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Class 12 Notes 2017 Boolean Algebra : Computer Science (Army Public School)

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Somnath PaulChoudhury
St. Mary's ICSE School, Koparkhairane, Navi Mumbai

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B oolean Algebra was developed by George Boole, an English Mathematician. This algebra which is a set of rules, are also suitable for digital circuits. The principle behind Boolean algebra is that the answer to its equations are either True (or 1) or False (or 0). Propositions are statements that can be either true or false. Boolean algebra is also called propositional algebra or two-state algebra. Consider the following propositions 14 is more than 11. (T) Chennai is a big city. (T) DoCoMo is GSM a TATA brand AND IndiComm is CDMA also of TATA (T) The two propositions DoCoMo is GSM a TATA brand , IndiComm is CDMA also of TATA are connected to each other by AND. Propositions connected in such a manner are called compound propositions. S witching Circuits & the Boolean algebra Claude Shannon in 1938 used the symbolic logic of Boolean algebra to analyze switching circuits at Bell labs. A switching circuit is composed of interconnected switches. A switch can be either closed or open . If closed electric current will pass, when open current cannot pass through the switch. Closed switch Open switch AND / OR Operations B B AND Operation (Switches in series) Case 1: S1 = open and S2 = open Case 2: S1 = open and S2 = closed Case 3: S1 = closed and S2 = open Case 4: S1 = closed and S2 = closed the current will not flow through the bulb B the current will not flow through the bulb B the current will not flow through the bulb B the current will flow through the bulb B The result tabulated below looks like S1 closed S2 closed No No No Yes Yes No Yes Yes Current flowing through bulb B No No No Yes B Table 4.1 We can further simplify the results by replacing 1 for Yes and 0 for No S1 closed S2 closed 0 0 0 1 1 0 1 1 Current flowing through bulb B 0 0 0 1 Table 4.2 Mathematically B is true only when both S1 and S2 are true simultaneously. So we can write B = S1 AND S2 In the above compound proposition AND connective has been used. In symbolic logic AND operator is known as logical AND or logical product. Table 4.2 is known as Truth Table which is a combination of all the possible inputs and their effect on the output(s) of a switching circuit. We can develop the Truth Table for logical AND with two inputs A and B and output say OUT. A 0 0 1 1 B OUT 0 0 1 0 0 0 1 1 B OR Operation (Switches in parallel) Case 1: S1 = open and S2 = open Case 2: S1 = open and S2 = closed Case 3: S1 = closed and S2 = open Case 4: S1 = closed and S2 = closed the current will not flow through the bulb B the current will flow through the bulb B the current will flow through the bulb B the current will flow through the bulb B The result tabulated below looks like S1 closed S2 closed No No No Yes Yes No Yes Yes Current flowing through bulb B No Yes Yes Yes Table 4.3 We can further simplify the results by replacing 1 for Yes and 0 for No S1 closed S2 closed 0 0 0 1 1 0 1 1 Current flowing through bulb B 0 1 1 1 Table 4.4 Mathematically B is true when either S1 OR S2 are true or both true simultaneously. So we can write B = S1 OR S2 in the above compound proposition OR connective has been used. In symbolic logic OR operator is known as logical OR, or logical sum. Table 4.4 is known as Truth Table which is a combination of all the possible inputs and their effect on the output(s) of a switching circuit. We can develop the Truth Table for logical OR with two inputs A and B and output say OUT. A 0 0 1 1 NOT Operation B OUT 0 0 1 1 0 1 1 1 S Case 1: S = open Case 2: S = closed B the current will flow through the bulb B the current will not flow through the bulb B The result tabulated