7.3 FIRST INVERSE ANTECEDENT BAF(BAF) OPERATION

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1 7.3 FIRST INVERSE ANTECEDENT BAFBAF OPERATION The inverse of the problem of the Direct BAFBAF Operation of the precedent Sections, could be also considered as a problem of Boolean Equations. Thus, if we consider the Direct BAFBAF Operation in the classical form, A q B p = C p x r x r x q, where the second member represents the result, it is possible to define two Inverse Operations, because the corresponding Direct BAFBAF Operation, as a Logical Implication, does not have the mathematical commutative property of the given terms. Then, in this First Inverse Antecedent BAFBAF Operation, the given data are the corresponding Result C6 rx System, A 6 px q q and the Consequent System B6 rx p. The Antecedent, is the Incognito System of this Inverse Operation. 05

2 In terms of Boolean Geometry language, we have Fig. 9: LJ In terms of Boolean Algebra language, we have the following literal expressions: Given the following two systems, C6 rx q C Z = F X, X,..., X q Z = F X, X,..., X q... Z r = F r X, X,..., X q of Boolean Functions, A logical equalities: and B and B6 rx p, in a given order, Z = G,,..., p Z = G,,..., p, determine a third System... Z r = G r,,..., p = f X, X,..., X q = f X, X,..., X q, in such way as to have the following... p = f p X, X,..., X q G f X, X,..., Xq, f X, X,..., Xq,..., fp X, X,..., Xq = F X, X,..., Xq G f X, X,..., Xq, f X, X,..., Xq,..., fp X, X,..., Xq = F X, X,..., Xq... Gr f3x, X,..., Xq 8, f3x, X,..., Xq8,..., fp3x, X,..., Xq8 = Fr 3X, X,..., Xq In order to solve this First Inverse Antecedent BAFBAF Operation, it will be determined previously the Inverse Analytical System of the given Consequent System 6 B rx p, that is, we determine system B 4 9. Thus, we reduce this problem to the q px solution of the following equivalent Direct BAFBAF Operation: 06

3 6 6 C rx q B px r = A px q To solve the operation given by 6, we can apply the practical gadget of the TABLES of the correspondent Discreet Function of Discreet Functions of the ordinary Algebra, using the given numerical compact representation. 6 th EXAMPLE: Given the systems of BAFs: = B = NT Z [ 03 3].{ 3 ; 3} x Z and C = NT Z = [ 0 3].{ ; XX }, determine the following system of BAF, x Z 646 A 3 = NT [ 3 a a a 0 ] 3.{ ; } x a X X =, so that we have: A 3 B 3 = C x x x. NOTE: This example corresponds to the First Inverse Antecedent BAFBAF Operation referring to the st EXAMPLE considered in Section 7.. In terms of Boolean Geometry language, we have Fig. 0: LJ It is opportune to remember that the Systems of BAFs, in the compact and vertical notation, could be also the numerical representative of the TABLES of the practical gadget. This practical gadget corresponds to the Discreet Function of Discreet Function of ordinary Algebra. 07

4 Then, it will be: 66 yx zy = z[yx] = zx, where: 67 yx = NT - { A 3x }, is the antecedent function; 68 zy = NT - { B x 3 }, is the consequent function and 69 z[yx] = zx = NT - { C x }, it is the result of the operation of the indicated functions. Then, in the definition of the First Inverse Antecedent BAFBAF Operation, the anti-numerical Transform of the Inverse Analytical System of the given Consequent System, will correspond to the inverse analytical of the respective consequent function, that is: 70 NT - { [ B x 3 ] - } = NT - { B 3x } = yz Therefore, the anti-numerical Transform of expression 6 will be: = 7 NT C rx q NT B px r NT A px q, where we have, in reference to the Discreet Functions of the Classical Mathematics, the following: NT C rx q = zx; NT B px r = yz and NT A px q = yx, that is, zx yz = yx In this example, expression 7 becomes: = 73 NT C NT B NT A x 3x 3 x Therefore, firstly we must determine the Inverse Analytical System of the given Consequent System through the following TABLE: 08

