Chapter 6. Order Relations
and Structure
Weiqi Luo (骆伟祺)
School of Software
Sun Yat-Sen University
Email：weiqi.luo@yahoo.com Office：A309
Chapter six: Order Relations and Structures






6.1. Partially Ordered Sets
6.2. Extremal Elements of Partially Ordered Sets
6.3. Lattices
6.4. Finite Boolean Algebras
6.5. Functions on Boolean Algebras
6.6. Circuit Design
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6.1 Partially Ordered Sets
 Partial Order
A relation R on a set A is called a partial order if R is
reflexive, antisymmetric and transitive. The set A together
with the partial order R is called a partially ordered set, or
simply a poset, denoted by (A, R)
For instance,
1.Let A be a collection of subsets of a set S. The relation ⊆ of
set inclusion is a partial order on A, so (A, ⊆) is a poset.
2.Let Z+ be the set of positive integers. The usual relation ≤
is a partial order on Z+, as is “≥”
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6.1 Partially Ordered Sets
 Example 6
Let R be a partial order on a set A, and let R-1 be the
inverse relation of R. Then R-1 is also a partial order.
Proof:
(a) Reflexive
Δ⊆R
 Δ=Δ-1 ⊆ R-1
(b) Antisymmetric R ∩ R-1 ⊆ Δ  R-1 ∩ R ⊆ Δ
(c) Transitive
R2 ⊆ R
 (R-1)2 ⊆ R-1
Thus, R-1 is also a partial order.
The poset (A, R-1) is called the dual of the poset (A, R).
whenever (A, ≤) is a poset, we use “≥” for the partial order ≤-1
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6.1 Partially Ordered Sets
 Comparable
If (A, ≤) is a poset, elements a and b of A are comparable if
a ≤ b or b ≤ a
In some poset, e.g. the relation of divisibility (a R b iff a |
b), some pairs of elements are not comparable
2 | 7 and 7 | 2
Note: if every pair of elements in a poset A is comparable,
we say that A is linear ordered set, and the partial order is
called a linear order. We also say that A is a chain.
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6.1 Partially Ordered Sets
 Theorem 1
If (A, ≤ ) and (B, ≤ ) are posets, then (A×B, ≤) is a poset,
with partial order ≤ defined by
(a, b) ≤ (a’, b’) if a ≤ a’ in A and b≤b’ in B
Note: the ≤ is used to denote three different partial orders.
Proof:
(a) Reflexive
support (a, b) in A×B, then
(a, b) ≤ (a, b) since a ≤ a in A and b ≤ b in B ((A, ≤)
and (B, ≤) are posets)
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6.1 Partially Ordered Sets
(b) Antisymmetry
support (a, b) ≤ (a’, b’) and (a’, b’) ≤ (a, b), then
a ≤ a’ and a’ ≤ a in A ; b ≤ b’ and b’ ≤ b in B
since A and B are posets, a=a’, b=b’ (antisymmetry property
in A and B, respectively), which means that (a, b)=(a’, b’)
and thus ≤ satisfies the antisymmetry property in A×B
(c) Transitive
support (a, b) ≤ (a’, b’) and (a’, b’) ≤ (a’’, b’’), then
a ≤ a’ and a’ ≤ a’’ in A ; b ≤ b’ and b’ ≤ b’’ in B ,
since A and B are posets, a ≤ a’’ and b ≤ b’’ (transitive
property in A and B, respectively), which means that
(a, b) ≤ (a’’, b’’)
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6.1 Partially Ordered Sets
 Product partial order
The partial order ≤ defined on the Cartesian product
A× B is called the Product partial order
 The symbol <
If (A, ≤ ) is a poset, we say a<b if a ≤ b but a ≠ b
 Lexicographic (dictionary) order
Another useful partial order on A×B, denoted by ≺, is
defined as (a, b) ≺ (a’, b’) if a<a’ or a=a’ and b ≤ b’
why ≺ is a partial order?
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6.1 Partially Ordered Sets
 Example 8
Let A=R, with the usual order ≤. Then the plane
R2=R×R may be given lexicographic order
P1=(x1,y1)
p3
P2=(x2,y2)
p2
P1 ≺ P2
P1 ≺ P3
p1
P2 ≺ P3
P3=(x2,y3)
(x1 ≤ x2, x1 ≠x2)
(x2=x2, y2 ≤ y3)
x
x=x1
x=x2
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6.1 Partially Ordered Sets
 Lexicographic ordering is easily extended to Cartesian
products A1×A2 …×An as follows:
(a1, a2, …, an) ≺ (a’1, a’2, …, a’n) if and only if
a1 < a’1 or
a1 = a’1and a2 < a’2 or
a1 = a’1and a2 = a’2 and a3 < a’3 or …
a1 = a’1and a2 = a’2 … , an-1 = a’n-1, an ≤ a’n
Why?