below looks like S closed No Yes Current flowing through bulb B Yes No Table 4.5 We can further simplify the results by replacing 1 for Yes and 0 for No S closed 0 1 Current flowing through bulb B 1 0 Table 4.6 Mathematically B is true when S is false and vice versa. So we can write B = NOT S. In symbolic logic NOT operator is known as logical NOT or logical Negative. Table 4.6 is known as Truth Table which is a combination of all the possible inputs and their effect on the output(s) of a switching circuit. We can develop the Truth Table for logical NOT with input A and output say OUT. A OUT 0 1 1 0 If a proposition is represented by the letter A, then its negation is represented by A The composite truth table A B A AND B A OR B 0 0 1 1 0 1 0 1 0 0 0 1 0 1 1 1 A 1 1 0 0 B 1 0 1 0 Boolean Variables and expressions Boolean variables are binary in nature. They can have only two values: true or false, 1 or 0, closed or open switch, etc. A Boolean variable is also called a literal. A Boolean expression can be expressed as: p = a < Boolean operator > b The result of the expression p depends upon i. Truth of a ii. Truth of b iii. Boolean operator Let us consider another Boolean expression X = A AND B OR (NOT C) In symbolic logic AND operator is logical product, OR operator is logical sum, NOT operator is logical Negative, the above expression can also be written as X = A.B + C Evaluation of complex Boolean expression p = A + B .C+D +B .D To evaluate any Boolean expression as above the order of precedence if followed, which states: i. Evaluate left to right ii. Evaluate expressions in parenthesis iii. Perform all NOT operations iv. Perform all AND operations v. Perform all OR operations p = A + B .C+D +B .D 3 4 5 1 2 If parenthesis is used the order of evaluation will change p = A + B .(C+D )+B .D i. First (C+D ) will be evaluated, (= say M) ii. iii. iv. v. Then B .M (=say N) B .D is evaluated next (= say O) A+N is evaluated next (=say P) P+O is evaluated finally The Boolean AND operator is represented by the symbol The Boolean OR operator is represented by the symbol The Boolean NOT operator is represented by the symbol The + sign in Boolean algebra does not stand for the addition From the truth table of OR as shown below A 0 0 1 1 B OUT 0 0 1 1 0 1 1 1 We can write further 0+0=0 0+1=1 1+0=1 1+1=1 In logic algebra, A + B = C means that if A is true OR B is true, then C will be true. It does not mean sum of A and B equals C. The meaning of + in the context of Boolean algebra becomes clear if we understand that 1 + 1=2 - decimal addition 1 + 1 = 10 1+ 1=1 - binary addition - OR addition The laws of OR can be put in more general terms as A=0 A=1 1 1 1 1 A+1=1 A=0 A=1 0 1 0 0 A+0=A A=0 A=1 A A A=0 A=1 A+A=A From the truth table of AND as shown below A 0 0 1 1 B OUT 0 0 1 0 0 0 1 1 We can write further 0.0=0 0.1=0 1.0=0 1.1=1 The laws of Boolean multiplication in general form can be written as A . 1= A A . 0= 0 A . A = A A=0 A=1 0 1 1 1 So we can write A . 1 = A A=0 A=1 0 0 0 0 So we can write A . 0 = 0 A=0 A=1 0 A=0 1 A=1 So we can write A . A = A Basic postulates of Boolean Algebra Boolean algebra deals with only binary (true/false) states and therefore it is different from the algebra of numbers. It has a set of postulates and theorems with the help of which a complex Boolean expression can be simplified. Let B is a set of elements of Boolean algebra such that a, b, c, 0, 1 B Postulate 1. 0 & 1 are unique identity elements such that a + 0 = a (additive identity) a . 1 = a (multiplicative identity) a 0 1 a+0 0 1 a 0 1 a.1 0 1 Postulate 2. For every element a B, there is an element a such that a + a = 1 a . a = 0 a is complement of a. a 0 1 a 1 0 a + a 1 1 a 1 0 a 0 1 a . a 0 0 Postulate 3. The operations logical OR and logical AND are commutative a + b = b + a a . b = b . a a 0 0 1 1 b 0 1 0 1 a + b 0 1 1 1 b + a 0 1 1 1 a 0 0 1 1 b 0 1 0 1 a . b 0 0 0 1 b . a 0 0 0 1 The last two columns in both the truth tables are identical, hence it s verified. Postulate 4. The operations logical OR and logical AND are distributive a . ( b + c) = a . b + a . c a + ( b . c) = a + b . a + c a 0 0 0 0 1 1 1 1 b 0 0 1 1 0 0 1 1 c 0 1 0 1 0 1 0 1 b . c 0 0 0 1 0 0 0 1 a +( b . c ) 0 0 0 1 1 1 1 1 a + b 0 0 1 1 1 1 1 1 a + c 0 1 0 1 1 1 1 1 ( a + b ) .