Boolean Algebra
Sungho Kang
Yonsei University
Outline
Set, Relations, and Functions
Partial Orders
Boolean Functions
Dont Care Conditions
Incomplete Specifications
Computer Systems Lab.
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2
Set Notation
v ∈V
Element v is a member of set V
v ∉V
Element v is not a member of set V
|V|
Cardinality (number of members) of set V
S⊆V
Set S is a subset of
∅
V
S′
The empty set (a member of all sets)
The complement of set S U
U
The universe: S ′ = U − S
Computer Systems Lab.
S
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V
3
Set Notation
Inclusion (⊂)
Proper Inclusion (⊆)
Complementation
Intersection (∩)
Union (∪)
Difference
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Power Sets
2 = {S|S ⊆ V }
V
2
V
|2 | = 2
V
|V |
The power set of set V (the set of all
subsets of set V
The cardinality of a power set is a
power of 2
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5
Power Sets
V = {0,1,2}
3-member set
2 = {∅,
{0},{1},{2},
{0,1},{0,2},{1,2},
{0,1,2}}
V
1 subset with 0 members
3 subsets with 1 members
3 subsets with 2 members
1 subset with 3 members
|2V | = 2|V | = 23 = 8
Power sets are Boolean Algebras
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Cartesian Products
The Cartesian Product of sets A and B is denoted
AxB
Suppose A = {0,1,2}, B = {a , b} , then
A × B = {(0, a ),(0, b),(1, a ),(1, b),(2, a ),(2, b)}
A = {0,1,2} Set A is unordered
(1, b) ( ) denotes Ordered Set
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Binary Relations
The Cartesian Product of sets A and B is denoted A x B
A x B consists of all possible ordered pairs (a,b) such that
a ∈ A and b ∈ B
A subset R ⊆ A × B is called a Binary Relation
Graphs, Matrices, and Boolean Algebras can be viewed as
binary relations
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Binary Relations as Graphs or Matrices
E = {ab, cb, cd , cf } ⊆ A × B
A = {a, c} B = {b, d , f }
a
c
b
d
f
Bipartite Graph
Computer Systems Lab.
b
a 1
c 1
d
f
0
1
0
1
Rectangular Matrix
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Binary Relations as Graphs or Matrices
E = {ab, ac, ad , bd , cd}∪ {aa, bb, cc, dd} ⊆ V × V
V = {a, b, c, d}
a
c
b
d
Directed Graph
Computer Systems Lab.
a1111
b0101
c0011
d0001
Square Matrix
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Properties of Binary Relations
A binary relation R ⊆ V
reflexive, and/or
transitive, and/or
symmetric, and/or
antisymmetric
× V can be
We illustrate these properties on the next few slides
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Notation for Binary Relations
If R ⊆ A × B , we say that A is the domain
of the relation R , and that B is the range.
If ( a, b) ∈ R , we say that the pair
is in the relation R , or aRb .
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Missing Figure 3.2, Page 80 3.1.3
≤ ⊆ A2 = {1,2,3,4}2
1
2
3
1 2 3 4
≤ ≤ ≤ ≤
≤ ≤ ≤
4
≤ ≤
≤
Graph View
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Example: Less than or Equal
0 1 2 3
A = B = {0,1,2,},
R ⊆ A × B = " ≤"
Computer Systems Lab.
0
1
2
3
≤
0

0
0


≤
≤
0
0
≤
≤
≤
0
≤ 

≤

≤


≤


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Example: a times b = 12
V = {0,1,2,f},
R ⊆ V × V = {(u, v)| u × v = 12}
, ),(2,6),(3,4),( 4,3),(6,2),(12,1)}
R = {(112
Computer Systems Lab.
1
2
3
12
6
4
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Reflexive Binary Relations
V = {a, b, c, d}
a
R⊆V ×V
v ∈V ⇒ vRv
c
b
d
A binary relation R ⊆ V × V is
reflexive if and only if (v, v) ∈ R
for every vertex v ∈V
a
b
c
d
Computer Systems Lab.
1
0

0
0

1
1
0
0
1
0
1
0
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1
1

1
1
16
Non-Reflexive Binary Relations
V = {a, b, c, d}
a
R⊆V ×V
∃v ∈V ∋ ¬vRv
c
b
d
Non-Reflexivity implies that
there exists v ∈V such that
(v, v) ∉ R . Here d is such a v .
a
b
c
d
Computer Systems Lab.
1
0

