MA unber, Pt 3 BOOLEAN ALGEBRA ision rin Delinition fnition (POSETS) non-empty set P, together with a binary relation R is said to form: any one of th A non- yifthe tollo followingcondition holds: a partially ordered set or ra 0 and 1 nexivity: a Ra for all a EP OR Anti symmetric Rb and b Ra a=b, where a, bEP nity. Rb, b Rc aRc for all a, b, cEP nsitive: Ifai y addition Aposet is gener is generally denoted by (P, R). wtrk: For con venier would be understoo nvenience, we generally use the symbol s in place of R. Thus whenever we say that P is a understood that s is the relation defined on P, unless another symbol is mentioned. we than or equal to (although it may have nothing to do with usual less than or equal to that we S as less than. nbol is mentioned. We amiliar with). Thus (P, S) is a poset. atural numbers form a poset under the usual AnIple. The set N of natura the set of integers, rational numbers also form poset under usual Similarly, the set o The set N of natural numbers under divisibility forms a poset. innple. Thus here a Sb mean a/b (a divide b) De. (Least Upper Bound and Greatest Lower Bound) S be a non-empty subset of a pOset P. An element a E P is called an upper bound of S if x sa. Let S .eS Further if a is an upper bound of S such that a s for all other upper bounds b of S then a is d least upper bound (l.u.b) or Supremum of S. We write sup S for supermum S. dle Similarly, an element a EP will be called a lower bound of S if a sxV x¬S and a will be called kgeatest lower bound (g.l.b) or Infimum of S (Inf S) if bsa for all other lower bounds b of S. imark. () Greatest and least element belong to the set whereas /ub and g ! b may or may not belong to eg. Let (Z, s) be the poset of integers. Let S= {.., -2.- 1,0, 1,2} then Sup S=2 and S has no Inf as there is no elementa such that a Sx VxES. 445 SPECTRUM DIsCRE 446 .MAT Definition : (LATTICE) that poset P m {2, 3,4,6 ARset (P, ) is sand to e a lattice if ever pair oI elements have greates1 upper bund hekngs to P ates lower bound ana m divisibility Sup p {a, b} Icm {a, b divisibility is noa a la11ce ple 3. PrOve EP NoW 3, 4 EP Sup (3, 4} = Icm VE P AEP {3,4 Nofe: To prove P to be a lattice we check with the help of operation tables. One for ini. sx lattice if at least onea v bor a n b does not belong to P where a, b EP. join and other for 12 £ P So P is not a lattice. ion (Boolean Algebra as Lattice ILLUSTRATIVE EXAMPLE Defnitie A BOolean algebra is a ent (generaly a distributive, complemented lattice having at least two elemenmt namely the least Eample 1. Write down the operation table for V and A for L{1, 2, 3, 5, 30} under divisibility relation. SoL The operation table for V and A for the lattice L is given below : noted by 0) and th the greatest element (generally deviated by 1). of each element in a Boolean algebra is unique. Thus a complementation isa the set under discussion. Therefore we list it together with other two operaions is will make a distinction beween lattices and latices which are Boolean algenra nerally devoted by 8, V, ,0, 1), where B is a non-empty set having aleast two inV and meet A). This wilE under d n a Boo ener a es Roolean algebra is gen Since gDary operation over the element : lementhavino where avb=lu b fa, b} = /. c. m {a, b} and a Ab=glb ia, b} = 8. c. d {a, b} id u 5 30 3 30 owith three erations (join (V), meet (A) and complementation ("), 30 wItn can algebra B consists of a set B together with two binary operations A and v on B, a nB and two specific elements 0 and I of B such that the following laws hold. We write 1 ements 2 30 ary operation 'on I 3= (B, V, A, ',0, 1). A Associative Laws: For all a, b, cEB. and efin. 3 1 15 30 s 5 10 15 5 30 5 5 30 30 30 30 30 30 30 Example 2. The set A = {1,2, 3, 4, 6, 12} of factors 12 under divisibility form a lattice. Sol In divisibility V b) V c=av (b v c) and (an b) A c=a n (ba c) (h Commutative Laws: For alla, b, E B a Vb Sup {a, b} = lcm {a, b} anb=Inf {a, b} = gcd {a, b} Operation tables are (av b) = (b v a) and (a n b)=(b a ) () Distributive Laws : For all a, b, cEB av (b A c)= (a V b) A (a V c) and a A (bV c)= (a A b) v (a n c) T 6 12 2 1 2 1 2 2 2 311 3 3 3 () Identity