4 BOOLEAN ALGEBRA AND LOGIC SIMPLIFICATION
4 BOOLEAN ALGEBRAANDLOGIC SIMPLIFICATIONBOOLEAN OPERATIONS AND EXPRESSIONSVariable, complement, and literal are terms used in Boolean algebra. Avariable is a symbol used to represent a logical quantity. Any single variablecan have a 1 or a 0 value. The complement is the inverse of a variable and isindicated by a bar over variable (overbar). For example, the complement ofthe variable A is A. If A 1, then A 0. If A 0, then A 1. Thecomplement of the variable A is read as "not A" or "A bar." Sometimes aprime symbol rather than an overbar is used to denote the complement of avariable; for example, B' indicates the complement of B. A literal is avariable or the complement of a variable.Boolean AdditionRecall from part 3 that Boolean addition is equivalent to the ORoperation. In Boolean algebra, a sum term is a sum of literals. In logiccircuits, a sum term is produced by an OR operation with no ANDoperations involved. Some examples of sum terms are A B, A B, A B C, and A B C D.A sum term is equal to 1 when one or more of the literals in the term are 1. Asum term is equal to 0 only if each of the literals is 0.ExampleDetermine the values of A, B, C, and D that make the sum termA B C Dequal to 0.
Boolean MultiplicationAlso recall from part 3 that Boolean multiplication is equivalent to the ANDoperation. In Boolean algebra, a product term is the product of literals. Inlogic circuits, a product term is produced by an AND operation with no ORoperations involved. Some examples of product terms are AB, AB, ABC,and ABCD.A product term is equal to 1 only if each of the literals in the term is 1. Aproduct term is equal to 0 when one or more of the literals are 0.ExampleDetermine the values of A, B, C, and D that make the product term ABCDequal to 1.LAWS AND RULES OF BOOLEAN ALGEBRA Laws of Boolean AlgebraThe basic laws of Boolean algebra-the commutative laws for addition andmultiplication, the associative laws for addition and multiplication, and thedistributive law-are the same as in ordinary algebra.Commutative Laws The commutative law of addition for two variables is written asA B B AThis law states that the order in which the variables are ORed makes nodifference. Remember, in Boolean algebra as applied to logic circuits,addition and the OR operation are the same. Fig.(4-1) illustrates thecommutative law as applied to the OR gate and shows that it doesn't matterto which input each variable is applied. (The symbol means "equivalentto.").
Fig.(4-1) Application of commutative law of addition. The commutative law of multiplication for two variables isA.B B.AThis law states that the order in which the variables are ANDed makes nodifference. Fig.(4-2), il1ustrates this law as applied to the AND gate.Fig.(4-2) Application of commutative law of multiplication.Associative Laws : The associative law of addition is written as follows for three variables:A (B C) (A B) CThis law states that when ORing more than two variables, the result is thesame regardless of the grouping of the variables. Fig.(4-3), illustrates thislaw as applied to 2-input OR gates.Fig.(4-3) Application of associative law of addition. The associative law of multiplication is written as follows for threevariables:A(BC) (AB)C
This law states that it makes no difference in what order the variables aregrouped when ANDing more than two variables. Fig.(4-4) illustrates this lawas applied to 2-input AND gates.Fig.(4-4) Application of associative law of multiplication.Distributive Law: The distributive law is written for three variables as follows:A(B C) AB ACThis law states that ORing two or more variables and then ANDing the resultwith a single variable is equivalent to ANDing the single variable with eachof the two or more variables and then ORing the products. The distributivelaw also expresses the process of factoring in which the common variable Ais factored out of the product terms, for example,AB AC A(B C).Fig.(4-5)illustratesthedistributive law inimplementation.Fig.(4-5) Application of distributive law.terms of gate