0
0

1
1
0
0
1
0
1
0
1
1

1
0
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Transitive Binary Relations
If u, v, w ∈V , and uRv,
and vRw, then uRw.
a
b
d
c
e
A binary relation is transitive if and only
if every (u,v,w) path is triangulated by a
direct (u,w) edge.
This is the case here, so R is transitive.
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Non-Transitive Binary Relations
R⊆V ×V
∃u, v, w ∈V , such that
uRv, vRw, but ¬uRw
a
b
d
c
e
A binary relation is not transitive if there
exists a path from u to w, , through v that is not
triangulated by a direct (u,w) edge
Here (a,c,e) is such a path.
Computer Systems Lab.
Edge (a,e) is
missing
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Symmetric Binary Relations
R⊆V ×V
(u, v ) ∈ R ⇒ (v, u) ∈ R
(uRv ⇒ vRu)
A binary relation is
symmetric if and only
if every (u, v ) edge is
reciprocated by a
( v, u ) edge
Computer Systems Lab.
1
2
3
12
6
4
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Non-Symmetric Binary Relations
R⊆V ×V
∃(u, v ) ∈ R such that (v, u) ∉ R
A binary relation is
non-symmetric if there
exists an edge (u, v ) not
reciprocated by an
edge ( v, u )
Computer Systems Lab.
Edge (c,a) is
missing
a
b
d
c
e
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Antisymmetric Binary Relations
∀ (u , v ) ∈ R , (u R v , v R u ) ⇒ (v = u )
A binary relation is antisymmetric if and only
if no (u, v ) edge is
reciprocated by a ( v, u )
edge v = u
a
b
d
c
e
Not antisymmetric if any such edge is reciprocated :
b → d, d → b
here
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Functions
A function f are binary relations from set A (called the
domain) to B (called the range)
But, it is required that each a in A be associated with
exactly 1 b in B
For functions, it cannot be true that both (a,b) in R and
(a,c) in R, b different from c
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The Binary Relation
The Binary Relation of Relations to Synthesis/Verification
*DC=dont care
Partial
Orders
Equivalence
Relations
Compatibility
Relations
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Combinational Logic (no DCs)
=> (0,1) Boolean Algebra
Combinational Logic (with DCs)
=> Big Boolean Algebras
Sequential Logic (no DCs)
Sequential Logic (with DCs)
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The Path Relation *→
G = (V , E )
a
c
b
{a,b,c,d ,e}
* {c,d ,e}
* {c,e}
* {c,e}
* {c,e}
*→

→
→
d
e
This graph defines path
relation
Relation
a
b
c
d
e
*→

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→
→
is sometimes called Reachability
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Equivalence Relations
aa
c
dd
ee
bb
Note R is reflexive, symmetric, and transitive
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Equivalence Relations
R = {(u,v)|v
G =(V , E )
a
u, u * v}
G = (V , R)
c
b
d
e
This graph defines path
relation *
→
NOT an equivalence
relation: E
Computer Systems Lab.
*→