( a + c ) 0 0 0 1 1 1 1 1 Since the 5th column a + ( b . c ) and 8th column ( a + b ) . ( a + c ) are identical the law is valid. Postulate 5. The operations logical OR and logical AND obey principle of substitution in such a way that if a = b, then b can be substituted for a, and vice-versa. a + c = b + c a . c = b . c Postulate 6. The operations logical OR and logical AND obey associative laws. a + ( b + c) = ( a + b) + c a . ( b . c ) =( a . b ) . c a 0 0 0 0 1 1 1 1 b 0 0 1 1 0 0 1 1 c 0 1 0 1 0 1 0 1 b + c 0 1 1 1 0 1 1 1 a +( b + c ) 0 1 1 1 1 1 1 1 a + b 0 0 1 1 1 1 1 1 ( a + b ) + z 0 1 1 1 1 1 1 1 Comparing the column a + ( b + c ) and ( a + b ) + z we find all values are same a 0 0 0 0 1 1 1 1 b 0 0 1 1 0 0 1 1 c 0 1 0 1 0 1 0 1 b . c 0 0 0 1 0 0 0 1 a . ( b . c ) 0 0 0 0 0 0 0 1 a . b 0 0 0 0 0 0 1 1 ( a . b ) . z 0 0 0 0 0 0 0 1 The columns a . ( b . c ) and ( a . b ) . c are same. Hence the law is verified. Postulate 7. The operations logical OR, logical AND & logical NOT obey closure property such that. a + b B a . b B a B a 0 0 1 1 b 0 1 0 1 a + b 0 1 1 1 a 0 0 1 1 b 0 1 0 1 a . b 0 0 0 1 Here the result of the two operations, a + b & a . b is either 0 or 1. So the closure property holds good in Boolean algebra. The above postulates of Boolean algebra are also known as Hontington s postulates. Theorems of Boolean Algebra The following Boolean expressions are very important Theorem 1. The Boolean algebra is idempotent a + a = a a . a = a We can prove idempotent law algebraically also a + a = ( a + a ) . 1 Multiplicative identity =( a + a ) . ( a + a ) Law of complement = a + a . a Distributive law = a Law of complement a . a = 0 Theorem 2. The Boolean algebra allows the dominance of 0 and 1 such that a + 1= 1 a . 0= 0 Theorem 3. The Boolean algebra is absorptive a + ( a . b) = a a . ( a + b ) = a a 0 0 1 1 b a.b 0 0 1 0 0 0 1 1 a+(a.b) 0 0 1 1 a+b 0 1 1 1 a.(a+b) 0 0 1 1 From the columns above we see a + ( a . b ) = a & a . ( a + b ) = a Theorem 4. The Boolean algebra is involuted ( a ) = a a a (a ) 0 1 0 1 0 1 De Morgan, a friend of Boole, worked extensively in logic. He contributed following two important theorems: Theorem 5. The complement of a sum equals the product of the complements ( a + b) = a . b Theorem 6. The complement of the product equals the sum of the complements ( a . b) = a + b Duality Principle If we observe the Boolean postulates we will find that the postulates are symmetric with respect to the operators AND & OR, 0 & 1. So for any Boolean identity we can produce a dual identity by i. Changing . sign to + and vice-versa ii. Complementing all 0 s & 1 s iii. Leaving all NOT alone For example from postulate 2 above if we take a + a = 1 and follow the above steps we obtain a . a = 0 Also if we consider the first expression of the absorption law a + ( a . b) = a and follow the same above steps we obtain a . ( a + b ) = a Terminology and Standard Forms i. Product Term - When two or more Boolean variables are connected by AND operator. o For ex. a . b . c is a product term ii. Sum Term - When two or more Boolean variables are connected by OR operator. o For ex. a + b + c is a sum term iii. SOP When in a Boolean expression product terms are connected by OR operator o For ex a c + b c is a sum of products iv. POS - When in a Boolean expression sum terms are connected by AND operator o For ex ( a + c ) . ( b + c ) is a product of sums v. Minterm It is a product term of all the entries in a row of the truth table. The variables that are 