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6.1 Partially Ordered Sets
 Theorem 2
The digraph of a partial order has no cycle of length
larger than 1
Proof: support that the digraph of the partial order ≤ on
the set A contains a cycle of length n>=2. Then there
exist distinct elements a1,a2,…,an in A such that
a1 ≤a2, a2 ≤a3,…, an-1 ≤an, an ≤a1
by the transitivity of the partial order, used n-1 times,
a1 ≤an
by antisymmetry, an ≤a1and a1 ≤an then a1=an
(Contradiction)
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6.1 Partially Ordered Sets
 Hasse Diagrams
Just a reduced version of the diagram of the partial order of
the poset.
a) Reflexive
Every vertex has a cycle of length 1 (delete all cycles)
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6.1 Partially Ordered Sets
 Transitive
a ≤ b, and b ≤c, then a ≤c (delete the edge from a to c)
c
c
c
b
b
b
a
a
a
Vertex  dot
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Remove arrow (all edges
pointing upward)
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6.1 Partially Ordered Sets
 Example 11
Let A={1,2,3,4,12}. Consider the partial order of
divisibility on A. Draw the corresponding Hasse
diagram.
12
12
4
4
3
3
2
2
1
1
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6.1 Partially Ordered Sets
 Example 12
Let S={a,b,c} and A=P(S). Draw the Hasse diagram of
the poset A with the partial order ⊆
{a,b,c}
{b,c}
{a,b}
{a,c}
{c}
{b}
{a}
ф
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6.1 Partially Ordered Sets
 The Hasse diagram of a finite linearly order set
Let A={a1,a2,…,an} be a finite set, and ai ≤aj if i ≤j
an
an-1
a2
a1
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6.1 Partially Ordered Sets
 Example 13
Fig. a shows the Hasse diagram of a poset (A, ≤), where
A={a, b, c, d, e, f}
Fig. b shows the Hasse diagram of the dual poset (A, ≥)
a
f
d
e
a
b
b
d
c
c
e
f
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6.1 Partially Ordered Sets
 Topological Sorting
If A is a poset with partial order ≤, we sometimes need
to find a linear order ≺ for the set A that will merely be
an extension of the given partial order in the sense that
if a ≤ b, then a ≺b. The process of constructing a linear
order such as ≺ is called topological sorting.
(refer to p.229 for the details of the algorithm)
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6.1 Partially Ordered Sets
 Example 14
Give a topological sorting for the poset whose
Hasse diagram as follows
f
e
g
b
d
c
a
f
f
f
g
e
g
e
d
c
d
b
e
c
g
b
b
a
c
a
d
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Usually, the
topological
sorting is not
unique.
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6.1 Partially Ordered Sets
 Isomorphism
Let (A, ≤) and (A’, ≤’) be posets and let f: AA’ be a one
to one correspondence between A and A’. The function f is
called an isomorphism from (A, ≤) to (A’, ≤’) if, for any a
and b in A,
a ≤ b if and only if f(a) ≤’ f(b)
If f: AA’ is an isomorphism, we say that (A, ≤ ) and (A’,
≤’) are isomorphic posets.
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6.1 Partially Ordered Sets
 Example 15
Let A be the set Z+ of positive integers, and let ≤ be the
usual partial order on A. Let A’ be the set of positive even
integers, and let ≤’ be the usual partial order on A’. The
function f: AA’ given by
f(a) = 2 a
is an isomorphism form (A, ≤) to (A’, ≤’)
Proof: First, it is very to show that f is one to one,
everywhere defined and onto (one to one correspondence).
Finally, if a and b are elements of A, then it is clear that
a ≤ b if and only if 2a ≤’ 2b. Thus f is an isomorphism.
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6.1 Partially Ordered Sets
 Theorem 3
Suppose that f : AA’ is an isomorphism from a poset (A,
≤) to a poset (A’, ≤’). Suppose also that B is a subset of A,
and B’=f(B) is the corresponding subset of A’. The
following principle must hold.
If the elements of B have any property relating to one
another or to other elements of A, and if this property can be
defined entirely in terms of the relation ≤, then the elements
of B’ must possess exactly the same property, defined in
terms of ≤’.
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6.1 Partially Ordered Sets
 Example 16
Let (A, ≤ ) be the poset whose Hasse diagram is shown below,
and suppose that f is an isomorphism from (A, ≤) to some
other poset (A’, ≤’). Note d ≤ x for any x in A, then the
corresponding element f(d) in A’ must satisfy the property
f(d) ≤y for all y in A’.
b
a
c
As another example, a ≤ b and b ≤ a.