→
a
c
d
e
b
This graph defines R
An equivalence relati
on:
R
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The Cycle Relation
C ={(u,v)|v
G =(V , E )
a
*→
←
G = (V , R)
c
b
d
e
This graph defines path
relation *
u}
a
c
d
e
b
←→
These are called the Strongly
Connected Components of G
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Refinement
Given any set B a partition of B is a set of subsets Bi ⊆B
with two properties
B B = for all i j
φ
≠
i∩
j
∪i Bi = B
Given two partitions P1 and P2 of a set S, P1 is a
refinement of P2 if each block B1i of P1 is a subset of some
block of π2
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Other Binary Relations
• Partial Orders (Includes Lattices, Boolean Algebras)
•Reflexive
•Transitive
•Antisymmetric
• Compatibility Relations
•Reflexive
•Not Transitive--Almost an equivalence relation
•Symmetric
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Functions
A function f from A to B written f : A → B is a rule that
associates exactly one element of B to each element of A
A relation from A to B is a function if it is right-unique and if every
element of A appears in one pair of the relation
A is called the domain of the function
B is called the co-domain (range)
If y=f(x) : A → B, y is image of x
Given a domain subset C A
⊆
IMG(f,C) = { y ∈ B | ∃x ∈ C ∋ y = f(x) }
preimage of C under f
PRE(f,C) = { x ∈ A | ∃y ∈ C ∋ y = f(x) }
A function f is one-to-one (injective) if x ≠ y implies f(x) ≠
f(y)
A function f is onto (surjective) if for every y∈B, there
exists an element x∈A, such that f(x) = y
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Algebraic Systems
The pair (V , ≤ ) is called an algebraic system
V is a set, called the carrier of the system
≤ is a relation on V × V = V 2
( ⊆ , ⇒ are similar to ≤ )
This algebraic system is called a partially ordered set,
or poset for short
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Partially Ordered Sets
A poset has two operations, • and + , called
meet and join (like AND and OR)
Sometimes written (V , ≤ ,⋅, + ) or (V 2 , ≤ ,⋅, + )
even ( V ,⋅, + ) , since ≤ is implied
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Posets and Hasse Diagrams
Integers (Totally Ordered):
Missing Figure 3.2
≤ ⊆ A2 = {1,2,3,4}2
1
2
3
4
1 2 3 4
≤ ≤ ≤ ≤
≤ ≤ ≤
4
3
2
≤ ≤
≤
Matrix View
1
Graph View
Hasse Diagram
Hasse Diagram obtained deleting
arrowheads and redundant edges
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Posets and Hasse Diagrams
(V , ≤ ,⋅, + )
a b ≤\ c , c ≤\ b
V = {a , b, c, d }
≤= {( a , a ), ( a , b ), ( a , c ), ( a , d ),
c
b
( b , b ), ( b , d ),
d
b c
a
d
( c , c ), ( c , d ), ( d , d )}
Relation ≤
Hasse Diagram
≤ :distance from top
≤ is refelxive, antisymmetric,
and transitive: a partial order
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Meet
An element m of a poset P is a lower bound of elements
a and b of P, if m ≤ a and m ≤ b .
m is the greatest lower bound or meet of elements a and b
if m is a lower bound of a and b and, for any m′ such
that m′≤ a and also m ′ ≤ b, m ′ ≤ m.
b
a
m
m′
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Join: the Dual of Meet
An element m of a poset P is a upper bound of elements a
and b of P , if a ≤ m and b ≤ m .
m is the least upper bound or join of elements a and b
if m is an upper bound of a and b and, for any m ′ such
that a ≤ m ′ and also b ≤ m ′, m ≤ m ′ .
m′
m
a
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b
37
⋅
Meet ( ) and Join (+)
posets:
⋅
d
b
d
c
b
a
a⋅b = a
a+b=b
Computer Systems Lab.
a
c⋅b = a
c+b=?
d
d
b
c
b
a
a
c⋅b = a
c⋅a = ?
c+b= d
d+b=?
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c
38
POSET Properties
Theorem 3.2.1 If x and y have a greatest
lower bound (meet), then
j
x ≥ x⋅y
x
y
Similarly, if x and y have a least
m
upper bound (join), then
x≤x+y
Proof: Since meet exists, x ≥ x ⋅ y by definition.
Also, since join exists, x ≤ x + y by definition.
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More POSET Properties
Theorem 3.2.2 x ≤ y ⇔ x ⋅ y = x
Proof (⇒):
This means assume x ≤ y .
• x is a lower bound of x and y (by Def.)