0 for the row are complemented and the variables which are 1 for the row are not disturbed vi. Maxterm It is a sum term of all the entries in a row of the truth table. The variables that are 1 for the row are complemented and the variables which are 0 for the row are not disturbed Row x 0 0 y Minterm Maxterm 0 x y x + y 1 2 3 0 1 1 1 0 1 x y x y x y x + y x + y x + y Q> Write the SOP form of a Boolean Function F, which is represented by the following truth table: A> The Boolean expression represented purely as sum of min terms or product terms is known to be Canonical Sum-of-Products A 0 0 0 0 1 1 1 1 B 0 0 1 1 0 0 1 1 C 0 1 0 1 0 1 0 1 F 1 0 0 1 0 0 1 1 Since the last column of the truth table gives the values of F. For each occurrence of 1 in this column we get a min term corresponding to that row. So there are 4 min terms for rows 1, 4, 7 & 8. SOP = A B C + A B C + A B C + A B C Canonical form of Boolean expressions For a given truth table, as shown below, we can write its corresponding Boolean expression. This is done in two ways. A 0 0 1 1 Method 1. B 0 1 0 1 Consider the 0th row with a 1 in the final (output) column Write the minterm for this row Repeat this for all rows which is having 1 in the last column OR all these minterms i. 0th row is equivalent to A B y 1 0 1 1 ii. 2nd row is equivalent to AB iii. 3rd row is equivalent to AB Thus the final Boolean expression is = A B + AB + AB The above form of expression (logical sum of minterms) is known as canonical sum of products. This is also known as AND-OR form and can be expressed as 0, 2, 3. The representation 0, 2, 3 means that the minterms of rows 0, 2, 3 are to be ORed together. Q> Given the following truth table, write the SOP form of the function F(x,y,z) x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 F 0 1 1 0 1 0 0 1 A> For the expression F we get 4 min terms for row numbers 2,3,5 & 8 giving the min terms canonical form or SOP form F = x y z + x y z + x y z + x y z Method 2. A 0 0 1 1 B 0 1 0 1 Consider the first row with a 0 in the final (output) column Write the maxterm for this row Repeat this for all rows which is having 0 in the last column AND all these minterms i. 1st row is equivalent to A+B ii. 3rd row is equivalent to A +B y 1 0 1 1 Thus the final Boolean expression is = (A+B ) . (A +B ) The above form of expression (logical product of maxterms) is known as canonical product of sums. This is also known as OR- AND form and can be expressed as (1, 3). The representation (1, 3) means that the maxterms of rows 1 & 3 are ANDed together. Q> Write the POS form of a Boolean Function F, which is represented by the following truth table. X Y Z 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 F 1 1 0 1 0 1 0 0 A> The last column of the truth table gives the vales of F. For each occurrence of 0 in this column we get a max term corresponding to that row. Thus for the expression F, we get 4 max terms for row numbers 3, 5, 7 & 8 giving the product of sums form. F = (X + Y + Z) . (X + Y + Z) . (X + Y + Z) . (X + Y + Z ) Q> Given the following truth table, write the products of sums form of the function F(x, y, z) x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 F 0 1 1 0 1 0 0 1 Writing the truth table for any Boolean expression Let us take the expression y = AB + AB i. ii. iii. iv. Draw columns for each separate propositions in the expression Fill the first row with 0s for the input Boolean variables Fill the remaining rows by counting in binary until all 1s appear In the expression find each ANDed section. Consider all TRUE proposition to 1s and false propositions to 0s. v. Identify the rows that correspond to these combinations of 1s and 0s. vi. Place 1 in the final column in the rows where there is a match vii. Place 0 in the final column elsewhere Step i A B y A 0 B 0 y A 0 0 1 1 B 0 1 0 1 y A 0 0 1 1 B 0 1 0 1 y 1 1 A 0 0 1 1 B 0 1 0 1 y 0 0 1 1 Step ii Step