Laws: For all a E B 4 6 12 (a V 0)= a and (a n 1)=a 6 12 6 2 (e) Complement Laws: For all a EB (a V a')=l and (a A a') =0 Lample 1. Let B {0, 1} be a set. The operation V, A and ' on B are given by 4 4 12 4 12 12 4 I2 1 4 2 4 6 6 6 6 1 2 3 2 6 12 12 12 12 12 12 12 12 12 1 2 3 4 6 12 12 From tables a, bEA 0 0 1 00 0 aVbEA Then (B, V, A, ', 0, 1) satisfies all the properties listed in 2.1.2. This is one of the simplest example Ma two-element Boolean algebra. and a A bEA So set A under divisibility is a lattice. sPECTRUM DISC 448 EAample 2. Let P) denote the Power set ol peraton on Pi\)by X Define two binary operation v and A amd orime such that p"n, where a> I, then P (X) VR A UB, A AB A NB and AX - A for all A, B Epo (have a complement i n D, because if ( d d)=1 f.d=[p', 4(P". d =IP'. d.1 (p. d where U, N and X - A denote the union, intersection and complementos (PNU, n. .o. N) fon a Boolean algebra. 449 o cannot \ for some dE n eory in a set the dED. tten oolean algebra is an Thus Note. If X has n elements, then P(X) has 2" elements and the diagram of the Bool usion an n : ab-a. b) (a. b)) Sn nc Partial order relation on P(X) corresponding to the operation , n is the incue The following are the diagram of the Boolean algebra P{X) when X has one el three elements respectively. nt, two elemens i.e. p.d n Hence P asP' dsn Thus lp, d =Pd (p,d) pd<n and so d cannot be a complement of p. a complement in D, and so D, is not a Boolean algebra, a contradiction. sop does not have a compl are free i.e., n is a product of distinct primes. Hence n is a square 1 above example, we can easily checked that Di. D Da1 D 2re Boolean algebra, Renark. In View of D. Ds, D2o D2s, are not Boolean algebra. denote the set of all statements formula involving a single variable. The algebraic system uhere a Boolean algebra for set having three elements A, , 1) 1s a Boolean ectively and the unary operation denote by negation. The element F and T denote the Boolean algebra for set having two elements pyample 4. Let S dena algebra. Here the binary operation V and A denote the d Boolean algebra for singleton set which are adiction and tautologies respectively. Let X be any topological space and Let CO(X) be the family of sets that are sim junction respectively and disjunction and mula Erample 5. Le oosed and open (i.e. Example 3. Let n be a positive integer and D, denote the set of all positive divisiors of n then (D, ctor of ,1, n) form a Boolean algebra if and only if n is square free, in the sense that it has no fa n Simultaneously e. clopen sets). The family CO(X) is a Boolean algebra w.rt the operation intersection p, where p is a prime. eet, union as join, a' = X\a,0=o and 1 = X. em. In a Boolean algebra (B, V, A, ,0, 1) Sol. We know that D, is a bounded distributive lattice. Therefore, we ned to show that it i s compleme lattice. if aVb| and a Ab=0, then b= d Let n =p1P2 Pm be the product of distinct primes ie the complement of an element in a Boolean algebra is unique. Let dbe any divisor ofn. Proof. Suppose that a V b=1 and a A b=0 Let d= P, Pi P where p, E Pi» P2»..... Pms Now b =b V 0 [ldentity law Take d= P P, be the product of all the prime divisors of n not dividing d, then =b V (a A a') [Complement law) =(b V a) A (6v a') [Distributive law dd' = (a v b) n (6 V a') Commutative law] gc.d. d, d'} =I and Lc.m {d, d'} = dd =n gcd{d,d'} =I A (bV d') [Given] = b Va [ldentity law d is the complement of d ..1) Hence D, is a complemented lattice and so it is a Boolean Algebra. Thus b =b Va' [ldentity law Conversely. Let D, be a Boolean Algebra. Again a=a' v0 Given To show that n is square free. =a' V (a A b) SPECTRUM DIsCR MATIEM lacbra (B, V, A, ,0, , the following properies hold a Boolean algebra ( 451 s. For all a EB rem h 1dempotent 1 Distributive Law] V) (V) IA (4V 450 and Complement Law] aAa=a a Va =a ( Bound Laws. For. aVI=I ldentity lavw] Laws. For all a E B Commulative law] and a A 0 0 V Absorption Laws. For aV (aA b)=a and all a, b, E B a nb=0 then b = a' also a Thus a (a V b)= a from () and (2), we get =a b mvolution Law (a=a Corolary. In a Boolean algebra (B. V., A. ',0, ). ifa Vb=l and a A b =0 then 2 wolution Laws. For all a E B hus