Such a pair is called incomparable in A, then
f(a) and f(b) are also incomparable in A’
d
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6.1 Partially Ordered Sets
Let (A, ≤ ) and (A’, ≤ ’) be finite posets, let f: AA’ be a
one-to-one correspondence, and let H be any Hasse diagram
of (A, ≤ ). Then
If f is an isomorphism and each label a of H is replaced by
f(a), then H will become a Hasse diagram for (A, ≤ ’)
Conversely
If H becomes a Hasse diagram for (A’, ≤ ’), whenever each
label a is replaced by f(a), then f is an isomorphism.
Two finite isomorphic posets have the same Hasse diagrams
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6.1 Partially Ordered Sets
 Example 17
Let A={1,2,3,6} and let ≤ be the relation | .
Let A’= {ф, {a}, {b}, {a, b}} and let ≤’ be set containment, ⊆.
If f(1)= ф, f(2)={a}, f(3)={b}, f(6)={a, b}, then f is an
isomorphism. They have the same Hasse diagrams.
6
{a, b}
2
3
{a}
{b}
ф
1
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6.1 Partially Ordered Sets
 Homework
Ex. 2, Ex. 14, Ex. 16, Ex. 18, Ex. 19 , Ex. 30, Ex. 32
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6.2 Extremal Elements of Partially Ordered Sets
Consider a poset (A, ≤ )
 Maximal Element
An element a in A is called a maximal element of A if
there is no element c in A such that a<c.
 Minimal Element
An element b in A is called a minimal element of A if
there is no element c in A such that c<b.
an element a in A is a maximal (minimal) element of (A, ≥ )
if and only if a is a minimal (maximal) element of (A, ≤ )
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6.2 Extremal Elements of Partially Ordered Sets
 Example 1
Find the maximal and minimal elements in the
following Hasse diagram
a3
a1
Maximal elements
a2
b1
Note: a1, a2, a3 are incomparable
b1, b2, b3 are incomparable
b2
Minimal element
b3
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6.2 Extremal Elements of Partially Ordered Sets
 Example 2
Let A be the poset of nonnegative real number with the
usual partial order ≤ . Then 0 is a minimal element of A.
There are no maximal elements of A
 Example 3
The poset Z with the usual partial order ≤ has no
maximal elements and has no minimal elements
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6.2 Extremal Elements of Partially Ordered Sets
 Theorem 1
Let A be a finite nonempty poset with partial order ≤.
Then A has at least one maximal element and at least
one minimal element.
Proof: Let a be any element of A. If a is not maximal, we can
find an element a1 in A such that a<a1. If a1 is not maximal,
we can find an element a2 in A such that a1<a2. This
argument can not be continued indefinitely, since A is a
finite set. Thus we eventually obtain the finite chain
a <a1<a2<…<ak-1<ak
which cannot be extended. Hence we cannot have ak < b for
any b in A, so Ak is a maximal element of (A, ≤).
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6.2 Extremal Elements of Partially Ordered Sets
 Algorithm
For finding a topological sorting of a finite poset (A ≤).
Step 1: Choose a minimal element a of A
Step 2: Make a the next entry of SORT and replace A
with A-{a}
Step 3: Repeat step 1 and 2 until A={ }.
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6.2 Extremal Elements of Partially Ordered Sets
 Example 4
b a
a
a
b a
c
c
d
b
a
c
e
e
SORT:
b
d
e c
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b
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6.2 Extremal Elements of Partially Ordered Sets
 Greatest element
An element a in A is called a greatest element of A if
x ≤ a for all x in A.
 Least element
An element a in A is called a least element of A if
a ≤ x for all x in A.
Note: an element a of (A, ≤ ) is a greatest (or least) element
if and only if it is a least (or greatest) element of (A, ≥ )
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6.2 Extremal Elements of Partially Ordered Sets
 Example 5
Let A be the poset of nonnegative real number with the
usual partial order ≤ . Then 0 is a least element of A.
There are no greatest elements of A
 Example 7
The poset Z with usual partial order has neither a least
nor a greatest element.
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6.2 Extremal Elements of Partially Ordered Sets
 Theorem 2
A poset has at most one greatest element and at most
one least element.
Proof: Support that a and b are greatest elements of a poset A.
since b is a greatest element, we have a ≤ b;
since a is a greatest element, we have b ≤ a; thus
a=b by the antisymmetry property. so, if a poset has a
greatest element, it only has one such element.
This is true for all posets, the dual poset (A, ≥) has at most one
greatest element, so (A, ≤) also has at most one least element.
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6.2 Extremal Elements of Partially Ordered Sets
 Unit element
The greatest element of a poset, if it exists, is denoted
by I and is often called the unit element.
 Zero element
The least element of a poset, if it exists, is denoted by 0
and is often called the zero element.