5 zy yz B x 3 B 3x b k b k SOLUTIONS: 7 3 4;0 sols. Nr. of solutions. 6 6;3; 3 sols. =x3xx= 5 0 7; sols. full solutions sol Therefore, we have: 74 B B NT ;. ; ZZ x x 3! $ = = = 6;3; 7; 5 x3xx = full solutions = B Then, we apply the practical gadget of TABLES of the correspondent Discreet Function of Discreet Function of the ordinary Algebra, we have: 75 C B A x 3x = 3 x NOTE: It is important to observe that it is an Inverse Operation, where the respective Direct Operation does not have the mathematical commutative property of the given terms. Therefore, this Inverse Operation is not UNIQUE. Then, there are several solutions that will be the same, which was considered in the st EXAMPLE of the First Inverse Antecedent BAFBAF Operation of Section 7.. Therefore, we have: GIVEN DATA RESULTS C x B 3 x A 3 x c k b k a k ; ;3; 6;3; 7; 7; ;0 09

6 This problem offers a result of x3xx = solutions, that is, the numerical vertical compact representation of the systems of BAF are the following, 76 A 3 x = NT 5 =! X X 3. ; 6 < A = form, 77 A63 x = NT = X XB ; 3! $ $ 6 6 3, or, as binary Boolean Matrix x3xx = full solutions In that global numerical expression of all those full solutions i.e. without any time-restriction, the abscissa component marked with corresponds to the solution that is the same, as it was considered in the st EXAMPLE of the First Inverse Antecedent BAFBAF Operation of the Section 7., that is, the numerical expression 4: A NT [ 6 4].{ ; XX } 3 x = 5 3 = 3 7 th EXAMPLE: Given the systems of BAFs: Z Z 78 B NT Z [ ].{ ; } 5 x 3 = =, and Z4 Z 5 0

7 Z Z 79 C NT Z [ ].{ ; XX } 5 x = 3 5 = 5 5 9, determine the following system of Z4 Z 5 BAF, 80 A NT [ a a a a 0 ].{ ; X X } 3 x = 3 3 =, so that we have: 3 8 A B C 3 x 5 x 3 = 5 x NOTE: This example corresponds to the Second Inverse Antecedent BAFBAF Operation referring to the nd EXAMPLE considered in Section 7.. In terms of Boolean Geometry language, we have the following Fig. : LJ It is opportune to remember that the Systems of BAFs, in the compact and vertical notation, could be also the numerical representative of the TABLES of the practical gadget. This practical gadget corresponds to the Discreet Function of Discreet Function of ordinary Algebra. Then, it will be: 8 yx zy = z[yx] = zx, where: 83 yx = NT - { A 3x }, it is the antecedent function; 84 zy = NT - { B x 3 }, it is the consequent function and 85 z[yx] = zx = NT - { C x }, it is the result of the operation of the indicated functions.

8 Then, in the definition of the First Inverse Antecedent BAFBAF Operation, the anti-numerical Transform of the Inverse Analytical System of the given Consequent System, will correspond to the inverse analytical of the respective consequent function, that is: 86 NT - { [ B x 3 ] - } = NT - { B 3x } = yz Therefore as we have seen, the anti-numerical Transform of expression 6 will be: = NT C rx q NT B px r NT A px q, where we have, in reference to the Discreet Functions of the Classical Mathematics, the following: NT C rx q = zx; NT B px r = yz and NT A px q = yx, that is, zx yz = yx In this example, that expression becomes: = NT C NT B NT A x 3x 3 x Therefore, firstly we must determine the Inverse Analytical System of the given Consequent System, through the following TABLE:

9 zy yz B 5x 3 B 3x 5 Col. k b k b k SOLUTIONS: {i 5 } {i 5 } one solution {i 5 } {i 5 } 5 6 {i 5 } 5 one solution 0 4 {i 5 } 3 {i 5 } {i 5 } {i 5 } 0 {i 5 } 9 {i 5 } 8 5 one solution 7 {i 5 } 6 {i 5 } 5 6;7 two solutions 4 {i 5 } 3 {i 5 } {i 5 } ;0 two solutions 0 {i 5 } 09 {i 5 } 08 {i 5 } 07 {i 5 } 06 {i 5 } 05 {i 5 } 04 {i 5 } 03 {i 5 } 0 {i 5 } 0 4 one solution 00 {i 5 } NOTE: The symbol {i 5 } means that, either we have a time-restriction at the indicated epoch or we have what corresponds to the Complementary System 0 5 x 5 referred as their own variables. 3

10 Z Z that is: {i 5 } = = ; 0 = 5 5 NT Z ZZ ZZ x 3 4Z5 Z4 Z 5 Therefore, we have: = B B = NT = i i i i i i i. ; Z Z Z Z Z 3 x ;7 53 ; ! $ 3 3 x = 4 restricted solutions 4 = B Then, we apply the practical gadget of TABLES of the correspondent Discreet Function of Discreet Function of the ordinary Algebra, we have: 87 C B A 5x 3x 5 = 3 x NOTE: It is important to observe that it is an Inverse Operation, where the respective Direct Operation does not have the mathematical commutative property of the given terms. Therefore, this Inverse Operation is not UNIQUE. Then, there are several solutions and one of them will be the same which as the one considered in the nd EXAMPLE of the First Inverse Antecedent BAFBAF Operation of Section 7.. Therefore, we have: 4

