Some Properties of Finite Boolean Algebra by Md

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Global Journal of Science Frontier Research Mathematics and Decision Sciences Volume 13 Issue 9 Version 1.0 Year 2013 Type : Double Blind Peer Reviewed International Research Journal Publisher: Global Journals Inc. (USA) Online ISSN: 2249-4626 & Print ISSN: 0975-5896

Some Properties of Finite Boolean Algebra By Md. Ekramul Islam, Md. Rezwan Ahamed Fahim, Arjuman Ara & Md. Hannan Miah Pabna University of Science and Technology, Bangladesh Abstract - In this paper we examine the relationship between the Ideal and Boolean Algebra of Lattice. Here the main result is that principal ideal (atom), principal dual ideal (filter) and also their product are Boolean algebra. Keywords : principal ideal, principal dual ideal, boolean algebra. GJSFR-F Classification : MSC 2010: 03G05, 94C10

Some Properties of Finite Boolean Algebra

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© 2013. Md. Ekramul Islam, Md. Rezwan Ahamed Fahim, Arjuman Ara & Md. Hannan Miah. This is a research/review paper, distributed under the terms of the Creative Commons Attribution-Noncommercial 3.0 Unported License http://creativecommons.org/licenses/by-nc/3.0/), permitting all non commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.

Notes Some Properties of Finite Boolean Algebra 13

0 2 α σ ρ Ѡ Md. Ekramul Islam , Md. Rezwan Ahamed Fahim , Arjuman Ara & Md. Hannan Miah r ea Y

Abstract - In this paper we examine the relationship between the Ideal and Boolean Algebra of Lattice. Here the main 61 result is that principal ideal (atom), principal dual ideal (filter) and also their product are Boolean algebra. Keywords : principal ideal, principal dual ideal, boolean algebra.

I. Introduction V Boolean algebra, as developed in 1854 by George Boole in his book An IX Investigation of the Laws of Thought, is a variant of ordinary elementary algebra ue ersion I

differing in its values, operations, and laws. Instead of the usual algebra of numbers, s s

Boolean algebra is the algebra of truth values 0 and 1, or equivalently of subsets of a I

given set. The operations are usually taken to be conjunction ∧, disjunction ∨, and XIII negation ¬, with constants 0 and 1. And the laws are definable as those equations that hold for all values of their variables, for example x∨(y∧x) = x. Applications include

mathematical logic, digital logic, computer programming, set theory, and statistics. ) F ) Boole's algebra predated the modern developments in abstract algebra and mathematical logic; it is however seen as connected to the origins of both fields. In an abstract setting, Boolean algebra was perfected in the late 19th century by Jevons,

Schröder, Huntington, and others until it reached the modern conception of an (abstract) Research Volume mathematical structure. For example, the empirical observation that one can manipulate expressions in the algebra of sets by translating them into expressions in Boole's algebra is explained in modern terms by saying that the algebra of sets is a Boolean algebra (note Frontier the indefinite article). In fact, M. H. Stone proved in 1936 that every Boolean algebra is isomorphic to a field of sets. In the 1930s, while studying switching circuits, Claude Shannon observed that one Science could also apply the rules of Boole's algebra in this setting, and he introduced switching of algebra as a way to analyze and design circuits by algebraic means in terms of logic gates.

Shannon already had at his disposal the abstract mathematical apparatus, thus he cast Journal his switching algebra as the two-element Boolean algebra. In circuit engineering settings today, there is little need to consider other Boolean algebras, thus "switching algebra" and "Boolean algebra" are often used interchangeably. Efficient implementation of Global Boolean functions is a fundamental problem in the design of combinatorial logic circuits.

Authors α σ : Department of Mathematics, Pabna University of Science and Technology, Bangladesh.

Author ρ : D epartment of Textile Engineering, City University, Bonani, Dhaka-1213, Bangladesh. Author Ѡ : Department of Mathematics, Begum Rokeya University, Rangpur, Bangladesh.

©201 Global Journals Inc. (US) II. Material and Method 2.1 Definition: A sub-lattice H of a lattice L is called a convex sub-lattice if for any ,a b ∈ H , a c << b implies c ∈ H . Example: in the fig below (fig-1) c,a,0{ }, c,b,0{ } are convex lattices.