Q: does unit/zero element exist for a finite
nonempty poset?
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6.2 Extremal Elements of Partially Ordered Sets
Consider a poset A and a subset B of A
 Upper bound
An element a in A is called an upper bound of B if b ≤ a
for all b in B
 Lower bound
An element a in A is called a lower bound of B if a ≤ b
for all b in B
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6.2 Extremal Elements of Partially Ordered Sets
 Example 8
Find all upper and lower bounds of the following subset
of A: (a) B1={a, b}; B2={c, d, e}
h
f
g
d
e
c
a
B1 has no lower bounds; The upper
bounds of B1 are c, d, e, f, g and h
The lower bounds of B2 are c, a and b
The upper bounds of B2 are f, g and h
b
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6.2 Extremal Elements of Partially Ordered Sets
Let A be a poset and B a subset of A,
 Least upper bound
An element a in A is called a least upper bound of B,
denoted by (LUB(B)), if a is an upper bound of B and
a ≤a’, whenever a’ is an upper bound of B.
 Greatest lower bound
An element a in A is called a greatest lower bound of B,
denoted by (GLB(B)), if a is a lower bound of B and
a’ ≤ a, whenever a’ is a lower bound of B.
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6.2 Extremal Elements of Partially Ordered Sets
Some properties of dual of poset
 The upper bounds in (A, ≤ ) correspond to lower
bounds in (A, ≥) (for the same set of elements)
 The lower bounds in (A, ≤ ) correspond to upper
bounds in (A, ≥) (for the same set of elements)
 Similar statements hold for greatest lower bounds and
least upper bounds.
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6.2 Extremal Elements of Partially Ordered Sets
 Example 9
Find all least upper bounds and all greatest lower
bounds of (a) B1={a, b} (b) B2={c, d, e}
h
f
g
d
e
c
a
(a) Since B1 has no lower bounds, it has no
greatest lower bounds; However,
LUB(B1)=c
(b)Since the lower bounds of B2 are c, a and b, we
find that GLB(B2)=c
The upper bounds of B2 are f, g and h. Since f and g
are not comparable, we conclude that B2 has no least
upper bound.
b
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6.2 Extremal Elements of Partially Ordered Sets
 Theorem 3
Let (A, ≤) be a poset. Then a subset B of A has at most
one LUB and at most one GLB
Please refer to the proof of Theorem 2.
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6.2 Extremal Elements of Partially Ordered Sets
 Example 10
Let A={1,2,3,…,11} be the poset whose Hasse diagram
is shown below. Find the LUB and GLB of B={6,7,10},
if they exist.
11
The upper bounds of B are 10, 11, and
LUB(B) is 10 (the first vertex that can be
Reached from {6,7,10} by upward paths )
10
9
5
8
7
6
3
The lower bounds of B are 1,4, and
GLB(B) is 4 (the first vertex that can be
Reached from {6,7,10} by downward paths )
4
2
1
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6.2 Extremal Elements of Partially Ordered Sets
 Theorem 4
Suppose that (A, ≤) and (A, ≤’) are isomorphic posets under
the isomorphic f: AA’
1. If a is a maximal (minimal) element of (A, ≤), then f(a) is
a maximal (minimal) element of (A’, ≤’)
2. If a is the greatest (least) element of (A, ≤), then f(a) is the
greatest (least) element of (A’,≤’)
3. If a is an upper (lower, least upper, greatest lower) bound
of a subset B of A, then f(a) is an upper (lower, least upper,
greatest lower) bound for subset f(B) of A’
4. If every subset of (A, ≤) has a LUB (GLB), then every
subset of (A’, ≤’) has a LUB (GLB)
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6.2 Extremal Elements of Partially Ordered Sets
 Example 11
Show that the posets (A, ≤) and (A’, ≤’), whose Hasse
diagrams are shown below are not isomorphic.
a
b
a’
c
b’
c’
(A, ≤) has a greatest element a, while (A’, ≤’) does not have a greatest element
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6.2 Extremal Elements of Partially Ordered Sets
 Homework
Ex. 2, Ex. 18, Ex. 22, Ex. 28, Ex. 34, Ex. 37
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6.3 Lattices
 Lattice
A lattice is a poset (L, ≤) in which every subset {a, b}
consisting of two elements has a least upper bound and
a greatest lower bound. we denote
LUB({a, b}) by a∨ b (the join of a and b)
GLB({a, b}) by a ∧b (the meet of a and b)
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6.3 Lattices
 Example 1
Let S be a set and let L=P(S). As we have seen, ⊆,
containment, is a partial order on L. Let A and B belong
to the poset (L, ⊆). Then
a∨ b =A U B & a ∧b = A ∩ B
Why?