11 GIVEN DATA Result C 5 x B 3 x 5 A 3 x c k b k a k 3 3 to 30 {i 5 } 3 0; ;7 5 8 to 6 {i 5 } 6; to 9 {i 5 } to 6 {i 5 } 5 6;7 4 to {i 5 } ;0 0 to {i 5 } 4 0 {i 5 } This problem offers a result of xxx = 8 solutions, that is, the numerical vertical compact representation of the systems of BAF is the following, A 3 x = NT = 3! 3 6 = B $ 00 form, 89 A NT 3 x = = X X 3. ; = ; XX B 0 0! $ xxx = 8 full solutions, or, as binary Boolean Matrix In that global numerical expression of all those 8 full solutions i.e. without any time-restriction, the abscissa component marked with corresponds to the solution that is the same, as it was considered in the nd EXAMPLE of the First Inverse Antecedent BAFBAF Operation of the Section 7., that is, the numerical expression: A NT [ 6 7 3].{ ; XX } 3 x = 3 = 3 5

12 8 th EXAMPLE: Given the systems of BAFs: Z 90 B NT Z x 5 = = Z 3 and 3.{ 5 ; XXX3}, 9 C = 3 x 3 NT Z Z = Z 3 XX X 5. = 3; 3 B, determine the following system of BAF, 9 A NT [ a a... a ].{ ; X X X } 3 x = = 3, so that we have: 93 C 3 x 3 = A x 3 B 3 x 5 NOTE: This example corresponds to the First Inverse Antecedent BAFBAF Operation referring to the 3 rd EXAMPLE considered in Section 7.. In terms of Boolean Geometry language, we have the following Fig. : LJ It is opportune to remember that the Systems of BAFs, in the compact and vertical notation, could be also the numerical representative of the TABLES of the practical gadget. This practical gadget corresponds to the Discreet Function of Discreet Function of ordinary Algebra. Then, it will be: 94 yx zy = z[yx] = zx, where: 6

13 95 yx = NT - { A x 3 }, is the antecedent function; 96 zy = NT - { B 3 x 5 }, is the consequent function and 97 z[yx] = zx = NT - { C 3 x 3 }, is the result of the operation of the indicated functions. Then, in the definition of the First Inverse Antecedent BAFBAF Operation, the anti-numerical Transform of the Inverse Analytical System of the given Consequent System will correspond to the inverse analytical of the respective consequent function, that is: 98 NT - { [ B 3 x 5 ]- } = NT - { B 5 x 3 } = yz Therefore as we have seen, the anti-numerical Transform of expression 6 will be: = NT C rx q NT B px r NT A px q, where we have, referring to the Discreet Functions of the Classical Mathematics, the following: NT C rx q = zx; NT B px r = yz and NT A px q = yx, that is, zx yz = yx In this example, that expression becomes: NT C NT B = NT A 3 x 3 5 x 3 5 x 3, Therefore, firstly we must determine the Inverse Analytical System of the given Consequent System, through the following TABLE: 7

14 zy yz B 3 x 5 B 5 x 3 Col. k b k b k Therefore, it will be: 8

15 B 5x = NT X X X3 = [ ]. 3; ZZZ 3 5 Then, we have: GIVEN DATA C 3 x 3 B 5x 3 A c k b k RESULT 5 x 3 a k This problem offers a result of 4 8 = 6 = 65,536 full solutions, that is, the numerical vertical compact representation of the systems of BAF are the following, A [ 5 3 = NT X = x X X ]. 3; XX X3 5 < A also with 4 8 = full solutions But, this numerical expression 99 shows that the final solution is conditioned by the existence of the following Complementary System relative to the input variables {X X X 3 } of the Antecedent System: 9

16 X 00 NT X = [ ] 3.{3; X X X 3 } X 3 0 NT X X X3 =! XX X $.{ ; } Then, to verify this condition within the 4 8 = full solutions above, expression 99 can be written in the following classical binary Boolean representation given by expression 0: A = NT X =. 3; XX X x 3 X X ! $ 06 < A following: 036 The full solutions that satisfy those conditions are marked by and are the A 5 3 = NT ; x X XX X3 = X X ! $ = [ ] 5.{3;X X X 3 } or, x = 4 full solutions to the example = B In the numerical expression of these four solutions, the set of BAVs, {X, X, X 3 }, appears in both sides of the numerical expression give by 03 and then, it can be deleted. Therefore, we have: = 0

