Boolean Algebra the basic fundamentals of computer science
- 1. XI Computer Science UNIT-1 Chapter-3 Boolean Algebra • Introduction to Boolean Algebra • Logical Operators NOT, AND, OR, and Boolean Logic • Principle of Duality • Basic Postulates of Boolean Algebra • DeMorgan’s Theorem • Universal Gates (NAND, NOR) , XOR • Truth Tables
- 2. operation called NOT denoted by prime(’)/not/bar symbol. Boolean Algebra uses a set of Laws and Rules to define the operation of a digital logic circuit. WHAT IS BOOLEAN ALGEbRA? WHO INTRODUcED IT? In 1854, Mathematician George Bool introduced a systematic treatment of logic and develop for this program and algebraic system called BOOLEAN ALGEBRA. In 1938, Sennon demonstrated that the properties of by- stable switching circuits can be represented by this algebra and called it SWITCHING ALGEBRA. BY WHAT IT IS MADE OY??? A Switching Algebra or Boolean Algebra is an algebraic system consisting of the set {0,1} the binary operations ‘OR’ & ‘AND’ denoted the symbol ‘+’ and ‘.’ respectively and unary
- 3. LOGICAL OPERATORS • Boolean algebra also deals with functions which have their values in the set {0, 1}. The basic operations of Boolean algebra are as follows: • Conjunction ( AND operation) • Disjunction ( OR operation) • Negation ( Not operation) TRUTH TABLE: • A Truth Table is a table which represents all the possible input combinations of logical variables along with all the possible results/output of the given combinations of values. LOGIC GATES: • Logic gates are the basic building blocks of any digital system. It is an electronic circuit having one or more than one input and only one output. The relationship between the input and the output is based on a certain logic. Based on this, logic gates are named as AND gate, OR gate, NOT gate etc.
- 4. AND OPERATOR Operation performed by AND operator is called Logical Multiplication. Thus X.Y means X AND Y. 0.0=0 0.1=0 1.0=0 1.1=1 The Symbol and Truth table of AND gate X Y F= ( X.Y) 0 0 0 0 1 0 1 0 0 1 1 1 X Y F
- 5. X F Y X Y F=(X+Y) 0 0 0 0 1 1 1 0 1 1 1 1 OR OPERATOR Operation performed by OR operator is called Logical Addition. Thus X+Y means X OR Y. 0+0=0 0+1=1 1+0=1 1+1=1 The Symbol and Truth table of OR gate
- 6. X X’ 0 1 1 0 This operator operates on single variable and operation performed by NOT operator is called complementation. Thus X’ means complement of X. 0’=1 1’=0 The Symbol and Truth table of NOT gate X X’ NOT OPERATOR
- 7. W h a t is B a s i c Postulates Of B o o l e a n Al ge b ra? Boolean algebra, consists of fundamental laws that are used to build a workable, cohesive framework upon which are based the theorem of Boolean algebra. These fundamental laws are known as Basic Postulates Of Boolean Algebra. POSTULATES i. If X!=0 then X=1; If X!=1 then X=0 ii. OR relations 0+0=0 0+1=1 1+0=1 1+1=1 iii. AND relations 0.0=0 0.1=0 1.0=0 1.1=1 iv. NOT relations 0’=1 1’=0
- 8. WHAT IS THE PRINcIPLE OY DUALITY? This principle states that starting with a Boolean relation, another Boolean relation can be derived by: 1. Changing each OR sign(+) to an AND sign(.) 2. Changing each AND sign(.) to an OR sign(+) 3. Replacing each 0 by 1 and each 1 by 0. EXAMPLE:- If X.Y=1 then its dual will be X+Y=0 , where X & Y belongs to 0,1. EXAMPLE:- If (X+Z)(Y+1)=1 then its dual will be X.Z+Y.0=0 , where X & Y belongs to 0,1. NOTE:- The derived relation using duality principle is called
- 10. 1.Properties of 0 and 1 a) 0+X=X b) 1+X=1 c) 0.X=0 d) 1.X=X 0 X R=X 0 0 0 0 1 1 1 X R=1 1 0 1 1 1 1 0 X R=0 0 0 0 0 1 0 1 X R=X 1 0 0 1 1 1 0 X X 1 X 1 0 X 0 1 X X