Rules of Boolean AlgebraTable 4-1 lists 12 basic rules that are useful in manipulating and simplifyingBoolean expressions. Rules 1 through 9 will be viewed in terms of theirapplication to logic gates. Rules 10 through 12 will be derived in terms ofthe simpler rules and the laws previously discussed.Table 4-1 Basic rules of Boolean algebra.Rule 1.A 0 AA variable ORed with 0 is always equal to the variable. If the input variableA is 1, the output variable X is 1, which is equal to A. If A is 0, the output is0, which is also equal to A. This rule is illustrated in Fig.(4-6), where thelower input is fixed at 0.Fig.(4-6)
Rule 2.A 1 1A variable ORed with 1 is always equal to 1. A 1 on an input to an OR gateproduces a 1 on the output, regardless of the value of the variable on theother input. This rule is illustrated in Fig.(4-7), where the lower input is fixedat 1.Fig.(4-7)Rule 3.A.0 0A variable ANDed with 0 is always equal to 0. Any time one input to anAND gate is 0, the output is 0, regardless of the value of the variable on theother input. This rule is illustrated in Fig.(4-8), where the lower input is fixedat 0.Fig.(4-8)Rule 4.A.1 AA variable ANDed with 1 is always equal to the variable. If A is 0 the outputof the AND gate is 0. If A is 1, the output of the AND gate is 1 because bothinputs are now 1s. This rule is shown in Fig.(4-9), where the lower input isfixed at 1.Fig.(4-9)
Rule 5.A A AA variable ORed with itself is always equal to the variable. If A is 0, then 0 0 0; and if A is 1, then 1 1 1. This is shown in Fig.(4-10), where bothinputs are the same variable.Fig.(4-10)Rule 6.A A 1A variable ORed with its complement is always equal to 1. If A is 0, then 0 0 0 1 1. If A is l, then 1 1 1 0 1. See Fig.(4-11), where oneinput is the complement of the other.Fig.(4-11)Rule 7.A.A AA variable ANDed with itself is always equal to the variable. If A 0,then 0.0 0; and if A 1. then 1.1 1. Fig.(4-12) illustrates this rule.Fig.(4-12)
Rule 8.A.A 0A variable ANDed with its complement is always equal to 0. Either A or Awill always be 0: and when a 0 is applied to the input of an AND gate. theoutput will be 0 also. Fig.(4-13) illustrates this rule.Fig.(4-13)Rule 9A AThe double complement of a variable is always equal to the variable. If youstart with the variable A and complement (invert) it once, you get A. If youthen take A and complement (invert) it, you get A, which is the originalvariable. This rule is shown in Fig.(4-14) using inverters.Fig.(4-14)Rule 10.A AB AThis rule can be proved by applying the distributive law, rule 2, and rule 4 asfollows:A AB A( 1 B)Factoring (distributive law) A.lRule 2: (1 B) 1 ARule 4: A . 1 AThe proof is shown in Table 4-2, which shows the truth table and theresulting logic circuit simplification.
Table 4-2Rule 11.A AB A BThis rule can be proved as follows:A AB (A AB) AB (AA AB) ABRule 10: A A ABRule 7: A AA AA AB AA ABRule 8: adding AA 0 (A A)(A B)Factoring 1. (A B)Rule 6: A A 1 A BRule 4: drop the 1The proof is shown in Table 4-3, which shows the truth table and theresulting logic circuit simplification.Table 4-3
Rule 12.(A B)(A C) A BCThis rule can be proved as follows:(A B)(A C) AA AC AB BCDistributive law A AC AB BCRule 7: AA A A( 1 C) AB BCRule 2: 1 C 1 A. 1 AB BCFactoring (distributive law) A(1 B) BCRule 2: 1 B 1 A. 1 BCRule 4: A . 1 A A BCThe proof is shown in Table 4-4, which shows the truth table and theresulting logic circuitsimplification.Table 4-4
DEMORGAN'S THEOREMSDeMorgan, a mathematician who knew Boole, proposed two theorems thatare an important part of Boolean algebra. In practical terms. DeMorgan'stheorems provide mathematical verification of the equivalency of the NANDand negative-OR gates and the equivalency of the NOR and negative-ANDgates, which were discussed in part 3.One of DeMorgan's theorems is stated as follows:The complement of a product of variables is equal to the sum of thecomplements of the variables,Stated another way,The complement of two or more ANDed variables is equivalent to the ORof the complements of the individual variables.The formula for expressing this theorem for two variables isXY X YDeMorgan's second theorem is stated as follows:The complement of a sum of variables is equal to the product of thecomplements of the variables.Stated another way,The complement of two or more ORed variables is equivalent to the ANDof the complements of the individual variables,The formula for expressing this theorem for two variables isX Y XYFig.(4-15) shows the gate equivalencies and truth tables for the twoequations above.