Assuming C is a upper bound of {a, b}, then
A ⊆ C and B ⊆ C thus A U B ⊆ C
Assuming C is a lower bound of {a, b}, then
C ⊆ A and C ⊆ B thus C ⊆ A ∩ B
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6.3 Lattices
 Example 2
Consider the poset (Z+, ≤), where for a and b in Z+, a ≤
b if and only if a | b , then
a∨b = LCM(a,b)
a∧b = GCD(a,b)
LCM: least common multiple
GCD: greatest common divisor
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6.3 Lattices
 Example 3
Let n be a positive integer and Dn be the set of all
positive divisors of n. Then Dn is a lattice under the
relation of divisibility. For instance,
D20= {1,2,4,5,10,20}
D30= {1,2,3,5,6,10,15,20}
30
20
4
10
2
5
6
10
15
2
3
5
1
1
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6.3 Lattices
 Example 4
Which of the Hasse diagrams represent lattices?
f
g
e
d
d
e
c
e
c
c
b
d
d
b
c
d
b
b
a
a
b
f
c
a
a
a
a
d
e
b
c
d
b
c
a
a
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6.3 Lattices
 Example 6
Let S be a set and L =P(S). Then (L, ⊆ ) is a lattice, and
its dual lattice is (L, ⊇), where “⊆” is “contained in”,
and “⊇” is “contains”. Then, in the poset (L, ⊇ )
join:
A∨B=A∩B ,
meet:
A∧B=A∪B.
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6.3 Lattices
 Theorem 1
If (L1, ≤ ) and (L2, ≤ ) are lattices, then (L, ≤ ) is a lattices,
where L= L1 ×L2, and the partial order ≤ of L is the product
partial order.
Proof: we denote
the join and meet in is L1by ∨1 and ∧1
the join and meet in is L2by ∨2 and ∧2
We know that L is a poset (Theorem 1 in p.219)
for (a1,b1) and (a2,b2) in L. then
(a1,b1) ∨ (a2,b2) = (a1 ∨1 a2, b1 ∨2 b2 ) in L
(a1,b1) ∧ (a2,b2) = (a1 ∧1 a2, b1 ∧2 b2 ) in L
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6.3 Lattices
 Example 7
(I1 , I2)
(I1 ,a)
I1
I2
×
01
L1
(01 ,I2)
=
b
a
02
(01 ,a)
(I1 ,02 )
(01 ,02 )
L1×L2
L2
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(I1 ,b)
(01 ,b )
6.3 Lattices
 Sublattice
Let (L, ≤) be a lattice. A nonempty subset S of L is
called a sublattice of L if a ∨ b in S and a ∧ b in S
whenever a and b in S
For instance
Example 3 is one of sublattices of Example 2
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6.3 Lattices
 Example 9
I
I
e
c
f
a
I
e
b
f
a
0
b
e
c
f
a sublattice
Not a lattice
a
b
a
c
b
0
0
a lattice, not a Sublattice
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6.3 Lattices
 Isomorphic Lattices
If f: L1L2 is an isomorphism form the poset (L1, ≤1 )
to the poset (L2, ≤ 2) , then L1 is a lattice if and only if
L2 is a lattice. In fact, if a and b are elements of L1, then
f(a ∨ b)= f(a) ∨ f(b) & f (a ∧ b)=f(a) ∧ f(b).
If two lattices are isomorphic, as posets, we say they are
isomorphic lattices.
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6.3 Lattices
 Example 10 (P.225 Ex.17)
Let A={1,2,3,6} and let ≤ be the relation | .
Let A’= {ф, {a}, {b}, {a, b}} and let ≤’ be set containment, ⊆.
If f(1)= ф, f(2)={a}, f(3)={b}, f(6)={a, b}, then f is an
isomorphism. They have the same Hasse diagrams.
6
{a, b}
2
3
{a}
{b}
ф
1
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6.3 Lattices
 a ∨ b (LUB{a, b})
1. a ≤ a∨b and b ≤ a∨b; a∨b is an upper bound of a and b
2. If a ≤ c and b ≤ c, then a∨ b ≤ c ; a∨ b is the least upper
bound of a and b
 a ∧ b (GLB{a, b})
3. a∧b ≤ a and a ∧ b ≤ b; a ∧ b is a lower bound of a and b
4. If c≤ a and c ≤ b, then c≤ a ∧ b ; a ∧ b is the greatest lower
bound of a and b
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6.3 Lattices
 Theorem 2
Let L be a lattice. Then for every a and b in L
(a) a ∨ b =b if and only if a ≤ b
(b) a ∧ b =a if and only if a ≤ b
(c) a ∧ b =a if and only if a∨ b =b
Proof:
(a) if a∨b =b, since a≤ a∨b, thus a ≤ b
if a ≤ b, since b ≤ b , thus b is a upper bound of a and b, by
definition of least upper bound we have a∨b ≤ b. since a∨b
is an upper bound of a and b, b ≤ a∨b, so a∨b =b
(b) Similar to (a); (c) the proof follows from (a) & (b)
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6.3 Lattices
 Example 12
Let L be a linearly ordered set. If a and b in L, then
either a ≤ b or b ≤ a. It follows form Theorem 2 that L is
a lattice, since every pair of elements has a least upper
bound and a greatest lower bound.