in a Boolean algebra (B, V, A, ,0, 1), we have 0-I andl'=0 rool. This follows from above Theorem and commutatively laws in B. a a Va Proof. Since 0 V I = 1l and 0 A l =0 a v (a A a') [ldentity Law] ool (a) By Corollary (1), we have 0' = 1 and 1'' =0 = (a V a) A (a V a') Complement Law] Theorem (DeMorgan Law's] = (a V a) A T Distributive Law] tor any a and b in a Boolean algebra (B, V, A, , 0, 1), we have ) (a V 6) =a' Ab' a Va [Complement Law] () (a A 6)' =a' V bb a Va a [ldentity Law] Proof. We have Thus a a VV (a V b) V(a' A b') = [(a v b) V a'] A [(a v 6) V b]=[(a V a') v b1 a Again [1 V b] A [a V 1] an (a V d') [ldentity Law] = I A = (a A a) v (a A a') Complement Law] =1 (a A a) v1 [Distributive Law] V b) Ad (a v b) A (a'Ab) =[a n (a' A b)) V [B A ("A b))=[(a ^ a') n 6'] v [(hv also = a Aa [Complement Law] [0 A b] v [0 A a'] [ldentity Law] = 0 VO aAa a = 0 (6) Now a VI =(a V 1) A l By Theorem, we have (a V 6)' = a' b [ldentity Law] = (a v 1) A (a V a') Similarly, (a n b) v (a' v b') = [a v (a' v b')] n [b A (a' V b)] [Complement Law] [(a V 1) A al v [(a v 1) A a'] [ (a V a') Ab] A [(6 V b) V a'] =[I Vb] a[I v a'] [Distributive Law] =1 Al [(a A a) v (1 A a)] v l(a n a') n (1A d)] =1 Also (a b) A (a' v b') = [(a A b) A a'] V [(a A b) A b'] = [(a A a') ^ b] v [a n (6 A 6) la V al v [0 V a'] [ldempotent, identity and Complement law] =a V a = [0 A b] v [a n 0] [Complement Law] 0 v0 Thus aVl = = 0 By Theorem , we have Again a A 0 =(a n 0) v 0 (a A b) =a vb' [ldentity Law] (a n 0) v (a A a') [Complement Law] [(a a 0) v al a [(a a 0) V a'] [Distributive Law] MA SPECTRUM DIsCRE 452 6 ata a leV a)A (O V a)) A (a V a') A (0 V a' a [ldempotent, identity and and Complemen aa la A al A [IA a'l 7y 453 an0 0 (8 aVl=1 v (a ^ b) =a a (aV bya aAa (a v b)' =a'A b' 9 a nby - d'v b 10) (a)'=a 1) The above stateme (10y (ay-a () Now dV (d A ) = (a 1) V (a ^ b) an (l Vb) Thus A0 0 taenmity Disribsuive Commulaive tement 1 to 10 and (1) to (10) are known as Boolean ldetities (or BasIC s). rk() 1he nota ons are used. for operations in Boolean algebra is derived from the algebra of logic. However. ed. These are summarized in the following chart: ean algcbra laws). =a A (b VI1) r nolalions are Test Fanl Boundet la Set Computer (Mathematician's Notation) Notation Read as Designer's Thus aV (a A b) =a Notation aA (a V b) = (a V 0) A (a V 6) dentity La Again Join or Sum Distribustive La a V (0 A b) meet or and or Product Commulativea C a V (b A0) Complement =a V0 Bounded la iian most frequency use ne notaion of text and on occasion use the set notation for Mathemetican most frequency a alean algebra. Computer designers use the notation (+) and (() as most of the computers did not have Thus an (a V b) =a in the letter notation o0LEAN IDENTITIES looks like as 1) a+b=b+a mbol V, A. (d)Since r'Vr =r Vr=l and rAx=x A x' =0 Thusi Thus rVr = l and x'Ar=l (1 ab=b *a Therefore by Complement law, we have () a+ (b+c) = (a + b)+c 2 a (b e)=(a b) c (3 a+b c=(a+b) (a+ c) (3) a (b+c)=a* bta c (4) a 1=l a=a 3.4. Principle of Duality for Boolean Algebra (4) a+ 0 =0 +a=a (5) aa =0 Let (B, V, A, ',0, 1) be a Boolean algebra (under s) and S be a true statement for B.lf S* is obtaine atd =1 (5) a from S by replacing s by 2, v by A, A by V, 0 by I and I by 0, then S* is aso a true statement. Weg that the statement S* is dual of the statement S and Vice Versa. (6) a ' a=a (6) at a=a (7) a 0=0 For example: If(B, V, A, ", 0, 1) be a Boolean algebra, then following statement l'to 10' are dual at5 =1 (8) a+a b=a (8) a (a+ b) = a the statement I to 10 and vice versa. (9 (a b) =a'+b6 (1) aVb=b Va (1) aAb=b Aa (9) (a+ b) = a': b' (10' (a')'=a (10) (a') =a ) aV (b V c)=(a V b) V c (2) a A (b A c) = (a ^ b) A c laample 1. Prove the following BOOLEAN IDENTITIES: (b) a (a' +b) =a *b (3) an (b V e)= (a n b) v (a A c) (3) a V (b A c)= (a V b) A (a V c) (4) a V 0=0 Va=a (a) at (a b)=atb (4) a A I=1 Aa=a (d) (a b c)+ (a b)=a b ) (a b)+ (a b) = a (5) a V a' = 1 (5) a M a' = 0 PECTRUM DisCRETE M. ale 3. Simpliy the fol (ab)+ (a + b) (ac)te+ [(6 +6') nlify the following Boolean e pression 454 455 (6) (ab' cy + (a-b c) (a-bC) (d) (1a)+ (0 ad) Sol. a(a 6) - (a* a') (a+ ) + wlarb -c +(ab' c)+ (a-b'.c') +(a + b)' =[a. bla+b) Thus a (a 6) -a +b a (a 6) =a * a' ta b = 0+a D (a (b' c))) + (a*5 c)=[(a' + a) bc) + (a-b'c)) = (1-6c) - (a-bc') a [(a'(6°*c) + (a-(b'.c () ac'b' = (c + ac')b' (c+ ac')b =(c* a) (c *c)5 =(¢+ a)-1.b' = (c +a) b = cb -arb' = a cb' ()+ (a 6) =a (6+6) = a a'c +[(b +b)-c)=« ac+c+[1 c) () +ctCa'c+c= (a+ 1)-c = 1 c=o (1)+( a) =a +a' a'ct Thus (a 6)+ (a 6) =a (d(a 6 c)+(a b) = (a b) c+(a * b) (a b)(c+1) .Prove by using Boolean algebra that pample 4 (9 A+B.C=(A +B). (A+C) (a 6)1 (i) A+ A.C=A+C is the complement ofA,1 for all A, B,C in a Boolean algebra =a b where A is L.H.S.