Notes 3

201 Year

62 Fig-1 : Convex Lattice 2.2 Definition: Let L be a lattice and I a non-empty subset of L. then I is called an ideal of L if V i. I is a sub -lattice of L ii. for any ∈∧∈∈ Iax,Lx,Ia . IX Example: in fig below (fig-2) a,0{ }, ,0{ b}, }c,b,a,0{ are ideals: XIII Issue ersion I )

) Fig-2 : Ide al 2.3 Definition: An ideal of a lattice L is generated by a single element is called a principal

Research Volume ideal. If ∈ La then the ideal generated by a principal ideal denoted by (a]. Infact x:Lx{]a( ≤∈= )a . The set of all ideals of a lattice L is denoted by (I )L .

Frontier 2.4 Definition: A non-empty subset D of a lattice L is called a dual ideal or filters if i. For all ba,Db,a ∈∧∈ D ii. For t,ta,Da ∈≤∈ L implies t ∈ D .

Science Example: In the lattice below (fig-3) 1{ }, c,1{ }, a,c,1{ }, b,c,1{ }, 0,b,a,c,1{ } are dual lattice: of Journal Global

Fig-3 : Dual Lattice

2.5 Definition: Let L be a lattice and a ∈ L . Then a:Lx{)a[ ≤∈= x} form a dual ideal. This ideal is called a principal dual ideal.

© 2 013 Global Journals Inc (US) 2.6 Definition: Let L1 and L 2 be two lattices then the mapping :f L → L 21 is called a meet homomorphism if for all ,a b ∈ L , ∧=∧ (f)a(f)ba(f b) and is called join homomorphism if for all ,a b ∈ L , ∨=∨ (f)a(f)ba(f b) and f called a homomorphism if f is both meet and join homomorphism.

2.7 Definition: A mapping :f L1 → L 2 is called an isomorphism if it is homomorphism,

one-one and onto. It is denoted by 1 ≅ LL 2 . otes N 2.8 Definition: Let L be a bounded lattice that is a lattice L with 0 and 1. for an element 13

a ∈ L , an element b ∈ L is called a compliment of a if a b =∧ 0 and a b =∨ 1 and is 2

denoted by b = ¬a . Obviously ¬0 = 1 and ¬1 = 0. r ea Y 2.9 Definition: In a lattice if every element has a complimented, then it is called a complimented lattice. 63 2.10 Definition: A complimented distributive lattice is called a Boolean lattice. 2.11 Algebraic definition of Boolean algebra: An algebra ∨∧=< 1,0,,,,LB >¬ , where L is a V non-empty set together with two binary operations ∧ and ∨ and a unary operation ¬ and null operations 0 and 1, is called a Boolean algebra if it satisfies the following IX

conditions: ue ersion I s s

a) ∧,∨ are independent I

b) ∧,∨ are associative XIII c) ∧,∨ are commutative d) ∧,∨ are satisfy absorption law ) F ) e) For all ,b,a c ∈ L , a ( ∨∧=∨∧ a()ba()cb ∧ c) . f) for all a ∈ L , there exists ¬a ∈ L such that a a =¬∧ 0 and a a =¬∨ 1. Example : Let = ,b,a,0{B }1 . Define ∧ ∨,, ¬ by following: Research Volume

∨ 0 a b 1 ∧ 0 a b ¬

Frontier 0 0 a b 1 0 0 0 0 0 1

a a a 1 1 a 0 a 0 a b Science of

b b 1 b 1 b 0 0 b b a

Journal 1 1 1 1 1 1 0 a b 1 0

Global Then B form a Boolean algebra under those operations.

III. Results and Discussion

3.1 Theorem: A principal ideal of a Boolean algebra is again a Boolean algebra.

Proof: Let ∧¬≤=< ∨ 1,0,,,,,BB > be a Boolean algebra and a ∈ B .

A principal ideal of B generated by a , (a], is: x:Bx{]a( ≤∈= a}

1. ≤(a] =≤ ∂(a]

2. (a] ∧¬<=¬ > ∈ (x:ax,x{ a]}

3. ∧ ( ]a = ∧∂(a]

4. ∨ (a] = ∨∂(a] Notes 3

201 5. (a] = 00 Year 6. ( ]a = a1 64 Now if ∈ (y,x a], then ,x y ≤ a hence yx( ) ≤∧ a and yx( ) ≤∨ a , hence ∈∧ ()yx( a]

and ∈∨ ()yx( a]. Hence ∧≤< ,,],a( ∨ ,0(]a(]a(]a( ]a > is a substructure of ∧≤< ∨ 0,,,,B > and

V ],a( ,∧≤< , (]a(]a( a] >∨ is a sub-lattice of ∧≤< ,,,B ∨ > . IX Again 0(a] is the minimum of (a]. 1(a] = a , which is, obviously, the maximum of (a].