- 11. 2. Indempotence Law a) X+X=X X X R=X 0 0 0 1 1 1 X X X b) X.X=X X X R=X 0 0 0 1 1 1 X X X i.e., 0+0=0 1+1=1 i.e., 0.0=0 1.1=1 The law states that,
- 12. 3.Involution Law (X’)’=X X X’ (X’)’=X X X’ (X’)’ = X 0 1 0 1 0 1 The law states that,
- 13. 4.Complementarity Law a) X+X’=1 X X’ R=X+X’ 0 1 1 1 0 1 X X’ b) X.X’=0 X X’ R=X.X’ 0 1 0 1 0 0 X.X’=0 i.e., 0+1=1 1+0=1 i.e., 0.1=0 1.0=0 X+X’=1 X’ X The law states that,
- 14. 5.Commutative Law X Y R Y X R a) X+Y=Y+X The law states that, X Y R=X+Y R=Y+X 0 0 0 0 0 1 1 1 1 0 1 1 1 1 1 1 + is
- 15. R Y X b) X.Y=Y.X X Y R=X.Y R=Y.X 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 X Y R 5.Commutative Law . is
- 16. 6.Associative Law a) X+(Y+Z)=(X+Y) +Z Y Z X Y+Z R X Y R Z X+Y X Y Z Y+Z X+Y R=X+(Y+Z) R=(X+Y)+Z 0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 1 0 1 1 1 1 0 1 1 1 1 1 1 1 0 0 0 1 1 1 The law states that,
- 17. X Y Z YZ XY R=X(YZ) R=(XY) Z 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 1 1 1 1 1 1 1 Y Z Y R R Z X X XY YZ b) X (Y.Z)=(X.Y) Z 6.Associative Law The law states that,
- 18. Y Z R X Y+Z Y X R Z XY XZ X Y Z Y+Z XY XZ R=X(Y+Z) R=XY+XZ 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0 0 a) X(Y+Z)=XY+XZ 7. Distributive Law (First) The law states that,
- 19. Y Z R X Y.Z Y X R Z X+Y X+Z b)X+YZ=(X+Y).(X+Z) R.H.S = (X+Y).(X+Z) =X.(X+Z)+Y.(X+Z) = X.X+XZ+XY+YZ =X+XZ+XY+YZ =(X.1)+XZ+XY+YZ =X(1+Z+Y) + YZ =X.1+YZ =X+YZ = Hence (Distributive Law1 X(Y+Z)=XY+XZ) (Properties of 0 and 1) (Properties of 0 and 1 , X=X.1) (Properties of 0 and 1) 7. Distributive Law (Second) The law states that, PROOF ( By Algebraic Method)
- 20. X’Y X’ Y X X X Y R L.H.S. = X+X’Y =X.1+X’Y =X(1+Y)+X’Y =X+XY+X’Y =X+Y(X+X’) =X+Y.1 =X+Y =R.H.S. Hence PROOF ( By Algebraic Method) (Complementarity Law) (Properties of 0 and 1) (Properties of 0 and 1) (Properties of 0 and 1) c) X+X’Y=X+Y 7. Distributive Law (Third) The law states that,
- 21. X X Y a) X+XY=X XY X Y X.Y X+XY 0 0 0 0 0 1 0 0 1 0 0 1 1 1 1 1 =R.H.S. PROOF ( By Algebraic Method) L.H.S. = X+XY =X(1+Y) =X.1 = X (By Properties of 0 and 1) Hence 8. Absorption Law The law states that,
- 22. X The law states that, X Y X.Y PROOF ( By Algebraic Method) b) X(X+Y)=X L.H.S =X(X+Y) =X.X+X.Y =X+X.Y =X(1+Y) =X =R.H.S Hence (Distributive Law ) (Indempotence Law) (Properties of 0 and 1), 1+Y=1 8. Absorption Law
- 23. DEMORGAN’S /IRST THEOREM (X+Y)’=X’.Y’ X Y R X+Y X’ Y’ X Y Let us assume that P=X+Y where P,X,Y are logical variables. Then, according to complementarity law, P+P’=1 and P.P’=0. i.e., if P=X+Y then P’=X’Y’ , so, P+P’= (X+Y)+X’Y’ must be equal to 1 and P.P’ = (X+Y).(X’Y’) must be equal to 0 It states that NOTE: BOTH THE DEMORGANS THEOREM CAN BE PROVED USING TRUTH TABLE ALSO X+Y R X’.Y’
- 24. Hence First Part Proved Let us prove the first part of 1st theorem, i.e., (X+Y)+(X’Y’)=1 (Distributive Law: A+BC=(A+B). (A+C) ) (Complementarity Law: X+X’=1) (Properties of 0 and 1) (Properties of 0 and 1) L.H.S. = (X+Y)+(X’Y’) = (X+Y+X’).(X+Y+Y’) = (1+Y).(X+1) = 1.1 = 1
- 25. Hence Second Part Proved Now, let us prove the second part of 1st theorem, i.e., (X+Y).(X’Y’)=0 (Distributive Law): A+B).C=A.C+A.B (Associative Law): X.Y=Y.X (Complementarity Law): X.X’=0 (Properties of 0 and 1) (Properties of 0 and 1) L.H.S. = (X+Y).(X’Y’) = (XX’Y’)+(YX’Y’) = (XX’Y’)+(X’YY’) = (0.Y’) + (X’.0) = 0+0 = 0 Thus, De-Morgan’s 1st Law proved (X+Y)’=X’.Y