=A +B.C A.1+B.C Example 2. In any Boolean algebra, show that A (1 +C) + B.C =A.l+A.C+B .C=A.(1+B) +A .C+B.C=A+A.B+A.C-B.C a=b ab'+a'b = 0 (i) a=0 e ab' ta'b = b Sol. ) Now ab' + a'b = 0 1-C-1 ab' +bb' + a'a + a'b = 0 = A.A+A.B+A.C+B.C =A. (A + B) +(A +B).C=(A +B). A+(A+B).Cc da=0- (a+b) b'+ a'(a + b) = 0 [: A.A= A] (a+b)b'+(a + b) a' = 0 = (A + B).(A+C)=R.H.S. (a+ b) a' +b')=0 (a+ b) (ab)' = 0 ( L.HS. A+ AC= (A + A). (A+C) =1.(A +C) a+b= ab [Using (] [xx=0 always A A -1] a=b = A+C R.H.S. (i) Now ab' + a'b = b l: a+a=a=alample 5. Reduce the following using rules of Boolean algebra. ab' + bb' + a'a + a'b = b ( A.B+ABC + A (B + AB) (a+b)b' +a'(a + b) =b (a+b)6 + (a+ b) a' = b (a+6) (a' +b') =b-1 in) AB+AC+ABC (AB C) .() Now A (B + AB)=A (B +A).(B+B) atb=b and a' +b' = 1 A (B+ A). I [: B+ B 1] : b-1- AB+ A.A=AB+ A a0 and d'= 1 A (B+1) =A. l : B+ 1 = 1] :0+b= 21+=1 a=0 A Also AB +ABC = A (B + BC) = A (B +B) (B +C A.1( B+C) = A (B +C) = AB +AC SPECTRUM DisCRE wan : a+ by Xy'* y? + (**s +y).x* y): AR ABC AR AC ARAC Xy'? +X yz +*x +r ( (B)). (A T) +yx+ yy) xy? +x +* y +x -xy'*xyz +x'y ztxy 451 ( B)(A T) : la by - .b BT - L.H.S. :ad O Thus AB ABC A (B AB) = A+ A+ BC 1 +BC EXERCISE 3.1 - B +B+ BC = B +B(1+ C)= B +B f all positive divisions of n(nEN). Show that D, DD, D, are Boolean IED, denotes the set of ere as D4, Di2 Dia, are not Boolean algeb Hence A B - ABC A (B+ AB) = T=0. gebra. algebr Jbe a topological space such that A, A e 9for every AE I Show tha (J, Prove that in a Boolean algebra the following conditions are equivalent Gn AB AT ABC(AB C)= AB + AC+A BCAB +A BCC AB+AC+ AABBC+ABC Leg Boolean algebra. ,n,u..0. X) is AB+AC+0.0.C+ABC AV#0(i) x Vy=1 ()x Ay=xamd ()x Vy=yforal z,y EB b)= aV b and a A (a v b)=a A bin a boolean algebra AB+AC+0+ ABC AB+AC + Show that a V (a' AB Reduce thei ()xA V) (irA y) V (r A z) V (ya) following in boolean algebra. Example 6. Using Boolean algebra, show that abc+ abc'+ab' c+a' bc=ab +bc+ca. Gn *Ay Az)v( Ay A) V (E A y') Sol. L.H.S. = abc+ abc'+ab'c+a'bc=ab(c+ c) +ab'c+a' bc &Prove by using oolean algebra (0 x+xz=xtz =ab. 1+ab' c+a' bc=ab+ ab' c+a bc=a(b + b' c)+a' b c (i) y+ry=x a (b + b') (6+ c)+a' bc=a.l.(b + c)+abe (in) (x+y) (+y)=x (v) xyz +ry + xyz' =y Using Boolean identities, show that o)(atb+)' [b' + (ac'y1' =abe ab+ac+a' bc=ab+ (a + a b) c=ab+(a +a) (a + b) c ( xz+x'y=ax +y) (vi) (x+y) (r+y) =0 ab+1.(a + b) c= ab + ac+bc=R.H.S. Remark: We represent ab=ab' +a b in XOR-gate (6) #'[(6'+c) + (b)} +{(a+b) c)= a'b In any Boolean algebra, show that (0(a+b')6+ c)(¢+ a') = (d + b)(6+c)(¢ +o) Example 7. Using Boolean algebra, show that () xy+xz +yz =xy+ (x®y)z Sol. () R.H.S. = xy + (x Dy)z =xy+'y+xy) (i) x' y' z+x' yz +xy' z +xyz =x@y4z (i (a+ b) (a + c)= ac t d'b = ac+ a'b+ bc (i) asb *a+t bc=b(a+c) xy+r'yz +xy' z =xy (1 +2) +x yz + xy' z ANSWERSS Xy +xyz +x' yz +xy' z =xy +x' yz+x z (y +y) xy+r yz +xz ) xAy (i) (xA z) V (*A y)(i) (x Ay) v (* A2) :y+y- =y x+x' ) +xz=y(x +r) (r +2)+xz=y (x +z) +xz &Boolean Expressions or Boolean forms xy+yz +xz =L.H.S. Mintion. Boolean Polynomial (Boolean expression, Boolean form or Boolean formula) (i) R.H.S.=x9y9z 2. be a set of n variables (or symbols). A Boolean polynomial (Boolean expression, d Torm or Boolean formula) f,X, ., X) in the variables X 2 demed usively as follows: =(*y' +r y) ®:= (xy' +r' y) z' +(*y' +x' y)' z =xy' ?' +x'y? + (xy).(x' yy) z [(a +by =d.)