Hence ≤< ∧ ∨ 0,,,],a( ,1(]a(]a(]a(]a( ]a > is bounded by 0( ]a and 1( ]a . XIII Issue ersion I Now as ∧≤< ,,,B ∨ > is distributive, it follows that ],a( ,∧≤< ,∨ (]a(]a( a] > is distributive. Hence (a] is a bounded distributive lattice. So we only need to prove that (a] is complemented, i.e. that ¬ is complementation on (a].

) (a]

F )

First: (a] is closed under ¬( ]a . Because x a ≤∧¬ a , for any x ∈ B , hence also for any ≤ ax . Research Volume Secondly: ¬(a] holds the laws of 0(a] and 1(a] : Let ∈ (x a] then

(]a( a] ∧=¬∧ ¬x(x)x(x ∧ a) x()xx( ∧∧¬∧= a) x(0 ∧∧= a) = 0 = 0(a] Frontier

And (]a( ]a ∨=¬∨ ¬x(x)x(x ∧ a) x()xx( ∨∧¬∨= a) x(1 ∨∧= a) = x ∨ a = a = 1( ]a Science Thus, indeed ¬( ]a is complementation on (a]. (Proved) of 3.2 Theorem: Principal dual ideal of a Boolean algebra is again a Boolean algebra.

Journal Proof: Let ∧¬≤=< ∨ 1,0,,,,,BB > be a Boolean algebra and a ∈ B . A principal dual ideal of B generated by a ,[a) , is: a:Bx{)a[ ≤∈= x}. Now an algebra Global

[)a )a[ , [ )a ¬≤=< )a[ ∧ )a[ ∨ )a[ )a[ 1,0,,,, )a[ > ) with the relation of B as

1. ≤[a) =≤ ∂[a)

2. [a) ∨¬<=¬ > ∈[x:ax,x{ a)}

3. ∧[a) = ∧∂[a)

5. 0[a) = a

6. 1[a) = 1

Let ∈[y,x a) , then ≤ xa , y , hence x(a ∧≤ y) and x(a ∨≤ y) , hence ∈∧ [)yx( a) and ∈∨ [)yx( c) . Hence ∧≤< ,,),a[ ∨ ,1 > is a substructure of ∧≤< ∨ 1,,,,B > and Notes [)a[)a[)a[ a) 13 also ),a[ ,∧≤< ,∨ > is a sub-lattice of ∧≤< ,,,B ∨ > . Now 1 is the maximum of [a)

[)a[)a[ a) [a) 0 2

and [a) = a0 , which is, obviously, the minimum of [a). Hence ≤< ∨∧ 0,,,),a[ 1, [)a[)a[)a[)a[ a) > r ea Y is bounded by 0[a) and 1[a) . 65

As ∧≤< ,,,B ∨ > is distributive, it follows that ),a[ ,∧≤< ,∨[)a[)a[ )a > is distributive. Hence [a) is a bounded distributive lattice. So we only need to prove that [a) is

complemented, i.e. that ¬[a) is complementation on [a). V

First: [a) is closed under ¬[a) . This is obvious, since obviously a x ∨¬≤ a for any ue ersion I s ∈ Bx , hence also for any x such that a ≤ x . s I

Secondly, ¬[a) respects the laws of 0[a) and 1[a) : XIII

Let ∈[x a) then, ) [)a[ a) ∧=¬∧ ¬x(x)x(x ∨ )a x()xx( ∧∨¬∧= )a x(0 ∧∨= )a = x ∧ a = = 0a [a) F )

[)a[ a) ∨=¬∨ ¬x(x)x(x ∨ a) = ∨ x()xx( ∨∨¬ )a x(1 ∨∨= )a = 1 = 1[a)

Thus, indeed ¬[a) is complementation on [a). (proved). Research Volume

3.3 Lemma: If a ≠ 1 then ¬∩ )a[]a( = φ Proof: Let []a(x ¬∩∈ a) . Then Frontier x ≤ a and ¬a ≤ x . Then x ≤ a and ¬x ≤ a . Science

Then x x ≤¬∨ a a1 a =⇒≤⇒ 1 of Hence if ≠ 1a then ¬∩ )a[]a( = φ. Journal 3.4 Theorem: (a] and [¬a) are isomorphic.

