8. BOOLEAN ALGEBRAS x x

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1 8. BOOLEAN ALGEBRAS 8.1. Definition of a Boolean Algebra There are man sstems of interest to computing scientists that have a common underling structure. It makes sense to describe such a mathematical structure abstractl and then theorems proved about it will be directl applicable in all these eamples. The structure is that of a Boolean Algebra. It is a set, with two binar operations + and, a unar operation and two distinct constants 0, 1 such that certain aioms hold. As in ordinar algebra, we write a b as ab. Aioms for a Boolean Algebra A: (1) (Closure under Addition) For all a A, a + b A. (2) (Associativit under Addition) For all a, b, c A, a + (b + c) = (a + b) + c. (3) (Commutativit under Addition) For all a, b A, a + b = b + a. (4) (Identit under Addition) For all a A, a + 0 = a. (5) (Complement under Addition) For all a A, a + a = 1. (6) (Closure under Multiplication) For all a A, ab A. (7) (Associativit under Multiplication) For all a, b, c A, a(bc) = (ab)c. (8) (Commutativit under Multiplication) For all a, b A, ab = ba. (9) (Identit under Multiplication) For all a A, a1 = a. (10) (Complement under Multiplication) For all a A, a a = 0. (11) (Distributivit of Multiplication over Addition) For all a, b, c A, a(b + c) = ab + ac. (12) (Distributivit of Addition over Multiplication) For all a, b, c A, a + bc = (a +b)(a + c). You will notice that most of these aioms hold in the ordinar algebra of real numbers. The eceptions are the Complement Laws and Distributivit of Addition over Multiplication Eamples of Boolean Algebras Logic: This eample is the smallest Boolean Algebra that can eist. Let B 1 = {0, 1} and define addition, multiplication and complement as follows This elements in this algebra perform like the integers 0, 1 in ordinar arithmetic with the eception that = 1, instead of 2. In fact these elements behave like the tags TRUE and FALSE, where 0 represents TRUE and 0 represents FALSE. Addition is the or operator, normall written as and multiplication is the and operator, normall written as, and p represents not p. You should check that the 11 aioms hold for this sstem. Sets: Let S be a set and let B = (S) be the power set of S, that is, the set of all subsets of S. The constants 0 and 1 represent the empt set and the whole set, S itself. Addition represents the union of two sets, normall written as and multiplication represents intersection, normall written as. If S is a finite set with n elements then B is a Boolean Algebra with 2 n elements. 101

2 Boolean Vectors: There are 2 n vectors ( 1, 2,, n ) where each i = 0 or 1. Let B n be the set of all these vectors and define the operations of addition, multiplication and complement pointwise. This means we define these operations on the functions in terms of the corresponding operations in B 1 :,..., +,,..., = +, +,..., + (, 2 n ) ( 1 2 n ) ( n n ) (,..., )(.,,..., ) (,,..., ) 1 ; 1, 2 n 1 2 n = n n ; (,..., ) (,,..., ) 1, 2 n = 1 2 n 8.3. Properties of Boolean Algebras Theorem 1: If + = 1 and = 0 then =. Proof: Suppose + = 1 and = 0. Then = 1 b Aiom 9 = ( + ) b Aiom 5 = + b Aiom 11 = + b Aiom 8 = 0 + b assumption = + 0 b Aiom 8 = b Aiom 4 = b Aiom 10 = + 0 b Aiom 4 = + b Aiom 10 = + b Aiom 8 = ( + ) b Aiom 11 = ( + ) b Aiom 3 = 1 b assumption = b Aiom 9 Corollar 1: For all, = Proof: The equations + = 1, = 0 not onl sa that is the complement of, but the also sa that is the complement of. Corollar 2: 0 = 1 and 1 = 0. Proof: = 1 b Aiom 4 and 1 0 = 0 b Aiom 9. So 0 and 1 are complements of each other b Theorem 1. Theorem 2: For all, 0 = 0. Proof: 0 = b Aiom 4 = 0 + b Aiom 10 = (0 + ) b Aiom