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6.3 Lattices
 Theorem 3
Let L be a lattice. Then
1. Idempotent properties:
a∨a =a ;
a∧a =a
2. Commutative properties: a∨b= b∨a; a∧b = b∧a
3. Associative properties: (a) (a∨b)∨c= a∨(b∨c )
(b) (a∧b) ∧c= a∧(b∧c)
4. Absorption properties:
(a) a ∨ (a ∧b) =a
(b) a ∧ (a ∨ b) =a
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6.3 Lattices
Proof: 3. (a) (a∨b)∨c= a∨(b∨c)
a≤ a∨(b∨c) & b∨c≤ a∨(b∨c)
b≤ b∨c & c ≤ b∨c
(definition of LUB)
b≤ b∨c & c ≤ b∨c & b∨c≤ a∨(b∨c) 
b≤ a∨(b∨c) & c ≤ a∨(b∨c) (transitivity)
a≤ a∨(b∨c) & b≤ a∨(b∨c)  a∨(b∨c) is a upper of a and b
then we have a∨b ≤ a∨(b∨c) (why?)
a∨b ≤ a∨(b∨c) & c ≤ a∨(b∨c) 
a∨(b∨c) is a upper of a∨b and c
then we have (a∨b)∨c ≤ a∨(b∨c)
Similarly,
a∨(b∨c) ≤ (a∨b)∨c
Therefore (a∨b)∨c = a∨(b∨c) (why?)
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6.3 Lattices
 (a ∨ b) ∨c = a ∨ (b ∨c) = a ∨ b ∨c
 (a∧ b)∧ c = a ∧ (b ∧ c) = a ∧ b ∧ c
 LUB({a1,a2,…,an})= a1 ∨ a2 ∨ … ∨ an
 GLB({a1,a2,…,an}) =a1 ∧ a2 ∧ … ∧ an
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6.3 Lattices
 Theorem 4
Let L be a lattice. Then, for every a, b and c in L
1. If a ≤ b, then
(a) a ∨ c ≤ b ∨c
(b) a ∧c ≤ b ∧ c
2. a ≤ c and b ≤ c if and only if a ∨ b ≤ c
3. c ≤ a and c ≤ b if and only if c ≤ a ∧b
4. If a ≤b and c ≤d, then
(a) a∨c ≤ b∨d
(b) a ∧ c ≤ b∧d
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6.3 Lattices
 Proof
1. (a) If a ≤ b, then a ∨ c ≤ b ∨c
c ≤ b ∨c ;
b≤ b ∨c (definition of LUB)
a ≤ b ; b≤ b ∨c  a≤ b ∨c (transitivity)
therefore,
b ∨c is a upper bound of a and c , which means
a ∨ c ≤ b ∨c
(why? )
The proofs for others left as exercises.
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6.3 Lattices
 Bounded
A lattice L is said to be bounded if it has a greatest
element I and a least element 0
For instance:
Example 15: The lattice P(S) of all subsets of a set S, with
the relation containment is bounded. The greatest element is
S and the least element is empty set.
Example 13: The lattice Z+ under the partial order of
divisibility is not bounded, since it has a least element 1, but
no greatest element.
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6.3 Lattices
 If L is a bounded lattice, then for all a in A
0 ≤ a ≤ I
a ∨ 0 = a,
a∨I=I
a∧0=0,
a∧I=a
Note: I (0) and a are comparable, for all a in A.
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6.3 Lattices
 Theorem 5
Let L={a1,a2,…,an} be a finite lattice. Then L is
bounded.
Proof:
The greatest element of L is a1 ∨ a2 ∨ … ∨ an, and the
least element of L is a1 ∧ a2 ∧ … ∧ an
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6.3 Lattices
 Distributive
A lattice L is called distributive if for any elements a, b
and c in L we have the following distributive properties:
1. a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)
2. a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c)
If L is not distributive, we say that L is nondistributive.
Note: the distributive property holds when
a. any two of the elements a, b and c are equal or
b. when any one of the elements is 0 or I.
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6.3 Lattices
 Example 16
For a set S, the lattice P(S) is distributive, since union
and intersection each satisfy the distributive property.