17 A NT = = = 3; XX X 3 B x! $ 05 A NT. ; XX X x = = 3 3! $ = B NOTE: The solution marked by in the above expression, that is, the following numerical expression, 05A A 3 NT x 3 = =. ; XX X 3! $ 6 = B the same numerical expression, as the data given in the 3 rd EXAMPLE., or, corresponds to 9 th EXAMPLE: Given the systems of BAFs: 066 B x 3 = NT Z { 3 ; X3X4}, Z = and 07 C = x 4 NT following system of BAF, Z = Z 08 A NT [ a a a a ].{ ; X X } x = = 3 0, so that we have: 09 C x 4 = A 3 x 4 B x 3. 4 = ; XX XX 3 4B, determine the NOTE: This example corresponds to the First Inverse Antecedent BAFBAF Operation relative to the 4 th EXAMPLE considered in Section 7.. In terms of Boolean Geometry language, we have the following Fig. 3:

18 LJ It is opportune to remember that the Systems of BAFs, in the compact and vertical notation, could be also the numerical representative of the TABLES of the practical gadget. This practical gadget corresponds to the Discreet Function of Discreet Function of the ordinary Algebra. Then, it will be: 0 yx zy = z[yx] = zx, where: yx = NT - { A }, is the antecedent function; x zy = NT - { B }, is the consequent function and x 3 3 z[yx] = zx = NT - { C }, is the result of the operation of the indicated x 4 functions. Then, in the definition of the First Inverse Antecedent BAFBAF Operation, the anti-numerical Transform of the Inverse Analytical System of the given Consequent System will correspond to the inverse analytical of the respective consequent function, that is: 4 NT - { [ B x 3 ]- } = NT - { B 3x } = yz Therefore as we have seen, the anti-numerical Transform of expression 6 will be: = NT C rx q NT B px r NT A px q, where we have, referring to the Discreet Functions of the Classical Mathematics, the following:

19 NT C rx q = zx; NT B px r = yz and NT A px q = yx, that is, zx yz = yx In this example, that expression becomes: 6 NT C NT B = NT A x 4 3 x 3 x 4, Therefore, firstly we must determine the Inverse Analytical System of the given Consequent System, that is: zy yz B x 3 B 3x col. b k b k Therefore, it will be: 7 B = NT X =. ; ZZ 3x = B X! $ 4 3 Then, we have: 3

20 C x 4 GIVEN DATA c k RESULTS B 3x A 3x This problem offers a result of 57,464 full solutions, that is, the numerical vertical compact representation of the systems of BAF is the following, b k a k [ A = NT X = x X ]. ; 3 4 XX XX 3 4 < A number of full solutions: 3 x x x x 3 x 3 x x x 3 x 3 x x x 3 x x x = = 57,464 But, this numerical expression 8 shows that the final solution is conditioned by the existence of the following Complementary System relative to the input variables VAB s {X 3, X 4 } of the Antecedent System, is projected in the same BAFi: x = NT X ; XX XX 3 4@ X =! $. ; or, in compact vertical numerical expression: 4

21 0 0 x or, 6 4 = NT = x X 3 X4 6 4 NT 30 X 3 X4 = =! $. 4 ; XX XX 3 4, 5. 4 ; XXXX 3 4, Then, to verify this condition within the 57,464 full solutions above, expression 8 can be written in the following classical binary Boolean representation given by expression : A 3 4 = NT X x 3 X =! ;XXX3X $ 3 Therefore, these marked full solutions are the following: A 3 4 = NT X x 3 X4 =! 0 = B $ with 8 = 56, or, in vertical compact form: 4 =! A 3 4 = NT X x 3 X = 4;XXX3X4B $ 4 0 = B. 4; XX 3 XX

22 also, with 8 = 56 full solutions In the numerical expression of these solutions, the set of BAVs, {X 3, X 4 }, appears in both sides of the numerical expression 3 and then, it can be deleted. Therefore, we have: 5 = =! A NT x $ = B. 4; XX XX From these = 56 above solutions represented, only the one is marked by is independent from BAVs {X 3, X 4 }, if we apply the properties of permutation from the ordinality ω 4 = { XX XX 3 4} to the ordinality ω 4 = { X 3 X 4 X X } of the new Boolean Arithmetical Field BAFi. In fact, we have: A NT. x 4 = = ! $. < 4; X4XXX3A= ! $. < 4; X3X4XXA Therefore, we have: { 3 4 } 4 [ ] { 3 4 } {} A = [ ]. 4; X X X X = 00. 4; X X X X = TN x Simplifying this last numerical expression, we have: 6 ;@ 6 A = [ 00] X X NT x. ; = unique solution It means that this unique solution obtained above 6 corresponds to the same numerical expression, as the data given in the 4 th EXAMPLE. 0 th EXAMPLE: The systems of BAFs are given: 6

