
Truth table:
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* Property of STI
Page31of 33
4. X + X ’Y =X ’Y ’
X + X ’Y = X’X+X’Y Distributive Property
(Postulate 5b)
= 0+X’Y Postulate 6
= X’Y Postulate 3
Truth table:
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BooleanAlgebra
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* Property of STI
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3. X ’ + XY =X ’+Y
X ’ + XY = (X’+X’Y )+XY Theorem 6a
= X’+Y (X’+X) Postulate 5b
= X’+Y (1) Postulate 6
= X’+Y Postulate 3
Truth table:
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* Property of STI
Page32of 33
Manipulate the following Boolean expressions
to minimize the number of terms or variables
used. Follow Operator Precedence and write-
out a corresponding truth table that will showthe output values for every input combination.
5. F = (A’ + B’ + C) · (AB)’
6. F = X + (Y’ + XY’)
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BooleanAlgebra
LogicDesignand Switching
* Property of STI
Page33of 33
Solution:
5. F = (A ’ + B ’ + C ) · (AB )’
= ( A ’ + B ’ + C ) · ( A ’ + B ’) De Morgan’s
Theorem
= ( A ’ + B ’) · ( C + 1) Postulate 5b
= ( A ’ + B ’) · 1 Theorem 2a= A ’ + B ’ Postulate 3b
Truth table:
6. F = X + (Y ’ + XY ’)
= X + Y ’ Theorem 6a
Truth table:
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