3 = b Aioms 3, 4 = 0 b Aiom 10. Theorem 3: For all, + 1 = 1. Proof: + 1 = ( + 1) 1 b Aiom 9 = ( + 1)( + ) b Aiom 5 = + 1 b Aiom 12 = + b Aiom 9 = 1 b Aiom 5. Theorem 4 (DE MORGAN): For all, : + = Proof: We shall show that ( + ) = 0 and + ( + ) = 1. Then, b Theorem 1, it follows that is the complement of +. ( + ) = ( ) + ( ) b Aiom 11 = + b Aioms 2 and 3 = b Aiom 10 = b Theorem 2 = 0 b Aiom 4. + ( + ) = ( + ) + b Aioms 2, 3 = (( + ) + )(( + ) + ) b Aiom 12 = [ + ( + )] [ + ( + )] b Aioms 2, 3 = ( + 1)( + 1) b Aiom 5 = 1.1 b Theorem 3 = 1 b Aiom 9. Hence is the complement of +. Corollar: For all, we have = +. Proof: Substituting for and for in the theorem we get + = = b Theorem 1 (Corillar 1). Hence and + are complements of one another, that is, = Smmetric Difference We define a third binar operation in terms of the other two. We define a b = a b + a b. Theorem 5: For all a, b: a (b c) = (a b) c a 0 = a and a a = 0. a b c = a b c + a b Proof: ( ) ( ) ( c) = ( ) a( bc bc) a bc + bc + + = a ( bc)( bc) + a( bc + bc) = a ( b + c)( b + c) + abc + abc 103

4 = a ( b + c)( b + c) + abc + abc = a bb + abc + acb + acc + abc + abc = a bc + abc + abc + abc. ( a b) c = ( ab + ab) c + ( ab + ab)c = a bc + abc + ab ab c = a bc + abc + ( a + b)( a + b)c = a bc + abc + ( a + b)( a + b) c = a bc + abc + aa c + abc + bac + bbc = a bc + abc + abc + bac = a bc + abc + abc + abc a 0 = a0 + a0 = 0 + a1 = a. a a = aa + aa = = 0. These facts show that a Boolean Algebra is a group under this new operation. A group is an algebraic structure G with a binar operation that satisfies the properties: (1) For all a, b G, a b G. (2) For all a, b, c G, a (b c) = (a b) c. (3) There eists an element e G such that a e = a for all a G. (4) For all a G there eists an element b such that a b = e. The group that arises in this wa has two additional properties. Clearl a b = b a for all a, b B. This is the commutative law, and a group that satisfies this propert is called an Abelian group. Moreover a a = 0 for all a B. Using elementar group theor it follows that if B is finite, the number of elements of B is 2 n for some n, which is the size of B n. In fact, all finite Boolean algebras have essentiall the same structure as (in technical terms we sa are isomorphic to ) B n for some n Dualit No doubt ou noticed that the aioms for a Boolean Algebra came in pairs. If ou swap + and, and swap 0 and 1, in one of each pair ou will get the second. For eample Aiom 11 states that for all a, b B we have a(b + c) = ab + ac. Swapping + and this gives a + bc = (a + b)(a + c) which is Aiom 12. Aiom 5 states that for all a B, a + a = 1. Swapping +, and and swapping 0 and 1, this becomes a a = 0 which is Aiom