23 7 Z B NT Z. 3 x 5 = Z 3 = {; 5 XXX3X4}, and 8 C = 3 x 4 NT the following system of BAF, ;@ Z Z = = 4 ; XX XX 3 4B, determine Z 3 9 A NT [ a a a a ].{ ; X X } x = = 3 0, so that we have: 30 C 3 x 4 = A 5 x 4 B 3 x 5 NOTE: This example corresponds to the First Inverse Antecedent BAFBAF Operation relative to the 5 th EXAMPLE considered in Section 7.. In terms of Boolean Geometry language, we have the following Fig. 4: LJ It is opportune to remember that the Systems of BAFs, in the compact and vertical notation, could be also the numerical representative of the TABLES of the practical gadget. This practical gadget corresponds to the Discreet Function of Discreet Function of ordinary Algebra. 7

24 Then, it will be: 3 yx zy = z[yx] = zx, where: 3 yx = NT - { A }, is the antecedent function; x 33 zy = NT - { B }, is the consequent function and 3x 5 34 z[yx] = zx = NT - { C 3x 4 }, is the result of the operation of the indicated functions. Then, in the definition of the First Inverse Antecedent BAFBAF Operation, the anti-numerical Transform of the Inverse Analytical System of the given Consequent System, will correspond to the inverse analytical of the respective consequent function, that is: 35 NT - { [ B 3x 5 ]- } = NT - { B 5x 3 } = yz Therefore as we have seen, the anti-numerical Transform of expression 6 will be: = 36 NT C rx q NT B px r NT A px q, where we have, respectively to the Discreet Functions of the Classical Mathematics, the following: NT C rx q = zx; NT B px r = yz and NT A px q = yx, that is, zx yz = yx In this example, that expression becomes: 37 = 6 B 4 9 = NT C 4 3x NT B NT A 5x 5x 3 4 Therefore, firstly we must determine the Inverse Analytical System of the given Consequent System, that is: 8

25 zy yz B 3x 5 B 5x 3 b k b k Therefore, it will be: X B = NT X = 5 x ! X 3 X ; ZZ Z 5 3 < 3A $ 9

26 X B = TN X = x X 3 X { ; ZZZ} Then, we have: Col. k C 3x 4 c k GIVEN DATA RESULTS B 5 x 3 A 5 x b k a k This problem offers a result of 4 6 = full solutions, that is, the numerical vertical compact representation of the systems of BAF is the following, 30

27 X A = NT X = [ x X3 X ] 5.. 4; XX XX 3 4 But, this numerical expression 38 shows that the final solution is conditioned by the existence of the following Complementary System relative to the input variables VAB s { X, X, X 3, X 4 } of the Antecedent System, projected in the same BAFi: X X x 4 = NT ; 4 XX XX 3 4@ =. ; X X 4! $ or, in compact vertical form: X X x 4 = NT =. ; XXXX 3 4 X X 4 6 Inside these full solutions, there is only one marked by, which satisfies the condition of the existence of that Complementary System, relative to the input variables VAB s { X, X, X 3, X 4 } of the Antecedent System, that is: 4 A 5 x 4 = NT X X X3 X = ! ;XXX3X $ < A Then, the unique solution is: 3

28 5x A = { ; X X X X } In this numerical expression 4, by the properties given in Chapter 4, it is possible to cancel the Complementary System in both members of the given expression 39. Then, we have: 43 A6 x < XX XX A NT 4 = ; 3 4 = unique solution Therefore, applying the properties of permutation of the ordinality to ω = 4 { X X X X 3 4 } of the Boolean Arithmetical Field BAFi, we have: 44 6 < A A NT X X X X x 4 = = ; 4 3 = = ; X3X4XX = 006 < A! $. 4; X3X4XX < A Simplifying this last numerical expression, we have: 45 A = 00 { X X } = TN{} [ ]. ; unique solution x It means that this unique numerical solution obtained above 45, corresponds to the same numerical expression, as the data given in the 5 th EXAMPLE. = 3

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