Proof: If a =1, then a[]a( ) =¬= B. So clearly they are isomorphic. So let a ≠ 1. We Global define: []a(:f ¬→ a) by: for every ∈ )x(f:]a(x x ¬∨= a . 1. f is a function: Since for every x ∈ B, a ∨≤¬ ax , also for every a:]a(x x ∨≤¬∈ a . Hence for every ∈ [)x(f:]a(x ¬∈ a) , hence f is indeed a function from (a] into [¬a) .

Let [y ¬∈ a) . Then ya y ≤¬⇒≤¬ a ...... (1). Hence ∈¬ (y a ] .

Now [a) ¬=¬ ¬y(y ) ∧ a = y ∧ a . Hence ∈∧ ()ay( a].

Again ay()ay(f ) ¬∨∧=∧ a ∧¬∨= a()ay( ∨ ¬a) ay( ) ∧¬∨= 1 y ∨= ¬a = y [from (1)] [a) Notes 3

Hence f is onto. 201 3. f is one-one: Year Let = (f)x(f x ) . 66 1 2

Then 1 ax =¬∨ x2 ∨ ¬a 1 ¬∨¬⇒ ¬= x()ax( 2∨¬a)

1 ax =∧¬⇒ ¬x2 ∧ a [by De-Morgan’s law] V

x =¬⇒ ¬ a(1]a( ]x2 IX

Hence these are the relative complements of x1 and x2 in Boolean algebra (a].

Hence x1 = x2. XIII Issue ersion I Hence h is one-one. 4. f is homomorphism: )

for = , =¬∨= ¬ = F 0( ]a 0 a0)0(f a 0(¬ a] )

for 1(a] = a , ¬∨= aa)a(f = 1 = 1[¬a)

Research Volume yx()yx(f ¬∨∧=∧ a) y()ax( ¬∨∧¬∨= )a ∧= (f)x(f )y =∨ ∨ yx()yx(f ) ∨ ¬a = ∨ ¬ y()ax( ¬∨∨ a) ∨= (f)x(f y)

Frontier And (a] =¬ ¬x(f)x(f ∧ a) ax( ) ¬∨∧¬= a ax( ¬∨¬∨¬= a) ¬= (¬a] x( ∨ ¬a) = ¬(¬a] (f( x))

Hence f is homomorphism. Science

of Thus, indeed, h is an isomorphism. (Proved). 3.5 Lemma: If a is an atom in B, then for every :}0{Bx a ≤−∈ x or a ≤ ¬x Journal Proof: Let a be an atom in B. Suppose that a( ≤¬ x) . Then xa( ) ≠∧ a . But

Global a x ≤∧ a . Since a is an atom, that means that a x =∧ 0 . But that means that a ≤ ¬x . 3.6 Corollary: If a is an atom in B, then a[]a( =∪¬ B) . Proof: This follows from lemma 5: Let x ∈ B and x ∉ [a). Then a( ) ≤¬ x . Hence by lemma 5 a ≤ ¬x , and that means that x ≤ ¬a , hence (x ¬∈ ]a .

3.7 Theorem: Let A and B be two Boolean algebra, then the product of A and B, A × B , is a Boolean a lgebra.

Proof: Since A and B are Boolean algebras. The product of A and B, × BA , is given by:

≤=<× ¬ ∧ ∨××××× 0,,,,,BBA ,1×× > where

1. B× A ×= B

2. × 11 <><<=≤ 22 >> ≤ aa:b,a,b,a{ 2A1 and b1 ≤ B b 2 }

Notes 3. For every

¬=<><¬×>∈< a)b,a(:BAb,a ,¬ b > 0

× A B 2

4. For every r ea Y ><< < >>∈ × : <∧><< ∧>>=< ∧ > 1 1 2 b,a,b,a 2 A B 11 × 22 ,aab,ab,a b12A1 b2B 67

5. For every 1 1 ><< < 2 b,a,b,a 2 >>∈ A× B : 11 × <∨><< 22 ∨>>=< ,aab,ab,a b12A1 ∨ B b 2 >

6. × =< 00 A ,0B > V

7. × =< 11 A ,1B > IX

(a) ≤ is a partial order. ue ersion I × s s

I Reflexive: Since for every ∈ :Aa a ≤A a and for every ∈ :Bb b ≤B b , for every XIII