Fig.(4-15) Gate equivalencies and the corresponding truth tables thatillustrate DeMorgan's theorems.As stated, DeMorgan's theorems also apply to expressions in which there aremore than two variables. The following examples illustrate the application ofDeMorgan's theorems to 3-variable and 4-variable expressions.ExampleApply DeMorgan's theorems to the expressions XYZ and X Y z.XYZ X Y ZX y Z XYZExampleApply DeMorgan's theorems to the expressions WXYZ and W X y z.WXYZ W X y ZW X y Z WXYZ
Applying DeMorgan's TheoremsThe following procedure illustrates the application of DeMorgan's theoremsand Boolean algebra to the specific expressionStep l. Identify the terms to which you can apply DeMorgan's theorems, andthink of each term as a single variable. Let A BC X and D(E F) Y.Step 2. Since X Y X Y, (A BC) (D(E F))Step 3. Use rule 9 (A A) to cancel the double bars over the left term (this isnot part of DeMorgan's theorem).(A BC) (D(E F)) (A BC)(D(E F ))Step 4. Applying DeMorgan's theorem to the second term,(A BC)(D(E F)) (A BC)(D (E F ))Step 5. Use rule 9 (A A) to cancel the double bars over the E F part ofthe term.(A BC)(D E F) (A BC)(D E F)ExampleApply DeMorgan's theorems to each of the following expressions:(a) (A B C)D(b) ABC DEF(c) AB CD EF
ExampleThe Boolean expression for an exclusive-OR gate is AB AB. With this as astarting point, use DeMorgan's theorems and any other rules or laws that areapplicable to develop an expression for the exclusive-NOR gate.BOOLEAN ANALYSIS OF LOGIC CIRCUITSBoolean algebra provides a concise way to express the operation of a logiccircuit formed by a combination of logic gates so that the output can bedetermined for various combinations of input values.Boolean Expression for a Logic CircuitTo derive the Boolean expression for a given logic circuit, begin at the leftmost inputs and work toward the final output, writing the expression for eachgate. For the example circuit in Fig.(4-16), the Boolean expression isdetermined as follows: The expression for the left-most AND gate with inputs C and D is CD. The output of the left-most AND gate is one of the inputs to the ORgate and B is the other input. Therefore, the expression for the ORgate is B CD. The output of the OR gate is one of the inputs to the right-most ANDgate and A is the other input. Therefore, the expression for this ANDgate is A(B CD), which is the final output expression for the entirecircuit.
Fig.(4-16) A logic circuit showing the development of the Booleanexpression for the output.Constructing a Truth Table for a Logic CircuitOnce the Boolean expression for a given logic circuit has beendetermined, a truth table that shows the output for all possible values of theinput variables can be developed. The procedure requires that you evaluatethe Boolean expression for all possible combinations of values for the inputvariables. In the case of the circuit in Fig.(4-16), there are four inputvariables (A, B, C, and D) and therefore sixteen (24 16) combinations ofvalues are possible.Putting the Results in Truth Table formatThe first step is to list the sixteen input variable combinations of 1sand 0s in a binary sequence as shown in Table 4-5. Next, place a 1 in theoutput column for each combination of input variables that was determinedin the evaluation. Finally, place a 0 in the output column for all othercombinations of input variables. These results are shown in the truth table inTable 4-5.
Table 4-5SIMPLIFICATION USING BOOLEAN ALGEBRAA simplified Boolean expression uses the fewest gates possible toimplement a given expression.ExampleUsing Boolean algebra techniques, simplify this expression:AB A(B C) B(B C)SolutionStep 1: Apply the distributive law to the second and third terms in theexpression, as follows:AB AB AC BB BCStep 2: Apply rule 7 (BB B) to the fourth term.AB AB AC B BCStep 3: Apply rule 5 (AB AB AB) to the first two terms.AB AC B BCStep 4: Apply rule 10 (B BC B) to the last two terms.