5 The dual of a Boolean epression is obtained b swapping + and and swapping 0 and 1. So Aioms 6 to 10 are the duals of Aioms 1 to 5 (and vice versa). Aioms 11 and 12 are duals of each other. An important consequence of dualit is the fact that an theorem in Boolean algebra remains a theorem if the epressions are replaced b their duals. Hence Boolean algebra theorems come in dual pairs. Principle of Dualit: The dual of an theorem in Boolean algebra is also a theorem. Proof: One can provide an independent proof of the dual b replacing ever epression in the proof of the original theorem b its dual. We have done this eplicitl in several places. However it is simpler, having proved a theorem, to state its dual and to sa b the Principle of Dualit Logic Gates We can use Boolean algebra to describe an electronic circuit. Each component in such a circuit is called a gate. Each gate implements one of the three basic logic operations. Inverter: This implements the complement operation. OR gate: This implements the operation of addition. + AND gate: This implements the operation of multiplication. These gates can be combined to produce output according to an Boolean epression. Eample 1: The following circuit implements the function z ( + z) z + + z z( + + z ) z z z The function ( + z) ( + z) = z ( z) z + = z ( + z) + can be rewritten as follows. = z + z z 105

6 = z since = and z z = 0. So a simpler circuit for this function is the following. z z Eample 2: Design a circuit that controls a light from two switches, so that changing the state of either of them changes the state of the light. Solution: Let the two states of each switch be 0, representing OFF, and 1 representing ON. The output will be similarl 0 or 1. The function to be implemented b the circuit will be F(, ). Suppose that F(1, 1) = 1, so that when both switches are ON the light is ON. If the first switch is switched OFF the light must go OFF so F(0, 1) = 0. If the second switch is now tripped the light must go ON again, so F(0, 0) = 1. Finall, F(1, 0) = 0. So the Boolean function we must implement is: Clearl F(, ) = +. A suitable circuit is the following. Eample 3: Design a circuit to add two integers modulo 2. Solution: The function, F, to be implemented is the following Clearl F(, ) = +. A suitable circuit is the following Addition Circuits Suppose we wish to design a circuit to add two bits, taking into account carring as in a base 2 calculation. First, consider how integers, represented in binar, are added. For eample, the calculation = 13 in binar becomes: 106

7 We begin b adding the bits in the right-hand column: = 1. There is no carr to the net column. Adding in the net column we get 10, the binar equivalent of 2. We put down 0 in the second column and carr the 1. In the 3 rd column from the right we must add , the third 1 being the 1 carried over from the previous column. This gives 3, which in binar is 11. So we put down one 1 and carr the other 1. The column to the left of this one has two blanks, effectivel two zeros. So we add these two zeros, plus the carried 1, giving 1 in the left-most column. So at each stage if the sum of the three inputs is 0 or 1 we write that down and the carr bit is 0. If the sum of the three inputs is 2 or 3 we write down 0 or 1 respectivel and the carr bit is set to 1. This circuit will have three inputs and two outputs. The third input and the second output is the carr bit. The functions that lie behind this circuit are as follows. c S(,, c) C(,, c) S(,, c) = c + c + c + c and C(,, c) = c + c + c + c. To implement this directl would require 3 inverters (one for each of, and c), 8 ADD gates (for the 8 products) and 6 OR gates for the sums. Let us see if we can simplif these functions first. If we write addition modulo 2 as then S(,, c) = c = ( ) c. So we can achieve this with two copies of the circuit in eample 3. c Now C(,, c) = c + c + c + c = c + c + (c + c) = c + c +. If two terms differ in just one factor we can alwas combine them in this wa. But now comes the reall trick part. C(,, c) = c + c + = c + c + + c. Wh? Because + c = (1 + c) = 1 =. But wh would we want to add an etra term? Watch closel! 107