iii Step iv-vi Step vii The above is the truth table for the expression y = AB + AB Circuit Implementations of Boolean Functions The Boolean functions can be implemented on the components of the computer system. A Boolean operator is represented as a Gate. A logic gate is an electronic circuit that makes logic decisions. It has one output and one or more inputs. Logic gates are the basic building blocks from which most of the digital systems are built up. The Boolean AND operator is represented by the symbol The Boolean OR operator is represented by the symbol The Boolean NOT operator is represented by the symbol NAND gate is AND gate with its output inverted by a NOT gate. The circuit of NAND The symbol for NAND The truth table for 2-input NAND gate is given below a 0 0 1 1 b 0 1 0 1 c 1 1 1 0 NOR gate is OR gate with its output inverted by a NOT gate. The circuit of NOR The truth table for 2-input NAND gate is given below a 0 0 1 1 The symbol for NOR b 0 1 0 1 c 1 0 0 0 XOR gate Exclusive OR This gate gives a true output if either of its input are true, but not if both input are true. The corresponding truth table is as below a 0 0 1 1 b 0 1 0 1 c 0 1 1 0 XNOR gate Exclusive NOR This is a not-XOR gate. Its logic symbol and truth table are just reverse of XOR a 0 0 1 1 b 0 1 0 1 c 1 0 0 1 Examples 1. Draw the logic circuits for the following Boolean expressions a. y = a + b b. y = a . b + c . a Ans 1. a 1. b a.b a b c a a c.a Algebraic simplification Minimize the number of gates Minimize the number of inputs The Boolean expression p = xy + xy + y It can be reduced to x + y Karnaugh maps This graphic technique for reducing logic functions and circuitry was developed by Maurice Karnaugh in early 1950s. The Karnaugh map is used to simplify Boolean equations in which the function to be simplified is displayed diagrammatically on a set of squares. Each square maps one term of the function. The number of squares in a Karnaugh map is 2n where n is the number of variables in the equation to be simplified. For example, in the equation y = A B + A B, n is 2 so that the number of squares required is 4. The rows and columns of the map are labeled as shown A B A ... .. B Each square represent a different combination of the two variables as AB, A B, A B , A B . If a term is present 1 is written in the square else 0 is written. Since in the equation y = A B + A B only A B & A B are present the corresponding squares are filled with 1. A B B A 0..... 1.... 1 0 For simplifying an equation using Karnaugh map, adjacent squares containing 1 are looped together. Looping means that the corresponding terms in the equation being mapped have been combined and any terms of the form A A have been eliminated. A A B B 1..... 1.... 0 1 Let us consider another equation y = A B + A B + A B which has been mapped above. If we loop the adjacent squares as shown below, the equation is simplified to y = A (B + B ) + B(A + A ) = A + B A B 1 1..... A 1 1.... B 0 1 1 Use of K Map for Minimization of Boolean expressions (up to 4 variables) K Maps can be used to simplify Boolean equations. K- Map using 2 variables A 0 1 0 B A 00 B A 01 1 .BA .. 10 BA ... 11 B K- Map using 3 variables BA 00 0 C 1 01 11 10 C B A C B A C AB C BA CB A CBA CB A CAB BA K- Map using 4 variables 00 B 00 01 D CB A DCB A 10 11 10 D C B A D C B A D C AB D C BA 11 DC 01 DC B A D CB A D C AB D CBA DCB A DC AB DCBA DC B A DC AB DC BA The progression of values is 00, 01, 11, 10 and not 00, 01, 10, 11. This is also known as the truth map to distinguish it from the truth table. How to Map The truth table is shown in the left pane and the K-Map in the right pane The canonical SOP expression of the truth table is F = A B + A B or F = (2, 3) To map this function an empty 2 variable K-map is drawn. Next all output 1 is searched in the truth table and corresponding squares in K-map is entered 1, rest is filled with 0. Terms