• x is also the meet of x and y
y
Proof: by contradiction. Suppose
x ≠ x ⋅ y. Then ∃m ≠ x such that x ≤ m
x
where m = x ⋅ y . But since m was a lower
bound of x and y , m ≤ x as well. Thus
m
m = x , by the anti-symmetry of posets.
Proof (⇐): From Definition of meet
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Well-Ordered
If all pairs of elements of a poset are comparable, then the
set is totally ordered
If every non-empty subset of a totally ordered set has a
smallest element, then the set is well-ordered
e.g.) Natural numbers
Mathematical Induction
Given, for all n N, propositions P(n), if
∈
P(0) is true
for all n>0, if P(n-1) is true then P(n) is true
then, for all n N, P(n) is true
∈
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Lattices
Lattice: a poset with both meet and join for every pair of
elements of the carrier set
Boolean Algebra: a distributed and complemented lattice
Every lattice has a unique minimum element and a unique
maximum element
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Lattices and Not Lattices
⋅
d
b
d
c
b
a
a⋅b = a
a+b=b
(lattice)
d
a
c⋅b = a
c+b=?
d
b
c
b
a
a
c⋅b = a
c⋅a = ?
c+b= d
(lattice)
d+b=?
c
(Boolean algebra)
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Examples of Lattices
g=1
d =1
e
c
b
a=0
Computer Systems Lab.
e=1
f
d
c
b
a=0
b
c d
a=0
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44
More Notation
• There Exists a v in set V is denoted by ∃v ∈V
• The following are equivalent:
(ab + a ′b ′ )
a⇔b
a ⇒ b and a ⇐ b
(( a′ + b)( a + b ′ ))
a is true if and only if b is true
• Does ( a ′ + b)( a + b ′ )make sense in a poset?
No--Complement is defined for lattices but
not for posets
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Properties of Lattices
Meet, Join, Unique maximum (1), mi
nimum (0) element are always defin
ed
x+x= x
Commutative: x + y = y + x
Idempotent:
Associative:
Absorptive:
x + ( y + z ) = ( x + y) + z
x ⋅ ( x + y) = x
x⋅x = x
x⋅y = y⋅ x
x ⋅ ( y ⋅ z ) = ( x ⋅ y) ⋅ z
x + ( x ⋅ y) = x
Absorptive properties are fundamental to op
timization
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Duality
Every lattice identity is transformed into another
identity by interchanging:
• + and ⋅
• ≤ and ≥
• 0 and 1
Example: x ⋅ ( x + y ) = x → x + ( x ⋅ y) = x
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Complementation
If x+y=1 and xy=0 then x is the complement of y and vice
versa
A lattice is complemented if all elements have a
complement
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Examples of Lattices
Complemented?
yes
no
d =1
g=1
e
c
b
yes
e=1
f
d
b
c d
a=0
c
b
a=0
b⋅c = a = 0
b⋅c = a = 0
b⋅c = a = 0
b+c= d =1
b+c = d ≠1
b+c=e=1
Computer Systems Lab.
a=0
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49
Distributivity
Semi-distributivity:
x ⋅ ( y + z ) ≥ ( x ⋅ y) + ( x ⋅ z )
x + ( y ⋅ z ) ≤ ( x + y) ⋅ ( x + z )
Proof :
1. x ⋅ y ≤ x
(def. of meet)
2. x ⋅ y ≤ y ≤ y + z
(def. of meet, join)
3. x ⋅ y ≤ x ⋅ ( y + z )
(def. of meet)
4. x ⋅ z ≤ x ⋅ ( y + z )
(mutatis mutandis: y ↔ z)
5. x ⋅ ( y + z ) ≥ ( x ⋅ y) + ( x ⋅ z ) (def. of join)
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Distributivity
• Boolean Algebras have full distributivity:
x ⋅ ( y + z ) = ( x ⋅ y) + ( x ⋅ z )
x + ( y ⋅ z ) = ( x + y) ⋅ ( x + z )
• Boolean Algebras are complemented. That is,
x = y ′ ⇒ ( x ⋅ y = 0) and ( x + y = 1)
must hold for every x in the carrier of the poset
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Definition of Boolean Algebra
A complemented, distributive lattice is a Boolean lattice or
Boolean algebra
Idempotent
x+x=x
Commutative
x+y=y+x
Associative
x+(y+z)=(x+y)+z
Absorptive
x(x+y)=x
Distributive
x+(yz)=(x+y)(x+z)
Existence of the complement
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xx=x
xy = yx
x(yz) = (xy)z
x+(xy) = x
x(y+z) = xy +xz
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Are these Lattices Boolean Algebras?
Complemented and distributed?
yes
no
d =1
g=1
c
b
no
a=0
b⋅c = a = 0
b + c = d =1
a(b + c) =
a⋅b + a⋅c = a = 0
Computer Systems Lab.
e
e=1
f
d
c
b
a=0
b⋅c = a = 0
b+c = d ≠1