8 C(,, c) = c + c +, combining the first and last terms. We are back to 3 terms, but we have two terms with 2 factors instead of onl one, not to mention the fact that this will need one less inverter. Now we do it again. C(,, c) = c + c + c + = c + c +, an epression that is considerabl shorter than the one we started with. A suitable circuit to implement this is: c c c + c + c + c Each modulo 2 circuit has 2 inverters, 2 AND gates and 1 OR gate, so with two of them we have 4 inverters, 4 AND gates and 2 OR gates. Incorporating the carr circuit above we have 4 inverters, 7 AND gates and 4 OR gates, compared with 3 inverters, 8 AND gates and 6 OR gates if we had implemented it without simplification. This illustrates the usefulness of simplifing Boolean epressions before implementation. Suppose now we wish to design a circuit in which numbers from 0 to 15 can be added. Such numbers can be represented b a 4 bit binar string of the form b 3 b 2 b 1 b 0, ranging from 0000 to Let us represent the above circuit as follows. Here, are the bits being added, z is the bit that is output and c is the carr bit. We use 4 copies of this circuit. 0 a 0 b 0 a 1 b 1 a 2 b 2 a 3 b Disjunctive Normal Form A Boolean epression can be etremel complicated, with man laers of compleit. Consider the following epression: + ( + z) + zz + We can simplif such an epression using three rules. (1) a + b = ab ; (2) ab = a + b ; (3) a ( b + c) = ab + ac. c + + z c + + c 0 c 1 c 3 z c 2 108

9 The first two of these are the De Morgan Laws and the third is the Distributive Law. Eample 4: Simplif the epression ( + z) + zz + Solution: + ( + z) + zz + = ( + z) zz + = ( + z) ( z + + ( z + ) = ( + z)( z + + z + ) = ( + z)( z + + z) +. = z + + z + zz + z + z z = z + z. A literal in a Boolean epression is one of the variables or its complement. So the literals for a binar epression B( 1, 2,, n ) are 1, 2,, n, 1, 2,, n, A disjunctive normal form in the variables 1, 2,, n is an epression as a sum of products where each term is a product of literals and where each variable occurs eactl once, either as itself or its complement. Each term gives eactl one combination of the variables under which the epression evaluates to 1. Eample 5: + is not in disjunctive normal form since the term contains twice. z is not in disjunctive normal form since it contains both and. z + is not in disjunctive normal form since, z do not occur as factors of the term. B(,, z) = z + z + z is in disjunctive normal form. B(,, z) = 1 for eactl the following three instances: = 1, = 0, z = 1; = 0, = 0, z = 1; = 1, = 1, z = 0. An Boolean epression can be epressed in disjunctive normal form. Since Boolean values commute under multiplication we insist that the variables occur in order in each term. If we also insist on the terms being arranged in alphabetic order we get a unique disjunctive normal form for each Boolean function. We do not alwas do this, but in counting disjunctive normal form we count epressions where the terms are merel rearranged as the same epression. Theorem 6: Ever Boolean function can be epressed uniquel in disjunctive normal form. With n variables there are 2n literals and 2 n disjunctive normal forms. Each variable can either occur as itself or as its complement. We can use the disjunctive normal form as the basis for the construction of a logic circuit, but it is generall not the simplest possible circuit. There are several criteria one might use to define simplest. We ma mean a circuit involving the fewest possible number of gates. This will potentiall decrease the cost of manufacture. However there is a more important consideration. The speed of a circuit is dependent on its depth. Man operations ma occur in parallel, but the time taken for the whole operation depends on the largest number of steps that a given variable must pass. 109

10 8.9. Representing of Boolean Epressions We can epress a disjunctive normal form in binar b coding variables as 1 and their complements as 0. If we arrange the variables in each term in a fied order we can determine which variables are present as themselves, and which as their complements, b the positions of the 0 s and 1 s. Eample 6: The disjunctive normal form z + z + z can be coded as If there are three variables, as in this case, we can represent the possible terms b the vertices of a cube So a disjunctive normal form is represented b a set of these vertices. Eample 6 (continued): B(,, z) = z + z + z is coded as and so corresponds to the set of vertices coloured black in the following diagram Of course, if there are more than four variables we cannot use a geometric representation, at least not one we can draw, or even visualise. But mathematicians have no difficult in discussing n-dimensional space, and even proving theorems about it, even though the are no more able to picture what is going on. The do this b working algebraicall. After all, if points are described b their coordinates, there is no difficult in considering the point (1, 0, 0, 1, 0) in 5-dimensional space. So we can sa that a disjunctive normal form in n variables is represented b a set of points on a hpercube, that is an n-dimensional version of a cube. Just do not epect to see it in a diagram! Now a disjunctive normal form can often be simplified, with the simpler version still being a sum of products. It is just that some variables ma be missing from sum terms. Eample 7: z + z + z = ( + )z + z = z + z. We code a sum of products b using the smbol 2 to represent a variable that is absent. It acts as a wildcard. 110