 Example 17
The lattice whose Hasse diagram shown as follows is
I
distributive.
b
d
a
c
0
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6.3 Lattices
 Example 18
Show that the lattices as follows are nondistributive.
I
I
a
a
c
b
c
b
0
0
a∧( b ∨ c) = a ∧ I = a
(a∧ b)∨(a ∧ c) = b ∨ 0 = b
a∧( b ∨ c) = a ∧ I = a
(a∧ b)∨(a ∧ c) = 0 ∨ 0 = 0
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6.3 Lattices
 Theorem 6
A lattice L is nondistributive if and only if it contains a
sublattice that is isomorphic to one of the lattices whose
Hasse diagrams are as show.
I
I
a
a
c
b
c
b
0
0
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6.3 Lattices
 Complement
Let L be bounded lattice with greatest element I and
least element 0, and let a in L. An element a’ in L is
called a complement of a if
a ∨ a’ = I and a ∧ a’ =0
Note that 0’=I and I’=0
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6.3 Lattices
 Example 19
The lattice L=P(S) is such that every element has a
complement, since if A in L, then its set complement A
has the properties A ∨ A = S and A ∧ A=ф. That is, the
set complement is also the complement in L.
 Example 20
I
I
a
a
c
b
c
b
0
0
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6.3 Lattices
 Example 21
20
30
4
10
6
10
15
2
5
2
3
5
1
1
D20
D30
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6.3 Lattices
 Theorem 7
Let L be a bounded distributive lattice. If a complement
exists, it is unique.
Proof: Let a’ and a’’ be complements of the element a in L, then
a ∨ a’ = I, a ∨ a’’= I ;
a ∧ a’ = 0, a ∧ a’’ =0
using the distributive laws, we obtain
a’= a’ ∨ 0 = a’ ∨(a ∧ a’’ ) = (a’ ∨ a) ∧ (a’ ∨ a’’)
= I ∧ (a’ ∨ a’’) = a’ ∨ a’’
Also
a’’= a’’ ∨ 0 = a’’ ∨(a ∧ a’ ) = (a’’ ∨ a) ∧ (a’’ ∨ a’)
= I ∧ (a’ ∨ a’’) = a’ ∨ a’’
Hence a’=a’’
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6.3 Lattices
 Complemented
A lattice L is called complemented if it is
bounded and if every element in L has a
complement.
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6.3 Lattices
 Example 22
The lattice L=P(S) is complemented. Observe that in
this case each element of L has a unique complement,
which can be seen directly or is implied by Theorem 7.
 Example 23
I
I
a
a
c
b
c
b
0
0
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6.3 Lattices
 Homework
Ex. 6, Ex. 7, Ex. 12, Ex. 19, Ex. 24, Ex. 32, Ex. 40
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6.4 Finite Boolean Algebras
 Theorem 1
If S1={x1,x2,…,xn} and S2={y1,y2,…,yn} are any two finite
sets with n elements, then the lattices (P(S1), ⊆) and (P(S2),
⊆) are isomorphic. Consequently, the Hasse diagrams of
these lattices may be drawn identically.
Arrange the elements in S1 and S2
A
S1: x1 x2 x3 … xn
S1: x1 x2 x3 … xn
f(A)
S2: y1 y2 y3 … yn
S2: y1 y2 y3 … yn
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6.4 Finite Boolean Algebras
 Example 1:
S={a, b, c} and T={2,3,5}. Consider the Hasse diagrams of
the two lattices (P(S), ⊆) and (P(T), ⊆).
{a,b,c}
{a,b}
{2,3,5}
{b,c}
{2,3}
{3,5}
{a,c}
{2,5}
{b}
{a}
{3}
{c}
{2}
ф
{5}
ф
Note : the lattice depends only on the number of elements in set,
not on the elements.
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6.4 Finite Boolean Algebras
 Label the subsets
Let a set S={a1,a2,…,an}, then P(S) has 2n subsets. We
label subsets by sequences of 0’s and 1’s of length n.
For instance,
{a1,a2}  1 1 0 0 …0
{a1,an }  1 0 0 0 …1
ф
 0 0 0 0 …0
{a1,a2,…,an} 1 1 1 1 …1
…
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6.4 Finite Boolean Algebras
 Get the{2,3,5}
unique Hasse Diagram
{2,3}
{2,5}
{3}
{2}
{3,5}
111
110
{5}
011
101
ф
{a,b,c}
010
{a,b}
{a,c}
100
{b,c}
001
000
{b}
{a}
{c}
ф
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6.4 Finite Boolean Algebras
 Lattice Bn
If the Hasse diagram of the lattice corresponding to a set with
n elements is labeled by sequences of 0’s and 1’s of length n,
the resulting lattice is named Bn. The properties of the partial
order on Bn can be described directly as follows. If x=a1a2…an
and y=b1b2…bn are two element of Bn, then
1. x ≤ y iff ak ≤ bk (as numbers 0 or 1) for k=1,2,…,n
2. x ∧ y=c1c2…cn, where ck= min{ak,bk}
3. x ∨ y=c1c2…cn, where ck= max{ak,bk}
4. x has a complement x’=z1z2…zn, where zk=1 if xk=0 and
zk=0 if xk=1
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6.4 Finite Boolean Algebras
 Boolean algebra
A finite lattice is called a Boolean algebra if it is
isomorphic with Bn for some nonnegative integer n.