b
c d
a=0
b ⋅ (c + d ) = b ≥
(b ⋅ c) + (b ⋅ d ) = a
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Fundamental Theorem of Boolean Algebras
• Every poset which is a Boolean Algebra has
a power of 2 elements in its carrier
• All Boolean Algebras are isomorphic to the p
ower set of the carrier.
1
Example:
V = {1}, 2 = {∅,1} = {0,1}
( = 1)
V
Computer Systems Lab.
0 ( = ∅)
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Examples of Boolean Algebras
1
( = 1)
1-cube
0 ( = ∅)
V ={1}, 2V ={∅,1}= {0,1}
, },
V ={01
2V ={∅,01
, ,{01
, }}
, ,1}
={0,01
1
0
( = {0,1} = V )
1
2-cube
0 ( = ∅)
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Atoms of a Boolean Algebra
• A Boolean Algebra is a Distributive, Comple
mented Lattice
• The minimal non-zero elements of a Boolean
Algebra are called atoms
| V | = 2 n ⇔ n atoms
1
1 1-atom
0
1 2-atoms
0
0
Can a Boolean Algebra have 0 atoms?
Computer Systems Lab.
NO!
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56
Examples of Boolean Algebras
A = {a, b, c}
V = 2A
1
= {A,
{a, b},{a, c},{b, c},
a, b, c,
∅}
Computer Systems Lab.
3 atoms
{a, b}
{a}
(= V )
{a, c}
{b}
{b, c}
{c}
0 ( = ∅)
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57
Examples of Boolean Algebras
n =| A| = 4 atoms, n + 1 = 5 levels, 2 n = 16 elts
A = {a, b, c, d}
V
Level 4--1 elt*
Level 3--4 elts
Level 2--6 elts
Level 1--4 elts
Level 0--1 elts
Computer Systems Lab.
= 2A
= {A,
C44
{a, b, c},{a, b, d},{a, c, d},{b, c, d},
C34
{a, b},{a, c},{a, d},{b, c},{b, d},{c, d} C24
4
{a},{b},{c},{d}
C1
4
C0
∅}
* elt = element
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58
Boolean Algebras
Theorem 3.2.6
Complementation is unique.
proof: Suppose x ′ and y are both complements
of x ( x + y = 1, xy = 0 ). Hence
y = y( x + x ′) = x ′y + xy = x ′y
= x ′y + x ′x = x ′( y + x ) = x ′
Note we used distributivity. Similarly, we have
Theorem 3.2.7 (Involution): ( x ′ )′ = x
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Properties of Boolean Algebra
x + xy = x+y
x(x+y) = xy
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Lemma: The Isotone Property
x ≤ y ⇔ xz ≤ yz
Proof: By Theorem 3.2.2, x ≤ y ⇔ x = xy , so
we get xz = xyz = xyzz = ( xz )( yz ) . Note we used i
dempotence, commutativity, and associativity. Then
we use Theorem 3.2.2 again to prove the lemma.
y
xz ≤ yz ≅ xz ⊆ yz
x
z
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Duality for Boolean Algebras
Every Boolean Algebra identity is transformed
into another valid identity by interchanging:
• + and ⋅
• ≤ and ≥
• 0 and 1
• ()′ and () This rule not valid for lattices
Example:
Computer Systems Lab.
xx ′ = 0 → x ′ + x = 1
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Theorem
x ≤ y ⇔ xy′ = 0
⇔ x′ + y = 1
y
x
Proof: By the isotone property we have
x ≤ y ⇔ xy ′ ≤ yy ′ ⇔ xy′ ≤ 0 ⇔ xy ′ = 0
The second identity follows by duality.
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Other Theorems
DeMorgans Laws
( x + y)′ = x ′y ′
( xy)′ = x ′ + y ′
Consensus
xy + x ′z + yz = xy + x ′z
( x + y)( x ′ + z )( y + z ) = ( x + y)( x ′ + z )
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Consensus Examples
1. a ′bc + abd + bcd = abc + a ′bd
3. abe + bce + bde + ac' d ′
= abe + be(c + d ) + a(c + d )′ Note use of
DeMorgans Law
= be(c + d ) + a(c + d )′
= bce + bde + ac' d ′
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Consensus Example
5. Is wrong. Replace by
a′c ′d + b ′c ′d + acd + bcd
= ( a′bd ) + a ′c ′d + b ′c ′d + acd + bcd
= a′bd + b ′c ′d + acd
•Uphill Move: Note addition of redundant consensus
term enables deletion of two other terms by consensus
• This avoids local minima--a crucial part of the logic
minimization paradigm
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Functions and Boolean Functions
Ordinary functions of 1 variable:
f ( x ): D R ⇔ f ⊆ D × R
Ordinary functions of 2 real variables:
f ( x , y ): Dx × Dy R ⇔ f ⊆ ( Dx × Dy ) × R
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Functions and Boolean Functions
1
Boolean functions of n variables:
f ( x1 ,, xn ): B B ,
n
f ( x1 ,, xn ) ⊆ ( B ×× B ) × B
Computer Systems Lab.