AB AC BStep 5: Apply rule 10 (AB B B) to the first and third terms.B ACAt this point the expression is simplified as much as possible.Fig.(4-17) Gate circuits for example above.ExampleSimplify the Boolean expressions:1- AB A(B C) B(B C).2- [AB( C BD) A B]C3- ABC ABC A B C ABC ABC
Standard and Canonical FormsSTANDARD FORMS OF BOOLEAN EXPRESSIONSAll Boolean expressions, regardless of their form, can be converted intoeither of two standard forms: the sum-of-products form or the product-ofsums form. Standardization makes the evaluation, simplification, andimplementation of Boolean expressions much more systematic and easier.The Sum-of-Products (SOP) FormWhen two or more product terms are summed by Boolean addition,the resulting expression is a sum-of-products (SOP). Some examples are:AB ABCABC CDE BCDAB BCD ACAlso, an SOP expression can contain a single-variable term, as inA ABC BCD.In an SOP expression a single overbar cannot extend over more thanone variable.ExampleConvert each of the following Boolean expressions to SOP form:(a) AB B(CD EF)(b) (A B)(B C D)(c) (A B) C
Fig.(4-18) Implementation of the SOP expression AB BCD AC.Fig.(4-19) This NAND/NAND implementation is equivalentto the AND/OR in figure above.The Standard SOP FormSo far, you have seen SOP expressions in which some of the productterms do not contain all of the variables in the domain of the expression. Forexample, the expression ABC ABD ABCD has a domain made up of thevariables A, B, C. and D. However, notice that the complete set of variablesin the domain is not represented in the first two terms of the expression; thatis, D or D is missing from the first term and C or C is missing from thesecond term.A standard SOP expression is one in which all the variables in the domainappear in each product term in the expression. For example, ABCD ABCD ABCD is a standard SOP expression.
Converting Product Terms to Standard SOP:Each product term in an SOP expression that does not contain all thevariables in the domain can be expanded to standard SOP to include allvariables in the domain and their complements. As stated in the followingsteps, a nonstandard SOP expression is converted into standard form usingBoolean algebra rule 6 (A A 1) from Table 4-1: A variable added to itscomplement equals 1.Step 1. Multiply each nonstandard product term by a term made up of thesum of a missing variable and its complement. This results in two productterms. As you know, you can multiply anything by 1 without changing itsvalue.Step 2. Repeat Step 1 until all resulting product terms contain all variables inthe domain in either complemented or uncomplemented form. In convertinga product term to standard form, the number of product terms is doubled foreach missing variable.ExampleConvert the following Boolean expression into standard SOP form:ABC AB ABCDSolutionThe domain of this SOP expression A, B, C, D. Take one term at a time.The first term, ABC, is missing variable D or D, so multiply the first term by(D D) as follows:ABC ABC(D D) ABCD ABCDIn this case, two standard product terms are the result.The second term, AB, is missing variables C or C and D or D, so firstmultiply the second term by C C as follows:AB AB(C C) ABC ABC
The two resulting terms are missing variable D or D, so multiply both termsby (D D) as follows:ABC(D D) ABC(D D) A BCD ABCD ABCD ABCDIn this case, four standard product terms are the result.The third term, ABCD, is already in standard form. The complete standardSOP form of the original expression is as follows:ABC AB ABCD ABCD ABCD A BCD ABCD ABCD ABCD ABCDThe Product-of-Sums (POS) FormA sum term was defined before as a term consisting of the sum(Boolean addition) of literals (variables or their complements). When two ormore sum terms are multiplied, the resulting expression is a product-of-sums(POS). Some examples are(A B)(A B C)(A B C)( C D E)(B C D)(A B)(A B C)(A C)A POS expression can contain a single-variable term, as inA(A B C)(B C D).In a POS expression, a single overbar cannot extend over more than onevariable; however, more than one variable in a term can have an overbar. Forexample, a POS expression can have the term A B C but not A B C.Implementation of a POS Expression simply requires ANDing the outputs oftwo or more OR gates. A sum term is produced by an OR operation and theproduct of two or more sum terms is produced by an AND operation. Fig.(4-