Pair A group of two adjacent 1 s on a K map Quad - A group of four adjacent 1 s on a K map Octet - A group of eight adjacent 1 s on a K map Redundant block A block where all 1 s are embedded into another block Isolated 1 s that cannot be included anywhere Overlapping 1 s encircled more than once Map Rolling Rolling the K map, opposite edges horizontally and vertically will touch each other A pair represents a product where a variable and its complement drop out. It removes the variable that changes from 0 to 1 or 1 to 0. So, a pair always removes one variable. A quad represents a product where two variables and their complements drop out. An octet represents a product where three variables and its complement drop out. Q> Obtain the simplified form of the Boolean expression using K Map, F(x, y, z) = (2, 3, 6, 7) A> The Boolean expression F(x, y, z) = (2, 3, 6, 7) can be written as F(x, y, z) = x y z + x y z + x y z + x y z z z x y x y z ..0 0 .1 .1 xy .1 .1 xy .0 .0 xy The 1 s on the K Map correspond to 4 min terms The simplified Boolean expression becomes F(x, y, z) = y Summary of SOP reduction by K map 1. 2. 3. 4. 5. 6. 7. Prepare truth table Draw empty K-map Map the output of 1 s from truth table to squares of K-map Fill up remaining squares with 0 s Separate out adjacent 1 s in the form of pairs, octets, quads. Roll and overlap for corners Reduce the expression and OR them Q> Reduce F(a, b, c, d) = (0, 2, 7, 8, 10, 15) using K-map In the above K-map two groups have been marked, one pair and one quad Pair reduces to A BCD + ABCD = (A + A)BCD = BCD Quad reduces to A B C D + AB C D + A B CD + AB CD = B C D + B CD = B D Q> What is the simplified Boolean equation for the function: F(A, B, C, D) = (7, 9, 10, 11, 12, 13, 14, 15) A> A B A B AB AB 00 01 11 10 C D C D CD CD 00 01 11 10 00 01 03 02 04 05 17 06 112 113 115 114 08 19 111 110 Pair = 17 + 115 Quad 1 = 112 + 113 + 115 + 114 Quad 2= 113 + 115 + 19 + 111 Quad 3= 115 + 111 + 114 + 110 The Pair reduces to BCD as the value changes from A BCD to ABCD (A to A) Quad 1 reduces to AB as both C & D changes values Quad 2 reduces to AD as B & C changes Quad 3 reduces to AC Final equation is = BCD + AB + AD + AC Q> What is the simplified Boolean expression using K map F(a, b, c, d) = (0, 1, 2, 4, 5, 7, 8, 9, 10, 11, 14) A> K- Map using 4 variables CD AB A B 00 A B 01 AB 11 C D 00 B C D 01 CD 11 A B C D 0 A B C D 1 A B CD 3 A B CD 2 A B C D 4 A B C D A B CD A B CD AB C D AB C D 12 AB 10 The truth table AB C D 5 7 6 13 AB CD 15 AB CD 14 AB C D 8 CD 10 9 AB CD AB CD 11 10 A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 f 1 1 1 0 1 1 0 1 1 1 1 1 0 0 1 0 CD C D 00 C D 01 CD 11 CD 10 AB A B 01 1 1 0 A B 01 1 1 0 AB 10 1 0 7 6 0 1 13 15 14 1 8 2 1 0 12 1 3 5 4 AB 11 0 1 1 9 1 11 10 Quad 1 = [0,1,4,5] Quad 2 = [8,9,11,10] Quad 3 = [0,2,8,10] Pair 1 = [5,7] Pair 2 = [14,10] F(a, b, c, d) = Quad 1 + Quad 2 + Quad 3 + Pair 1 + Pair 2 Quad 1 = A B C D + A B C D + A B C D + A B C D = A B C + A B C = A C Similarly, Quad 2 = A B Quad 3 = B D Pair 1 = A B D Pair 2 = A C D F(a, b, c, d) = A C + A B + B D + A B D + A C D P-O-S reduction by K map 1. 2. 3. 4. 5. 6. 7. Prepare truth table Draw empty K-map Map the output of 0 s from truth table to squares of K-map Fill up remaining squares with 1 s Separate out adjacent 0 s in the form of pairs, octets, quads. Roll and overlap for corners Reduce the expression and AND them Q> What is the simplified Boolean expression using K map X(a, b, c, d) = (0, 1, 3, 5, 6, 7, 10, 14, 15) CD C+D 00 B C+D 01 C +D 11 C +D 10 AB A+B 00 ..0 .. ..0 0 . .1 .. 0 A+B 01 1 A +B 11 1 A +B 10 1 1 0 4 0 5 1 12 2 0 7 0 13 1 8 3 6 0 15 1 9 14 0 11 10 Pair 1 = A + B + C Pair 2 = A + C + D Quad 1 = A + D Quad 2 = B + C X(a, b, c, d) = (A + B + C)( A + C + D)( A + D )( B + C ) Computer application of logic circuits Adders Decoders Encoders

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