a
B = {0,,1}
b
0
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Boolean Formulae
Boolean Formulae: Meets and Joins of
Variables and Constants
F1 = 0
F2 = x1x2′ + x1′x2
F3 = x1x1x2′ + x1′x2
F3 = ax1 + b ???
Computer Systems Lab.
1
a
b
0
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69
n-variable Boolean Formulae
B = {0,l,1}
The elements of B are Boolean formulae.
• Each variable x1,l, x n is a Boolean formula.
• If g and h are Boolean formulae, then so are
• g+h
• g⋅h
• g′
• A string is a Boolean formula if and only if
it derives from finitely many applications of
these rules.
•
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Distinct Formulas, Same Function
B = {0, a , b,1}
2
Truth Table for f : B B
x1
x2
F = x1 + x2
G = x1 + x1′x2
Computer Systems Lab.
0000
0ab1
0ab1
0ab1
aaaa
0ab1
aa11
aa11
bbbb
0ab1
b1b1
b1b1
1111
0ab1
1111
1111
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Boolean Functions on Boolean Algebra
B = {0,l,1}
• f ( x1,l, x n ) = xi is a Boolean function
• f ( x1,l, x n ) = e ∈ B is also
• If g and h are Boolean, functions then so are
• g+h
• g⋅h
• g′
• A function is Boolean if and only if it derives
from finitely many applications of these rules.
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Cofactors
Positive Cofactor WRT x1: f x1 = f (1, x2 ,, x n )
Negative Cofactor WRT x1: f x ′ = f ( 0, x2 ,, x n )
1
Note: prime denotes
complement
f x1 = f (1, x2 ,, xn ) = f x1 =1
f x1′ = f (0, x2 ,, xn ) = f x1 =0
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Cofactors
Positive Cofactor WRT x1: f x1 = f (1, x2 ,, x n )
Negative Cofactor WRT x1: f x ′ = f ( 0, x2 ,, x n )
1
Example:
f = abc ′d + a ′cd ′ + bc
f a = bc ′d +
bc
fa′ =
+ cd ′ + bc
Computer Systems Lab.
This term drops
out when a is
replaced by 1
This term is
unaffected
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Cofactors
B = {0, a , b,1}, f ( x ): B B
n
f x1 = a = f ( a, x2 ,, x n )
f x1 = f (1, x2 ,, x n )
Example:
f = ax1′ + bx2
f x1 = a = aa ′ + bx2 = 0 + bx2 = bx2
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Booles Expansion Theorem
f ( x1, x2 ,, x n ) =
xi f xi + xi′ f xi′ = [ xi + f xi′ ] ⋅ [ xi′ + f xi ]
f x1
Example: f = ax1′ + bx2 ,
= a 0 + bx2 = bx2 , f x1′ = a1 + bx2 = a + bx2
Sum form:
f = x1( bx2 ) + x1′( a + bx2 )
Product form: f = [ x1 + ( a + bx2 )] ⋅ [ x1′ + ( bx2 )]
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Variables and Constants
• In the previous slide a and b were constants, and
the x1,l, x n were variables.
• But we can also use letters like a and b as variables,
without explicitly stating what the elements of the Bo
olean Algebra are.
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Booles Expansion (Sum Form)
f = abc ′d + a ′cd ′ + bc
f a = bc ′d +
bc
fa′ =
+ cd ′ + bc
f = af a + a ′f a ′ = a( bc ′d + bc ) + a ′( cd ′ + bc )
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78
Booles Expansion (Product Form)
f = abc ′d + a ′cd ′ + bc
f = [a + f a ′ ][a ′ + f a ]
= [a + ( cd ′ + bc )][a ′ + ( bc ′d + bc )]
= aa ′ + a( bc ′d + bc ) + a ′( cd ′ + bc ) +
( cd ′ + bc )( bc ′d + bc )
Note application
= abc ′d + a ′cd ′ + bcd ′ + bc of absorptive law
= abc ′d + a ′cd ′ + bc
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Minterm Canonical Form
The minterm canonical form is a canonical, or st
andard way of representing functions. From p10
f = x + y ′ +isz represented by millions of
0,
distinct Boolean formulas, but just 1 minterm
canonical form. Note
f = x + y ′ + z = ( x ′yz ′) ′
Thus some texts refer to f as {0,1,3,4,5,6,7}
f = x ′y ′z ′ + x ′y ′z + x ′yz + xy ′z +
(0,
1,
3,
4, )
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Minterm Canonical Form
f ( x1 , x2 ,l, xn )
= x1′ f x1′ + x1 f x1
n levels of recursive cofactoring
n
create 2 constants
x1′x2′ f x1′x2′ + x1′x2 f x1′x2 + x1x2′ f x1x2′ + x1x2 f x1x2
These elementary functions
are called minterms
r
= x1′m xn′ f x1′mxn′ + x1′m xn′ −1xn f x1′mxn′ −1xn +m
+ x1′x2 m xn f x1′x2 mxn + x1m xn f x1mxn
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Minterm Canonical Form