20) shows for the expression (A B)(B C D)(A C). The output X ofthe AND gate equals the POS expression.Fig.(4-20)The Standard POS FormSo far, you have seen POS expressions in which some of the sum terms donot contain all of the variables in the domain of the expression. For example,the expression(A B C) (A B D) (A B C D)has a domain made up of the variables A, B, C, and D. Notice that thecomplete set of variables in the domain is not represented in e first two termsof the expression; that is, D or D is missing from the first term and C or C ismissing from the second term.A standard POS expression is one in which all the variables in the domainappear in each sum term in the expression. For example,(A B C D)(A B C D)(A B C D)is a standard POS expression. Any nonstandard POS expression (referred tosimply as POS) can be converted to the standard form using Booleanalgebra.Converting a Sum Term to Standard POSEach sum term in a POS expression that does not contain all the variables inthe domain can be expanded to standard form to include all variables in thedomain and their complements. As stated in the following steps, a
nonstandard POS expression is converted into standard form using Booleanalgebra rule 8 (A A 0) from Table 4-1:Step 1. Add to each nonstandard product term a term made up of the productof the missing variable and its complement. This results in two sum terms.As you know, you can add 0 to anything without changing its value.Step 2. Apply rule 12 from Table 4-1: A BC (A B)(A C)Step 3. Repeat Step 1 until all resulting sum terms contain all variables in thedomain in either complemented or noncomplemented form.ExampleConvert the following Boolean expression into standard POS form:(A B C)(B C D)(A B C D)SolutionThe domain of this POS expression is A, B, C, D. Take one term at a time.The first term, A B C, is missing variable D or D, so add DD and applyrule 12 as follows:A B C A B C DD (A B C D)(A B C D)The second term, B C D, is missing variable A or A, so add AA andapply rule 12 as follows:B C D B C D AA (A B C D)(A B C D)The third term, A B C D, is already in standard form. The standardPOS form of the original expression is as follows:(A B C)(B C D)(A B C D) (A B C D)(A B C D) (A B C D)(A B C D) (A B C D)
Examples:-CANONICAL FORMS OF BOOLEAN EXPRESSIONSWith one variablex & x.With two variablesx y, x y, x y and x y.With three variablesx y z, x y z, x y z, x y z, x y z, x y z, x y z & x y z.These eight AND terms are called minterms.n variables can be combined to form 2n minterms.xyzminterm000xyzm0x y zM0001xyzm1x y zM1010xyzm2x y zM2011xyzm3x y zM3100xyzm4x y zM4101xyzm5x y zM5110xyzm6x y zM6111xyzm7x y zM7(AND terms)designationmaxtermdesignation(OR terms)Note that each maxterm is the complement of its correspondingminterm and vice versa.
For example the function FxyzF00000011010001101001101011001111F xyz xyz xyzF m1 m4 m7Any Boolean function can be expressed as a sum of minterms (sum ofproducts SOP) or product of maxterms (product of sums POS).F xyz xyz xyz xyz xyzThe complement of F F FF (x y z) (x y z) (x y z) (x y z) (x y z)F M0 M2 M3 M5 M6ExampleExpress the Boolean function F A BC in a sum of minterms (SOP).SolutionThe term A is missing two variables because the domain of F is (A, B, C)A A(B B) AB ABbecause B B 1
BCmissing A, soBC(A A) ABC ABCAB(C C) ABC ABCAB(C C) ABC ABCF ABC ABC ABC ABC ABC ABCBecause A A AF ABC ABC ABC ABC ABCF m7 m6 m5 m4 m1In short notationF(A, B, C) (1, 4, 5, 6, 7)F(A, B, C) (0, 2, 3)The complement of a function expressed as the sum of minterms equalto the sum of minterms missing from the original function.Truth table for F A 161100017111001
ExampleExpress F xy xz in a product of maxterms form.SolutionF xy xz (xy x)(xy z) (x x)(y x)(x z)(y z)rememberx x 1F (y x)(x z)(y z)F (x y zz)(x yy z )(xx y z)F (x y z)(x y z)(x y z)(x y z)(x y z)(x y z) ---------------------------------- F (x y z)(x y z)(x y z)(x y z)F M4 M5 M0 M2F(x, y, z) (0, 2, 4, 5)F(x, y, z) (1, 3, 6, 7)The complement of a function expressed as the product of maxtermsequal to the product of maxterms missing from the original function.To convert from one canonical form to another, interchange the symbols , and list those numbers missing from the original form.F M4 M5 M0 M2 m1 m3 m6 m7F(x, y, z) (0, 2, 4, 5) (1, 3, 6, 7)
ExampleDevelop a truth table for the standard SOP expression ABC ABC ABC.Converting POS Expressions to Truth Table FormatReca11 that a POS expression is equal to 0 only if at least one of thesum terms is equal to 0. To construct a truth table from a POS expression,list all the possible combinations of binary values of the variables just as wasdone for the SOP expression. Next, convert the POS expression to standardform if it is not already. Finally, place a 0 in the output column (X) for eachbinary value that makes the expression a 0 and place a 1 for all the remainingbinary values. This procedure is illustrated in Example below:ExampleDetermine the truth table for the following standard POS expression:
SolutionThere are three variables in the domain and the eight possible binaryvalues are listed in the left three columns of. The binary values that make thesum terms in the expression equal to 0 are A B C: 000; A B C: 010:A B C: 011; A B C: 10l; and A B C: 110. For each of thesebinary values, place a 0 in the output column as shown in the table. For eachof the remaining binary combinations, place a 1 in the output column.