11 0 1 variable absent 2 Eample 7 (continued): z + z is represented b In fact, z + z = ( + )z = ( + )( + ) (b Aiom 12) = +. This is an even simpler form. It would be represented as Terms with wildcards are not represented geometricall as points, but rather as lines, or planes etc. Again, for illustrative purposes, we stick to three variables. The term 121 is equivalent to These are two adjacent points on the cube but we represent it as the line joining them rather than as the pair of points. Remember that onl points on the cube have an significance so there is no point in considering an of the other points on this line. The term 212 is represented b the relevant plane (with four points 010, 011, 110, 111). Eample 7 (continued): The Boolean epression z + z + z is simplified to + which is coded as Being of onl three variables we can picture it as follows Don t confuse the Boolean epression 1 with the code 1. Since all variables are absent in 1 we write it as 222. However the Boolean epression 0 is difficult to represent in this wa. Strictl speaking it is a sum of no terms. If we ever have occasion to represent it, and we are ver unlikel to do so, we would give it the special smbol of. There are 27 strings of length 3 on the alphabet {0, 1, 2}. These correspond to the 8 vertices plus the 12 edges, plus the 6 faces plus the whole cube, making 27 geometric objects in all Simplifing Boolean Epressions Suppose we have a Boolean epression in disjunctive normal form. It can be simplified b use of the following operations. (1) If two terms A, B are such that wherever the differ A has a 2 then B ma be deleted. This because of a repeated application of the identit + =. 111

12 Eample 8: = (2) If two terms A, B differ in eactl one position, and neither smbol is 2, the ma be combined b putting a 2 in that position. This is because of the identit + = 1. Eample 9: = (3) If two terms A, B differ in eactl two positions and the smbols in these positions in A are 2, a 2 and in B are b 1, b 2 with b 2 < 2 then B ma changed b changing b 2 to 2. This is because of the Boolean identit + = +. Note that the positions in which these differences occur need not be consecutive. Nor need the occur in the order of the variables. That is the a 1 and b 1 ma occur for a later variable then the a 2 and b 2. Eample 10: = Here A = , B = , a 1 = 2, b 1 = 0, a 2 = 1, b 2 = 0. (4) If three terms A, B, C coincide ecept for three positions and the smbols in these positions are: A: 2, b, c B: a, 2, c C: a, b, 2 where a, b and c are all less than 2 then B and C ma be combined b replacing b and c b 2. Eample 11: Simplif B(u, v, w,,, z) = u v z + u v w z + u v w + u v w. Solution: Coding this epression we get = b rule 4 = b rule 2. So B(u, v, w,,, z) = u v z. Eample 12: Simplif B(,, z) = z + z + z + z + z. Solution: We code it as = b rule 2 applied twice = b rule 2 again Thus B(,, z) = z + z. Eample 13: Simplif B(a, 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. Solution: In code it becomes = b rule 2 = = = = = So B(a, b, c, d) = a c d + b c d + a d. These rules will simplif man Boolean epressions, but ma not produce the shortest sum of products. There are algorithms which guarantee that the final answer will be the shortest possible for a sum of products. The method of Karnaugh maps uses a version of the geometric representation while the Kline-McClusk method uses smbol manipulation along the lines of the above discussion. We don t discuss these here. 112