- 26. (X.Y)’=X’+Y’ X Y R XY R X’ Y’ X Y Let us assume that P=X.Y where P,X,Y are logical variables. Then, according to complementarity law, P.P’=1 and P.P’=0. i.e., if P=X.Y then P’=X’+Y’ so, we have to prove P+P’= XY+(X’+Y’) must be equal to 1 and P.P’= XY.(X’+Y’) must be equal to 0 DEMORGAN’S SEcOND THEOREM It states that, XY X’+Y’
- 27. L.H.S. = XY+(X’+Y’) =(X+X’+Y’).(Y+X’+Y’) = (X+X’+Y’).(X’+Y+Y’) = (1+Y’). (X’+1) = 1.1 = 1 Let us prove the first part of 2nd theorem, i.e., XY+(X’+Y’)=1 (Distributive Law: A+BC=(A+B). (A+C) ) (Commutative Law): X+Y=Y+X (Properties of 0 and 1): X+1=1 (Properties of 0 and 1): 1.1=1 Hence First Part Proved
- 28. L.H.S. = (XY).(X’+Y’) = XY.X’+XY.Y’ = X.X’.Y + X.Y.Y’ = 0.Y + X.0 = 0+0 = 0 Now, let us prove the second part of 2nd theorem, i.e., XY.(X’+Y’)=0 Hence Second Part Proved Thus, De-Morgan’s 2nd Law proved (XY)’=X’+Y’ (Distributive Law): A+B).C=A.C+A.B (Associative Law): X.Y=Y.X (Complementarity Law): X.X’=0 (Properties of 0 and 1) (Properties of 0 and 1)
- 29. Boolean Algebra Rules 1. 0+X=X Properties of 0 2. 0.X=0 3. 1+X=1 Properties of 1 4. 1.X=X 5. X+X=X Indempotence Law 6. X.X=X 7. (X’)’=X Involution Law 8. X+X’=1 Complementarity Law 9. X.X’=0
- 30. Boolean Algebra Rules 10. X+Y=Y+X Commutative Law 11. X.Y=Y.X 12. X+(Y+Z)=(X+Y)+Z Associative Law 13. (X.Y).Z=X.(Y.Z) 14. X(Y+Z)=XY+XZ Distributive Law 15. X+YZ=(X+Y).(X+Z) 16. X+XY=X Absorption Law 17. X(X+Y)=X 18. X+X’Y=X+Y Third Distributive Law 19. (X+Y)’=X’.Y’ Demorgan’s Theorem 20. (X.Y)’=X’+Y’
- 32. NAND Gate NAND gate is a universal gate which is a combination of AND with NOT 2 input NAND Gate X Y 3 input NAND Gate P Q R 4 i n p X Y X.Y F= X.Y 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 X Y F= X.Y X Y Z X.Y.Z X.Y.Z 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 0
- 33. NOR Gate NOR gate is a universal gate which is a combination of OR with NOT 2 input NOR Gate X Y 3 input NOR Gate P Q R 4 input NOR Gate A B C X Y X+Y F=X+Y 0 0 0 1 0 1 1 0 1 0 1 0 1 1 1 0 X Y F= X+Y X Y Z X+Y+Z F=X+Y+Z 0 0 0 0 1 0 0 1 1 0 0 1 0 1 0 0 1 1 1 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 0 F F F
- 34. XOR Gate XOR gate produces output 1 for only those input combinations that have ODD number of 1’s 3 input XOR Gate P Q R 4 input XOR Gate A B C No. of 1’s Even/Odd X Y X + Y Even 0 0 0 Odd 0 1 1 Odd 1 0 1 Even 1 1 0 X Y F = X + Y No. of 1’s X Y Z F = X + Y + Z Even 0 0 0 0 Odd 0 0 1 1 Odd 0 1 0 1 Even 0 1 1 0 Odd 1 0 0 1 Even 1 0 1 0 Even 1 1 0 0 Odd 1 1 1 1 F F XOR Addition can be summarized as: 0 + 0=0 1 + 0=1 0 + 1=1 1 + 1=0
- 36. Q) Verify the following expression using Truth Table: (A’+B’).(A+B) = A’.B + A.B’ Solution : A B A’ B’ A’+B’ A+B (A’+B’).(A+B) A’.B A.B’ (A’.B)+(A.B’) 0 0 1 1 1 0 0 0 0 0 0 1 1 0 1 1 1 1 0 1 1 0 0 1 1 1 1 0 1 1 1 1 0 0 0 1 0 0 0 0 Since, Column no. 7 and 10 are having same values as per the given question. Hence, proved.
- 38. Q) Draw a logic circuit for the following expression: F(X,Y,Z) = (X+Y’+Z).(X+Y’+Z’) Solution: X Y’ Z X Y’ Z’ F (X+Y’+Z).(X+Y’+Z’) X+Y’+Z X+Y’+Z’
- 39. Q) Write the equivalent Boolean expression for the following logic circuit: Solution : F(A,B,C) = A’B+AB+B’C F
- 40. Q) Draw a logic circuit for the following expression: F(A,B,C) = (AB’C)+(C’B) Using NAND-to-NAND Logic Gate only. Solution: A B C C B C’ B’ F(A,B, C) (AB’C). (C’B) = (AB’C)+ (C’B) = (AB’C)+(C’B) (A.B’.C) (C’.B) De-Morgan’s Theorem (applied). NOTE: No need to solve. Directly write the answer. This is done just to prove and show.
- 41. REFERENCES •Wikipedia •Tutorialspoint •XII Computer Science by Sumita Arora, Edition:2016
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