111
|Bn| =2n
11
1
110
011
101
01
10
010
100
0
n=0
n=1
001
00
000
n=2
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6.4 Finite Boolean Algebras
 (P(S), ⊆)
Each x and y in Bn correspond to subsets A and B of S.
Then x ≤ y, x ∧ y, x ∨ y and x’ correspond to A ⊆ B, A ∩ B,
A U B and A. Therefore,
(P(S), ⊆) is isomorphic with Bn, where n=|S|
 Example 3
Consider the lattice D6 consisting of all positive integer
divisors of 6 under the partial order of divisibility.
6
11
10
2
3
1
87
D6 is a Boolean algebras
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6.4 Finite Boolean Algebras
 Example 4
Consider the lattices D20 and D30 of all positive integer
divisors of 20 and 30, respectively.
20
30
D20 is not a Boolean algebra
(why? 6 is not 2n )
4
10
6
10
15
2
5
2
3
5
D30 is a Boolean algebra,
D30  B3
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1
1
6.4 Finite Boolean Algebras
 Theorem 2
Let n=p1p2…pk, where the pi are distinct primes. The Dn is a
Boolean algebra.
Proof:
Let S={p1 , p2 , … , pk}. If T ⊆ S and aT is the product of the
primes in T, then aT | n. Any divisor of n must be of the form aT
for some subset T of S (let aф=1) .
If V and T are subsets of S, V ⊆T if and only if aV | aT
aV ∩ T = aV ∧ aT = GCD(aV,aT) &
aV U T= aV ∨ aT = LCM(aV,aT)
Thus, the function f: P(S) Dn given by f(T)=aT is a isomorphism
form P(S) to Dn. Since P(S) is a Boolean algebra, so is Dn.
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6.4 Finite Boolean Algebras
 Example
Let S={2,3,5}, show the Hasse diagrams of (P(S), ⊆)
and D30 as follows.
2×3×5=30
{2,3,5}
{2,3}
{2}
{2,5}
{3,5}
{3}
{5}
30
3×5=15
ф
6
10
15
2
3
5
1
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6.4 Finite Boolean Algebras
 Example 5
Since 210=2×3×5×7, 66=2×3×11 and
646=2×17×19, then D210, D66 D646 are all Boolean
algebras.
 Example 9
Since 40=23×5, and 75=3×52, neither D40 and D75 are
Boolean algebras.
Note: If n is positive integer and p2 | n, where p is a
prime number, then Dn is not a Boolean algebra.
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6.4 Finite Boolean Algebras
 Theorem 3 (Substitution rule for Boolean algebra)
Any formula involving U or ∩ that holds for arbitrary subsets
of a set S will continue to hold for arbitrary elements of a
Boolean algebra L if is ∧ substituted for ∩ and ∨ for U.
Example 6 If L is any Boolean algebra and x,y and z are in L,
then the following three properties hold.
1. (x’)’=x
2. (x∧y)’ = x’ ∨ y’ 3. (x∨y)’ = x’ ∧ y’
This is true by theorem 3,
1. (A)=A
2. (A∩B) =A U B
3. (A U B ) = A ∩ B
hold for arbitrary subsets A and B of a set S.
More properties can be found in p. 247, 1 ~12
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6.4 Finite Boolean Algebras
 Example 7
Show the lattice whose Hasse diagram shown below is
not a Boolean algebra.
I
a and e are both complements of c
a
f
e
However, based on the 11.
Every element x has a unique complement x’
Every element A has a unique complement x
d
b
c
Theorem 3 (e.g. properties 1~14) is usually used
to show that a lattice L is not a Boolean algebra.
0
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6.4 Finite Boolean Algebras
Denote the Boolean algebra B1 simply as B. Thus B
contains only the two elements 0 and 1. It is a fact that any
of the Boolean algebras Bn can be described in terms of B.
The following theorem gives this description.
 Theorem 4
For any n>=1, Bn is the product B×B×…×B of B, n
factors, where B×B×…×B is given the product partial
order.
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6.4 Finite Boolean Algebras
 Homework
Ex. 6, Ex. 10, Ex. 12, Ex. 20, Ex. 24, Ex. 26, Ex. 29
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