( The symbols 0 and 1I are Boolean polynomials. sPECTRU DISCRETE M. MAIEM tion) 458 Boolean lunctio , ant &, , ,)ae uo Boolean )and ( ** ,0, 1) be a booles expression of n variables wo Boolean polynomials, then so are on (B, V, A, ya boolea, algebra. function from B" to B called boolean function it can 459 .are all Bovlean ohomials ) &A*2 mplete product or a Fundamental produet tion (innterm an expression of n variable oroduct oft n variables if be '*| ariables x ¥2 S said to be minterm or complete p f ,)sa Booean polynomial. then so is (4) If f,. mental produc dbtaine s of the forms be minterm or complete product or a A Boo 2 , , Other those 2 here , denotes eitherT () There each minterm is completely detemined by a sequence of 0's and l's of ength n, and any are no Boolean polynomial in the variable ariables using Booleane acvordane with ruk I thrugh 4. a number between 0 and 2" - lin binary representation. Thus Boviean eapression is an expression formed from the given Observe that. V. A and n will be denoted by min , or m if the associated sequence of its exponent gves sequence determine For exampke: For the variable 1. ), : the expressions J. ) (r Vy) A A particular minterm umber j in binary ed by o "1 ., m binary representatio (Here 0 S js2" -1). Thus, we have 2" minterms in n .For example, in three variables a ) =(a Vy') V OA 1) "2" -1 n variabless ns *X1 A2 As. become 5 in the y representation 1 Also, these (. , )-(a Vy) A (a A y) ninterms satisfy the following fundamental properties are Boolean expressions. 2"-1 Note that a Boolean expression in n variables may or may not contain all the n varint Defnition (Equivalent Boolean Expressions) Two Boolean expression (. and J2 1*2 )are said to be e iables. ii) V mV m V..V mn 0m, Am 0 ifi *j I = 0 =1. daition (Maxterm) oolean expression of n variables ,x)are said to be equivalent if asume the same value for every assignment of values to the n variables. 2 X, 15 said to be marterm if it is of the form For erample. The expression f.23) = (, A x2) v ( Ar3) TV .V X,, When X, denotes either ï, ora and f.,X;) =n (r, vr) are equivalent. Cimilarly the maxterm satisty the following fundamentals properties 2 o M, vM, =l ifiaj 2- vM M, a M, A.A Mn 0. i =1 0 0 0 0 0 0 0 0 0 There are 2" maxterm in n variables denoted by Mo. M, M, 0 0 0 0 -1 0 0 Minition. (Disjunctive Normal form or Sum of Product or SOP) or DNP 0 0 A boolean expression over two-valued Boolean algebra ({0, 1}, V, A, ', 0, 1) is said to be in junctive normal form (or sum of Product) if it is join of minterms. 0 0 For example: (xax, ax,) v (ax Arg)v (7 A¥z A*3) 0r ,43) and J; M,~2,43) assume the same value for every assignment of the valus xtxi xh xj +xj z2 *3 =Zm(1,0,7) the variables x1, X2 and x. So f= fa. 000lean expression in disjunctive normal form of three minterms. SPECTRUM DISCRE Mi xyA z, m, x Ayh z' interm are iton (Conjuncthe mormal form or P'roduct of Sum or POS) CNP Def ovean eyesw over m-1alue Boolean algeora (i0, 1}, v,. 461 ,0. ) 1s said AyANz 460 Normal form term Hence Minte v m, V m3 m V A he nomnal forn (or PnniNt of sum) if it s meet of mavtems. -(r Ay' A ) V (x Ay A z') v( (xAy 2) V (x A y A ?) V hy A 2) ify the Boolean expression or exampe. (, V V) ( V ; VX) Or Oean eresson in conjunctive nomal form of three variables. is a b f(x, y, 2) =xAy z) V (x A y A z) btaining Boolean expression in Disjunctive Normal torm and conjunctive nos )A Bowlean expression can be obtained in disjunctive normal form nction &v having a minterm coresponding to each ordered n-table c rmal form ple 2. Simplify the, 4)4, )++1) = nM(4, 6, 0) aupe rms. correspondin fx,y, 2) a Ay A Z) V (x AyA2) = (x AzAy') V (x A : Ay) [(x A z) A y V y)) = ((x A 2) A 1] find its conjective norm: I for which the vale of 0 and =x Az in conjunctive normal form corre tion is 0. the function is 1. )A Boolean expression can be obtained in conjunctnve normal forn FxAz Remark: Boolean function represented as a sum of mintems or product of maxter canonical form. Max Tunction by having a mssterm corresponding to 0 and I at which the value of functinrespon said to be Mo 0 M 0 Example 1. Simplify the Boolean expression M Say, ) =(r' A :) V A :) V 0A ) 0 and write in minterm normal form. 0 Sol fa, ). ) = (t' A ) V (y A ) V (v A z) MA (Using distributive la 0 A) Vv [y A (z V :)) M (rA )V 0'AD :zV M6 (xA )Vy x' Az = (x A ) V y | M min Gince Maxterm corresponds to each ordered triple of 0 and I for which the value of the function is 0. 