>≤<×>∈< × < ab,a:BAb,a ,b >

Anti-symmetric: Let < > ≤ < > and . < > ≤ < > a1 ,b1 × a ,b 22 a ,b 22 × a1 ,b1 ) F

)

Then a1 ≤ A a 2 and b1 ≤ B b 2 and a 2 ≤ A a1 and b 2 ≤ B b1 , Hence a1 = a 2 and b1 = b 2 , hence

a ,b11 >< = a b, 22 ><

Transitive: Let < a ,b > ≤ < a ,b > and. < a ,b > ≤ < a ,b > . Then a ≤ a and b Research Volume 1 1 × 2 2 2 2 × 3 3 1 A 2 1 ≤ B b 2 and a 2 ≤ A a 3 and b 2 ≤ B b3 , hence a1 ≤ A a 3 and b1 ≤ B b3 , hence a1 b, 1 >< ≤×

Frontier < a 3 ,b3 >

(b) < a1 ,b1 > ∧× < a 2 ,b2 > = < a1 ∧ A a 2 , b1 ∧ B b 2 > : Science

of a1 ∧ A a 2 ≤ A a1 , a1 ∧ A a 2 ≤ A a 2 , b1 ∧ B b2 ≤ B b1 , b1 ∧ B b2 ≤ B b2

Hence Journal < a1 ,b1 > ∧× < a 2 ,b 2 > ≤× < a1 ,b1 >

Global < a1 ,b1 > ∧× < a 2 ,b 2 > ≤× < a 2 ,b 2 >

Let < ,a b > ≤× < a1 ,b1 > and, < ,a b > ≤× < a 2 ,b 2 >

Then a ≤ A a1 and b ≤ B b1 and a ≤ A a 2 and b ≤ B b2 ,

hence a≤ A a1 ∧ A a 2 and b ≤ B b1 ∧ B b2 , hence < ,a b > ≤× a1 b, 1 >< ∧× a 2 b, 2 ><

Hence ∧× is meet in ≤× .

Similarly ∨× is join in ≤× .

(c). 0× =< 0A , 0B >:

Since for every a∈A, 0A ≤ A a and for every b∈B, 0B ≤ B b, for every ∈A × B,

< 0A , 0B > ≤× . Hence 0× is the minimum under ≤× .

Similarly 1× is the maximum under ≤ . A Notes 3 So A × B is a bounded lattice. 201 (d). < a1 ,b1 > ∧× ( < a 2 ,b 2 > ∨× a 3 b, 3 >< ) = < a1 ∧ A ( a 2 ∨ A a 3 ), b1 ∧ B ( b 2 ∨ B b3 ) = <(a1

Year ∧ A a 2 ) ∨ A ( a1 ∧ A a 3 ),( b1 ∧ B b2 ) ∨ B ( b1 ∧ B b3 )> = < a1 ∧ A a 2 , b1 ∧ B b2 > ∨× = ( < a1 ,b1 > ∧× < a 2 ,b 2 >) ∨× ( < a1 ,b1 > ∧× < a 3 ,b3 > ) So A × B is distributive.

V (e). ¬× satisfies the laws of 0× and 1× : IX ∧× ¬× () = ∧× < ¬A a, ¬B b> = =< 0A , 0B > = 0× .

∨ × ¬ × () = ∨× < ¬A a, ¬B b> = =<1A ,1B >= 1× .

XIII Issue ersion I So A × B is a Boolean algebra.

3.8 Theorem: Let B1 and B2 be isomorphic Boolean algebras such that B1 ∩ B2 = φ and h ) let h be an isomorphism between B1 and B2 . The product of B1 and B2 under h, B 1+2 ,

F ) is a Boolean algebra.

h Proof: Let us definite the product of B1 and B2 un der h, B 1+2 as,

Research Volume h h +21 +21 ¬≤=< ∧ ∨ ++++ 0,,,,,BB + ,1 +212121212121 > where:

h 1. B 1+2 B1 ∪= B2 Frontier

2. + 2121 <∪≤∪=≤≤ 21 > 1)b(h:b,b{ ≤2 b2}

Science 3. ¬ +21 is defined by: of

¬1 (h( b)) ,if b ∈ B1 Journal ∈ +21 ¬1+2 (:Bb b) =

−1 ¬1 (h( b)) ,if b ∈ B2 Global