5 KARNAUGH MAP MINIMIZATIONA Karnaugh map provides a systematic method for simplifying Booleanexpressions and, if properly used, will produce the simplest SOP or POSexpression possible, known as the minimum expression. As you have seen,the effectiveness of algebraic simplification depends on your familiarity withall the laws, rules, and theorems of Boolean algebra and on your ability toapply them. The Karnaugh map, on the other hand, provides a "cookbook"method for simplification.A Karnaugh map is similar to a truth table because it presents all of thepossible values of input variables and the resulting output for each value.Instead of being organized into columns and rows like a truth table, theKarnaugh map is an array of cells in which each cell represents a binaryvalue of the input variables. The cells are arranged in a way so thatsimplification of a given expression is simply a matter of properly groupingthe cells. Karnaugh maps can be used for expressions with two, three, four.and five variables. Another method, called the Quine-McClusky method canbe used for higher numbers of variables.The number of cells in a Karnaugh map is equal to the total number ofpossible input variable combinations as is the number of rows in a truthtable. For three variables, the number of cells is 23 8. For four variables,the number of cells is 24 16.The 3-Variable Karnaugh MapThe 3-variable Karnaugh map is an array of eight cells. as shown in Fig.(51)(a). In this case, A, B, and C are used for the variables although otherletters could be used. Binary values of A and B are along the left side (notice
the sequence) and the values of C are across the top. The value of a givencell is the binary values of A and B at the left in the same row combinedwith the value of C at the top in the same column. For example, the cell inthe upper left corner has a binary value of 000 and the cell in the lower rightcorner has a binary value of 101. Fig.(5-1)( b) shows the standard productterms that are represented by each cell in the Karnaugh map.(a)(b)Fig.(5-1) A 3-variable Karnaugh map showing product terms.The 4-Variable Karnaugh MapThe 4-variable Karnaugh map is an array of sixteen cells, as shown inFig.(5-2)(a). Binary values of A and B are along the left side and the valuesof C and D are across the top. The value of a given cell is the binary valuesof A and B at the left in the same row combined with the binary values of Cand D at the top in the same column. For example, the cell in the upper rightcorner has a binary value of 0010 and the cell in the lower right corner has a
binary value of 1010. Fig.(5-2)(b) shows the standard product terms that arerepresented by each cell in the 4-variable Karnaugh map.(a)(b)Fig.(5-2) A 4-variable Karnaugh map.Cell AdjacencyThe cells in a Karnaugh map are arranged so that there is only a singlevariable change between adjacent cells. Adjacency is defined by a singlevariable change. In the 3-variable map the 010 cell is adjacent to the 000cell, the 011 cell, and the 110 cell. The 010 cell is not adjacent to the 001cell, the 111 cell, the 100 cell, or the 101 cell.
Fig.(5-3) Adjacent cells on a Karnaugh map are those that differ by only onevariable. Arrows point between adjacent cells.KARNAUGH MAP SOP MINIMIZATIONFor an SOP expression in standard form, a 1 is placed on the Karnaugh mapfor each product term in the expression. Each 1 is placed in a cellcorresponding to the value of a product term. For example, for the productterm ABC, a 1 goes in the 10l cell on a 3-variable map.ExampleMap the following standard SOP expression on a Karnaugh map:see Fig.(5-4).ExampleMap the following standard SOP expression on a Karnaugh map:See Fig.(5-5).
Fig.(5-4)Fig.(5-5)ExampleMap the following SOP expression on a Karnaugh map:SolutionThe SOP expression is obviously not in standard form because each productterm does not have three variables. The first term is missing two variables,
the second term is missing one variable, and the third term is standard. Firstexpand the terms numerically as follows:ExampleMap the following SOP expression on a Karnaugh map:SolutionThe SOP expression is obviously
Rules of Boolean Algebra Table 4-1 lists 12 basic rules that are useful in manipulating and simplifying Boolean expressions. Rules 1 through 9 will be viewed in terms of their application to logic gates. Rules 10 through 12 will be derived in terms of the simpler rules and the laws previously discussed. Table 4-1 Basic rules of Boolean algebra.File Size: 2MB
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