Thus a Boolean function is uniquely determined by
its values at the corner points 0m0,0m01,l,1m1
f ( x1 , x2 ,l, xn )
= x1′ f x1′ + x1 f x1
x1′x2′ f x1′x2′
n
These 2 constants are aptly
called discrimants
+ x1′x2 f x1′x2 + x1x2′ f x1x2′ + x1 x2 f x1x2
r
= x1′m xn′ f x1′mxn′ + x1′m xn′ −1xn f x1′mxn′−1xn +m
+ x1′x2 m xn f x1′x2 mxn + x1m xn f x1mxn
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Minterm Canonical Form
2
f :B v B
B = {0, a , b,1}
f = ax1′ + bx2 , f x1′ = a + bx2 , f x1 = bx2
= x1′ x2′ a + x1′ x2 ( a + b) + x1x2′ 0 + x1x2b
This function is Boolean (from Boolean Formula)
x1
x2
F = ax1′ + bx2
x1′ x2′ a + x1′ x2 + x1x2b
0000
0ab1
aa11
aa11
⇑
⇑
aaaa
0ab1
00bb
00bb
bbbb
0ab1
aa11
aa11
1111
0ab1
00bb
00bb
⇑
⇑
Thus all 16 cofactors match--not just discriminants
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Now Suppose Truth Table is Given
• Here we change 15th cofactor, but leave
the 4 discriminants unchanged
• Since the given function doesnt match
at all 16 cofactors, F is not Boolean
x1
x2
F
x1′ x2′ a + x1′ x2 + x1x2b
Computer Systems Lab.
0000
0ab1
aa11
aa11
⇑
⇑
aaaa
0ab1
00bb
00bb
bbbb
0ab1
aa11
aa11
1111
0ab1
001b
00bb
⇑
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⇑
84
2-Variable Boolean Functions
Minterms
2-variable functions
15
f =1
4
11−14
3 f
= x ′ + y ′, x ′ + y , x + y ′, x + y
5−10
2 f
= x ′, y ′, x , y , xy ′ + x ′y , xy + x ′y ′
1− 4
1
f
= x ′y ′, xy ′, x ′y , xy
0
0
f =0
Note that the minterms just depend on the number
and names of the variables, independent of the
particular Boolean Algebra
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Boolean Function Algebras
The notation Fn(B) means the Boolean Algebra whose ca
rrier is the set of all n-variable Boolean Functions which
n
map B into Fn(B): Bn |-> B minterms
If
B = {0,1} the atoms of
Fn ( B ): B B
Fn(B) are its n-variable
B is called the Base Algebra of the Boolean Function al
gebra
The atoms of B are called Base Atoms
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Large Boolean Algebras?
• When you design an optimal circuit, each gate
must be optimized with respect to its Dont Cares
• Because of Dont Cares, 4 functions of ( x , y ) are
equivalence preserving replacements for gate g
• Optimal Design: pick best such replacement
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Boolean Algebra
F1 ({0, a, b,1})
Level (= # of atoms)
4
atoms of F1 {0, a , b,1}) are
base atoms ⋅ minterms
3
Each element has 3 Atoms
2
Each element has 2 Atoms
1
Computer Systems Lab.
Each element has 1 Atom
Note: base atoms act like 2
extra literals a ≈ y , a ′ = b ≈ y ′
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Boolean Algebra
F2 ({0,1})
Level (= # of atoms)
4
poset: { f ,, f }
fi≤ f j ⇔
minterms(f i ) ⊆ minterms(f j )
0
15
3
3 minterms (3 atoms)
2
2 minterms (2 atoms)
1
1 minterm (1 atom)
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89
Boolean Algebra
Boolean Algebra F1 ({0, a, b,1}) of
1-Variable Functions
atoms of F1 {0, a , b,1}) are
base atoms ⋅ minterms
3 Atoms
2 Atoms
1 Atom
Subalgebras are incompletely specified
Boolean Functions (most important)
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Boolean Subalgebra
Boolean Subalgebra: Interval of 2-Variable Function
Lattice
The 3-Atom element
x + y = xy ′ + xy + x ′y
is ONE of subalgebra
2-Atom elements are
atoms of subalgebra
1-Atom element xy′ is
ZERO of subalgebra
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More On Counting
Level (= # of atoms)
4
3
An algebra (or subalgebra)
with n + 1 levels has exactly
2 n elements, because n + 1
levels implies n atoms
2
1
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A larger Subalgebra
Interval of 2-Variable Function Lattice
ONE of algebra (4-Atoms) is
also one of this subalgebra
2-Atom elements are
atoms of subalgebra
1-Atom element is
ZERO of subalgebra
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Counting Elements of
Fn(B)
For each of the 2 n minterms, the discriminant can be
chosen as any of the | B| elements of B.
Therefore, the number of elements of Fn( B) is
Examples
( 2n )
| B|
| A ( B )| ( 2 n )
= (2
)