0 0 0 Maxterm are M,* Vy Vz, M=Vy Vz, M,=x Vy V :', M,=x' Vy V z 0 M=x Vy Vz, M=x Vy Vz' 0 0 Hence disjunctive normal form = MA M M, ^ M,A M, A M -VyV) A F' Vy V 2) A Vy V :) A Vy V :) A (xVy' v:) A (r V y V:) Ugorithm for obtaining complete Sum-of-Product Expression Let the given boolean expression be Í1*2 0 m4 0 0 0 m6 ep 1. Find a product P in fx, .., X)which does not contain the variable x, and then multiply P 0 ,+x),deleting any repeated products (as x+r' = l and P + P = P) Since minterms corresponds to each ordered triple of 0 and 1 for which the value of the runco the n-variables. Repeat Step I until every product in fxx ..,)is a minterm ie. every product P contains sPETRIM DISCRE1 MATHEM Agorithn for ohtaining Prodect of sum canonkcal orm L the given hndean eyrkerone 0, 462 h Show tha )+ }7x,)+ 63 L.HS t contains the variable x and then t - I(ri) which does not contains Step 1. Find a sum &in is a maxterm l.e. every Sum COntains S deleting any repeated sum (as n = 0 and SSS) Step2 Reat Ster 1 till every sum in /(,.a2 R.H.S. lowing Boolean expresSions are equivalent to one anstet Obain their Sum 1. Show that the Canonical form Varnadles tampeUsing Boolean algebra. construct the DNF of the boolean function z)y+2) i) S,{z, y, z) = (xz)+(7y)-y3) rduct Ca Sol. Here (x, V. 2) =x(V+:) =+ =n"1 t71 :+") + xz(v +y') =yz +Yz *yz +'z (x, y. 7) =( *y)(*+2) aluation of the given b0olean expression a (iv) fx,y, 2) =a+zy ( + nz) + y': +yz = z * y'7 + yz' which is in the DNF of the boolean function f (x, y, 2) yariables The binary valuation of th +y +z |ytz y Eampe 4 Express x, +x, and x,x, in its complete Sum-oft-Product term in three variakt Sol. () Now 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 Which is the complete Sum-of-Product form. (i) Also, ince the values of the boolean expressions for SS and fa are equal over every uriple of the two- which is the complete Sum-of-Product form. e element Boolean algebra. So these are equivalent. Example 5. Obtain Product of Sum Canonical form of boolean expression xx, in three vo variable To write them in Sum-of-Product cannonical form. We have S, y, 2) *z +xy=u:l +xry-1 =«y+y)+ry¢+#) Sol. Here 2 ( +0) ( +0) = I +(, *,r2 +G 7] xyz+xy' z +x'yz +r which is in the Sum-of-Product canonical form. 3,) +) +) tmple &. Find the Boolean Expression that defines the function f by S0,0,0) 0 S(0,0, 1) =0 S,0,0) = 1 S,1,0) =0 +t)r2 + t3) +* S0, 1,0) =1 S0, 1,) =0 which is the required product of Sum Canonical form. S1,0,1) =1 Sa,1.) =1 sPECTRUM DIsCRET MAIE ession f(x., y, 2)= 465 Gh2) v (y h 2) V (y ~ ) and write in min. Simplify the Boolean exns (XA 2) (yA2) V (yA - (ïA 2) V (ya 1) nle1 Simplify normar tom ((yAaV )) y'A V AyA :) Distributive law) De iminterms are 0 1 0), r0.0 0). f101). s0, 1.,1) = (FA ) Vy :VE =1 A:).( 1 A(ArA ). (T A A ) Distributive DNFIs /, )( AA :)Va Vr V:)V (TAy A 2)V (r A Az yibutve a Distriby can be siumplifiei as 0 0 0 (AIA:) V1A(V A:) VO A) V(V A :) [:Vz=I, yp 0 ( AA:) VrA [ A:' V :)} V(yA :)} Distributive 0 0 ( AA:) Va A(V V(OA 2)) AA:)VrA(r V1) A (y A ) Distributivelan 0 ( AA:) VaA (r' V:) 0 0 0 0 AA:) V [a Ay) v (r A )) 0 0 Example 9, Find the Boolean Expression in CN-form. 0 0 0 1 mg 1 0 1 Min. term are m2 x AyAZ 0 maXAyA Z, m4 x Vy A z 0 0 m XAyAz 0 0 mgX A y N z Min. term Normal form is ma V m3 V m4 V m, V mg CN form is (r Vy V:') A (x Vy'n :) A (x' Vy V :) A (r' Vy V:)A r Vy' V z) (TA 5A ) V (XAyA Z) V(TAyA z) V (rAyA E)V(x Ay A :) Example 10. Using Boolean laws show Max. term is (r A y) v (r Ay A z) V [r A (y V (r A y)} and x Ay (FV V z) A (F Vy V E)A (FVy V :) A (r Vy V E) A (r Vy V:) are equivalent. 