=2
B = {0,1}, | A( B )| = 1
( 2n )
n = 2 ⇒| B|
( 2n )
n = 3 ⇒| B|
Computer Systems Lab.
(| A ( B )|⋅2 n )
=2
22
23
= 2 4 = 16
= 2 = 2 = 256
8
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Counting Elements of Fn(B)
Examples with
B = {0, a , b,1}, A( B ) = {a , b}
n=2⇒2
n = 3⇒ 2
Examples with
(| A ( B )|⋅2 n )
(| A ( B )|⋅2 n )
2⋅2 3
= 216
2⋅2 4
= 2 32
=2
=2
| A( B )| = 4
Note 2(4) base atoms act like 1(2) extra variables
n=2⇒2
n = 3⇒ 2
Computer Systems Lab.
(| A ( B )|⋅2 n )
(| A ( B )|⋅2 n )
4⋅2 3
= 2 32
4⋅2 4
=2
=2
=2
64
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95
Boolean Difference (Sensitivity)
A Boolean function f depends on a if
and only if fa ≠ fa′ . Thus
∂f / ∂a = fa ⊕ fa′
is called the Boolean Difference, or
Sensitivity of f with respect to a
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Boolean Difference (Sensitivity)
Example:
f = abc + a′bc
∂f / ∂a = fa ⊕ fa′ = (bc) ⊕ (bc) = 0
∂f / ∂b = fb ⊕ fb′ = ( ac) ⊕ ( a ′c) = c
Note the formula depends on a , but the
implied function does not
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Dont Care
An interval [L, U] in a Boolean algebra B is the subset of B
defined by [L,U] = { x ∈ B : L ≤ x ≤ U }
Satisfiability dont cares
Observability dont cares
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Intervals and Dont Cares
D = D O b s + D S a t = x ′y ′
L = g − D = gD ′
= ( x ′ y + x y ′ )( x + y ) = ( x ′ y + x y ′ ) = g
U = g + D = ( x ′y + x y ′ ) + x ' y '
= x '+ y '
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Dont Cares
xy ∈ DObs if and only if
x
y
f = x, f = y ⇒ ∂z / ∂w = 0
for all possible xy.
Input
Filter
Output
Filter
xy ∈ DSat if and only if local
input xy never occurs
The complete dont care set for gate g is
D =D
g
Computer Systems Lab.
Sat
+D
Obs
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100
Dont Cares
Local inputs
x = u ′ + v ′, y = v
( y = 0 ) ⇒ ( x = 1)
For this circuit, local input combinations
x ′y′ ( x = 0, y = 0)
do not occur. That is, the local minterm
x ′y ′ is dont care.
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Dont Cares
z = w′ + t ′
∂z / ∂w = zw ⊕ zw′
= t′ ⊕ 1
=t
For this circuit, global input combination 10 sets
t ′ = uv ′ = 1
which makes z insensitive to w . However, local
input pair 10 (xy ′) is NOT dont care, since u ′v ′
also gives xy ′, and in this case tu′v ′ = 1.
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Computing ALL Dont Cares
u v
x
y w
t
z
∂z / ∂w
0 0 1 0
1
1 0
1
0 1 1 1
0
1 1
1
1 0 1 0
1
0 1
0
1 1 0 1
1
1 0
1
y
f
u
v
x
(
,
)
=
f
(u, v ) = y
if
and
only
and
xy ∈ D
does not occur for any row u, v in the truth table.
Here, DSat = x ′y ′ (00 does not occur)
Sat
Computer Systems Lab.
x
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103
Computing ALL Dont Cares
Note these differ!
u v
x
y w
t
z
∂z / ∂w
0 0 1 0
1
1 0
1
0 1 1 1
0
1 1
1
1 0 1 0
1
0 1
0
1 1 0 1
1
1 0
1
Similarly xy ∈ D if and only for every row u, v such
that f x (u, v) = x and f y (u, v ) = y ,
∂z / ∂w = zw ⊕ zw′ = 0
Here 10 (xy ′) is NOT dont care since ∂z / ∂w = 1 in
the first row.
Obs
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Computing ALL Dont Cares
Note these differ!
u v
x
y w
t
z
∂z / ∂w
0 0 1 0
1
1 0
1
0 1 1 1
0
1 1
1
1 0 1 0
1
0 1
0
1 1 0 1
1
1 0
1
Dg = DSat + DObs = x ′y ′
g = xy ′ + x ′y ( + x ′y ′ ) → x ′ + y ′
Thus the exclusive OR gate can be replaced by a NAND
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105
Dont Cares and Function Subalgebras
Suppose we are given a Boolean Function f and
a dont care set D. Then any function in the
interval (subalgebra)
[ fL , fU ] = [ fD′, f + D]
is an acceptable replacement for f in the environment
that produced D . Here fD′ is the 0 of the
subalgebra and f + D is the 1.
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Incompletely Specified Functions
Suppose we are given a Boolean Function g and
a dont care set D. Then the triple
( f , d, r )
where f = gD,′ d = D , and r = ( f + D)′ is called
an Incompletely specified function.
Note f + d + r = gD′ + D + ( g + D)′ = 1 .
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