16 Representation of Boolean Functions (rA y) v (r Ay A :) V r A (y v (r Ay)] rAr= Ch the function corresponds. Such a representation of a Boolean function will be found convenient for e purpose. The existence of one-one correspondence between every Boolean function f: B" B and a Boolean (x A y) V (x Ay A z) v [« A y) V (x A y)] avuapression in n variables. We can represent a Boolean function by any one of the Boolean expression to S way of representing a Boolean function is simply to give the Boolean equation for the values of *ynbol (s) in the output combination in terms of some subset of the elements of the input combinations. =(r A y) V (x Ay A z) V (r A y) ((x A y) V (r A y)) V (x AyA z) =(x Ay) V (r AyA z) = (x A y) V (1 A 2) (r A y) Vl=x Ay TE MATNEMAT sPECTRUM DisCRETE M. 467 A For eampke: Ihe tiuntihm /(. )* Irt(E.7)I gives the. ariables ) and output ems of the in BT table. (trnuth table) , Cnaustin eh all pos1he mnbination of the imput variabies and which for each revord a functional value such input configur which D ABKB AO XNOR (O) Another methonl of npresentatm of Bwlean tunction is to form This representation doe a circuit. sian does seem appropriate since Boolean functions can express the functioning o diagram actually shows which circuit are to be connected to which other circurts, n to make use of a circuit diagram to eliminate unnecessary connectives and nus For eample: Consaer the fiuncton )=I): then Because occasionaly da simpler circuit. possible or example: The cinn. A2)=*y2' circuit diagram of the following Boolean functions () hy, 2)=x +yz 0 0 +xy (v) fa(x, y, 2) = xy' +x'z are 1 0 0 0 0 0 Tbe other method for representing Boolean function is circuit diagram. Composed of AND. OR. NOT GATES. NOT ( ,y, 2)=X +y'z D OR AND () (*) Other standard Gates are NAND, NOR, XOR, XNOR GATE PA.B yo B A.B y ZO NAND ND NOT Ci) (x. y, 2) =xy'z +r'yz +xy' A (A +B) (A +B) yz B NOR OR NOT D" r'y2 D BT AB+ AB B XOR D () SPECTRU M DiseR ction and alise he the he'lp of NAN B. C D) = A BCD+TR Simplify the following Boolean 469 logc diagam of the reduced functio 468 , of NAND gate only +ABCD ABCD D+ABCD ABCD ECD+ BCD+ ABCD+ABCD-ABCD ABCD- ABCD ABKD ABCD A CD RO (T+C)+ ABC (D+D)+ ABC (D+D) ABC (D-D) FlA. B, CD)- BD+ABC+ ABC+ ABC A BD ABC +AC(B+ B) ABD+ABC AC Also (t. :)=t:r'y: +xy =r :(V *}) *iy BD+A(BC +C) = ABD A(C-B) (C-T) Eampk 1. Wnte the cinuit (gate) diagram of the following Boolean function. ABD A(C+ B). 1 = ABD AC +AB f., )=Xi. (F, (F 4)) AC+B (AD+A) = AC +B(A + +D) ((T I.})= (r.I +;).(2 + X3) *3 Sol. () The circuit diagram of the function (TiI2,3 , X4 )=Xi (2.(F.*)) - AC+B. 1 (A + D) = AC +BA +BD. IS as sthoWn below DDT Fwith the help of NAND GATE A Co or A O- DD B o D (i) The circuit diagram of the function Fwith the help of NAND GATE h a1,3,X) = (1.x+ X). (r + x3) + F- (AC+ BA +BD) = AC.BA.BD is as shown below: - (A+T).(B+ D) :(a. b)=a' +b', (a') =a DDD - (A+C).(BB + BD + = ( (B+ BD+ + AB+ ABD + + +CB+ BTD+ ABC+ADC D AB+ ABD + + +CB+ BD+ ADC " ata=a AB+ AD (1+T)+ BT(1+A)+ BCD SPECTRM DISCRET [(a Ab''N c)V(a'n bn Ac V[{(an -(a b' A c) (a Aba. :a 470 -i 471 AB AD BT. ABD ORD [(a A bA c)V(a'a bbA C)Jv((a v byn o AB (1D) BT(+ D)+ AD V{a v by b'hd [(a Ab' c')V(a'n bbn. AB BT AD - (a A b' C)V(a'nbac)y c)v(a'A AbAc)V(6 A a A c)v(a' n b a c)v(a'nbAc)V(b Aan c)V(a' nbA c)j a) v (an b)v (6Ab}nc (a Sinilarly Aa)V(6'AcA a)V(¢ A bA a)v (B' A ¢ A a D +) ta(6 n e' ( A C'A a)V(6A e° , 'AcAa') = (a A b'A c')V(a'n b Aan ev(a' AbNc) BOOLEAN RING +and defined as : above a+b) +c=( (b+ c) +a=a+(b+ c) Additive Identity: For all a E B. Theorem: Prove that a Boolean algebra forms a ring under the binary operations+ and Thus Proof: Let B be a Boolean algebra. Then for all a bEB. a+0 (a A 0) V (@ A 0)= (a n 1) v0=av0 a+b (a A b') V (a' A6) a a-b =a Ab 0+a (O A a') V (O A a) =0 v (1 A a)= 0Va ) Addition is closed a For a,bEB, a',b¬B Thereforeanb,a' AbEB Thus a+0 =a=0+a for all a EB (a n b') v (a' a b) EB Hence 0 is additive identity in B. a+b EB 0)Additive Inverse: For any a E B, we have (i) Addition is Commutative: For a, bEB, ata = (a A a') V (a' A a) atb (a A 6) v (a' A b)=(a' A b) V (a n b')=(b A a') V (6A a) = 0 v0 =b+a = 0 (ti) Addition is associative: For a, b, cEB, Thus a itself is additive inverse of a in B. (a+ b)+c= ((a+ b) A c'] v [(a+ b)' A c] (vi) Multipication is closed: For all a, b EB, [((anb')v(a'n b)} A c] v [{laa b) v (a' a b)}' A c] ab =a A bEB SPE TRUM DIsCRE MATL nt of B is order Each ele OR s2 in its a 472 ditive gyoup ( Multiplica tion is associatihe: For a, A, c E B, 473 nt a E B, prove that 2a =0 d ) AA)a A ( A ) = (0 A b)A C= (a:b) n or any element (pi) Multiplication is distributive over addition 2a a+a :For anyYaEB For a cEB, we have (a AbAc) v ( (a A a)V (a' A a) ()=a A (h+ c)=a A [(b A c) V (6' A c)] = (a = 0v0 Aiso d'c = (a A b) + (a A c) 0 la n b) A (a n c)] V [a A b) A (a A c)] ler addition order of a is 2. a A b) A (a' V c)] V [(a' Vb) A (a A c)] Thus under addition. plary. y 3. All elements ofB = (a A bA a') V (a Ab Ac) V(a Aan c)V (6' A a Ad elements ofB are idempotent OR (a A a')A b) V (a ^bAC) V(0 A c) V (b' A an c B. +, , we have aa for all a ¬B. = 0 V (a A bA c) VO V (a A b'A ) n the ring B, +, .>| f: For all a EB, we have a'a =(a AbAc) V (a A6 Ac) Thus a (b + c) =ab+ac Sirmilarly we can show that aA a (a+ b) c =ac+bc Thus all elements of B are idempotent. viaition: A ring is said to be Boolean ring if all its elements are idempotent Thus all the eight properties of a ring are satisfied. Hence B,+ > forms a ring Corollary 1. IfB is a Boolean algebra, then the ring <B, +, >15 commutative ring with unit element am above definition and result, it is clear that "Every Boolean algebra is a Boolean ring with Proof: For all a, b E B, we have Me: From ab =a A b =b A a=ba ama: In a Boolean ring <B,t, >, prove that Therefore B is commutative ( x+x=F0 (i) x+y=0 *=y Now for all a EB, ii) y=y (Boolean ring is commutative) of: We know that cancellation law holds in any ring. al =a ^I=a () For any x E B, x +x EB and l-a =1 Aa=a Since B is a Boolean ring Thus 1 is unit element of B. +x =x+x +x) (r+ x) =x +x SPECTR M DisCRE laws: For a, bE B 474 Comula aAb = a*b = b'a= bh 475 a a Vb =a+h b+ ab= b+ a+ ba- b v tive laws: For a, b, c E B A. Associative aA ( (h Ac)a n (bc) =a (bc) = (ab)e = (a A b) Ac t( 1) = (1)U a V (b V v cat (b V c)* a(6V c)= a+ b+c+bc+ a(b-c+bc) (We have a+b+c+ be + ab + ac + abc = (a +b+ ab) + c (a- b- ah) c =X+r = (a V b)+c+ (aV b) c= (a V c)V c laws: For a, b EB (u) For x. IEB. *JEB ) Absorptiona an (a V b) =a(a b)=a (a+b+ ab)= d + ab+ a (ab) Sunce B is a Boolean ring = at ab + (aa) b= a+ ab +db= a+ (ab+ ab) ()=+ (x-)r+y) =r*y a+0 a (-y)r(x + y)y =X+y a V (a A b) = a V (ab) +7+y+y =x+y =at abt a (a b) = a+ ab + (aa) b = a +ab + d'b= a+ (ab ab) r+jx + n+y =r+y a+0 (7 y) (x+) = (r+y) +0 yy =0 yx Thus B, +, > forms a lattice. uSing ( Theorem: Any Boolean ring with unity defines a Boolean algebra. Now for a, b, c E B, we have Proof: Let <B. +, >be a Boolean ring with unity. an (b V c) =a(b V c) Define two binary operations A and V on B by =a (btc+be)= ab t ac + abe= ab t ac + a'be as a =a aAb =ab ab + act a (ab) c= ab + ac+ a (ba) c aVb =a+b+ ab for all a, bE B ab+ ac+ (ab) (ac) = (ab) v (ac) ) Laws of idempotency : For a E B, (a A b) V (a A c) aAa =aa=d=a Thus a A (b V c)= (a A b) V (a A c) for all a, b, cEB a Va =a tata'a=0+d=a Therefore B is a distributive lattice. he product ol sums nnonical form in thre ohtam the pro ee variables x 76 477 or all ER for expressions (a) for all a EB Find the value of the Boolean xpression given below A attice B (lrnty yv(ra)]vl(rai (x)} for x = 1,y- for all a E B (a8 ring) and z 0 hercton 0Is the zeno elnent o ne vz for x = 0, y = 1,z = 1. (b) Als for all a E B obtain the value of the Boolean forms (b)x neretore l ts the unit element of the lattice t () vr,) (c) vl af Products and Product-of-Sums cannonical foms of the following L we show that the lattice B is complementea. btain the stum of Products For any a EB. we have aA (a+ 1) =a (a + ) =a" ta=ata=0 At+X1 +(2 ¥3)]7 +x,) and a V (a 1) at (a +1)+ a (a + 1) (a a)+ 1 +a"+ a ANSWERS 0-1+ (a ta) 10 Therefore a' =a - 1. Thus B is a distributive and complemented lattice with 0 and 0 0 Hence B is a Boolean algebra. 0 Note: Boolean algebra and Boolean ring with unity are equivalent systems. EXERCISE 3.2 1. Show that the Boolean functions t, y, 2) = (X V X2)V X, and ,4, y, 2)= x v (x, vri. cquivalent. 0 0 0 2. Construct the truth table for the following expressions 0 0 0 0 (a) ) 0 0 3. Find the truth value of f(,,)= ( vr)nx} vx)a(x, Vxi) 0 0 4. Write the following Boolean expressions in an equivalent sum of Product cannonical form in thie variables X, *3 I0 0 0 (6) X,VX3 () vx,)var) 0 0 SPETRUM DISCRETE (a) VA VA,)n(V VX3) 6. (a) 1 (6) 0 x a (*2 Vx2 X V( Ax2) 0 0 0 0 0 0 0 0 0 1 0 1