233
7.4 SECOND INVERSE CONSEQUENT BAF(BAF) OPERATION
As we have seen in Section 7.3 of this Chapter, the inverse of the problem
of the Direct “BAF(BAF)” Operation of the precedent Sections could also be
considered as a problem of Boolean Equations.
Thus, if we consider the Direct “BAF(BAF)” Operation in the classical form,
q2 x r)C(p2 x r
(B) q2 x p)A( =→ , where the second member represents the
result, it is possible to define two “Inverse Operations”, because the corresponding
Direct “BAF(BAF)” Operation, as a Logical Implication, do not have the
mathematical commutative property of the given terms.
Then, in this Second Inverse Consequent BAF(BAF) Operation, the given
data are the corresponding Result C r x q1 6 2 and the Antecedent System A px q1 6 2 .
The Consequent System is the Incognito System of this Inverse Operation.
In terms of Boolean Geometry language, we have the following Fig. 15:
In terms of Boolean Algebra language, we have the following literal
expressions:
...... ...
234
Given the following two systems, C r x q1 6 2 and B r x1 6 2p , in the given order,
( )
( , ,..., )
( , ,..., )
...
( , ,..., )
( )
( , ,..., )
( , ,..., )
...
( , ,..., )
C
Z F X X Xq
Z F X X Xq
Zr Fr X X Xq
A
Y f X X Xq
Y f X X Xq
Yp f p X X Xq
and
1 1 1 2
2 2 1 2
1 2
1 1 1 2
2 2 1 2
1 2
=
=
=
%
&KK
'KK
=
=
=
%
&KK
'KK
, determine the
third System of Boolean Functions, ( )
( , ,..., )
( , ,..., )
...
( , ,..., )
B
Z G Y Y Yp
Z G Y Y Yp
Zr Gr Y Y Yp
1 1 1 2
2 2 1 2
1 2
=
=
=
%
&KK
'KK
, in such way as to
have the following logical equalities:
G f X X X f X X X f X X X F X X X
G f X X X f X X X f X X X F X X X
G f X X X f X X X
q q p q q
q q p q q
r q
1 1 1 2 2 1 2 1 2 1 1 2
2 1 1 2 2 1 2 1 2 2 1 2
1 1 2 2 1 2
, ,..., , , ,..., ,..., , ,..., , ,...,
, ,..., , , ,..., ,..., , ,..., , ,...,
...
, ,..., , , ,...,
3 8 3 8 3 8 3 8
3 8 3 8 3 8 3 8
3 8
=
=
q p q r qf X X X F X X X3 8 3 8 3 8,..., , ,..., , ,...,1 2 1 2=
%
&
KKK
'
KKK
In order to solve this Second Inverse Consequent BAF(BAF) Operation, it
is necessary to determine previously the Inverse Analytical System of the given
Antecedent System, ( )Apx q2
, that is, we determine the system Aq x
−14 9 2p
. Thus,
we reduce this problem to the solution of the following equivalent Direct
“BAF(BAF)” Operation:
1462 2 2
11 6
( ) ( ) ( )Aqx p C
r x q Brx p
− → =
To solve the operation given by (146), we can apply the practical gadget of
TABLES of the correspondent “Discreet Function of Discreet Functions” of the
ordinary Algebra, using the given numerical compact representation.
Then, it will be:
235
(147) y(x) → z(y) = z[y(x)] = z(x),
where:
(148) y(x) = NT A NT Apx qxq p
− − − −%&'
()*
= %&'()*
12
1 1 1
21 6 4 9
(149) z(y) = NT C rx q−1
21 6= B
(150) z(x) = NT B rx p−1
21 6= B
Then, expression (147) becomes the following, that is similar to expression
(146):
1512 2 2
1 1 11 6 NT
-1 ( ) ( ) ( )Aqx p NT C
r x q NT Brx p
− − −%&'
()*
→ %&'()*
= %&'()*
11th EXAMPLE:
The Systems of Boolean Arithmetical Functions (BAFs) are given:
( ) . ;#
152 5 23 2
1
2
3
2 1 22 3A NT
Y
Y
Y
X Xx1 6 ; @=%&K
'K
()K
*K=
%&K
'K 6 1 4
and
( ) . ;#
153 0 22 21
22 1 22 2C NT
Z
ZX Xx1 6 ; @=
%&'
()*
=%&'
2 1 3 , determine the following system
of Boolean Arithmetical Functions (BAFs),
( ) . ;#
154 32 21
27 6 5 4 3 2 1 02 1 2 33 2B NT
Z
Zb b b b b b b b Y Y Yx1 6 ; @=
%&'
()*
=%&'
, so that we have:
( )155 3 2 2 2 2 22 3 2 A1 6 1 6 1 6x x xB C→ =
NOTE: This example corresponds to the Second Inverse Consequent BAF(BAF)
Operation, referring to the 1st EXAMPLE considered in Section 7.1.
In terms of Boolean Geometry language, we have the following Fig. 16:
236
It is opportune to remember that the Systems of BAFs, in the compact
vertical numerical notation, could be also represented by the TABLES of the
practical gadget. This practical gadget corresponds to the “Discreet Function of
Discreet Function” of the ordinary Algebra.
Then, applying expression (151) we have:
1582 23 2 22 2 23
1 1 11 6 NT
-1 ( ) ( ) ( )Ax
NT Cx
NT Bx
− − −%&'()*
→ %&'()*
= %&'()*
Therefore, firstly we must determine the Inverse Analytical System of the
given Antecedent System of BAFs, that is:
A1 63 22x ( )Ax
−1
2 23
col.k
ak#
col.k
ak#
−1
3 5 7 Φ2
2 6 6 21 1 5 30 4 4 0
3 Φ2
2 Φ2
1 10 Φ2
237
Therefore, we have:
(159) ( )
#
. ;Ax
NTX
XY Y Y− =
%&'K
()*K
=1 2 2 2 2
2 231
2 22 3 1 2 3Φ Φ Φ Φ 2 3 0 1
= B , where
(160) ΦΦΦ
2
2
21
13 2=
%&'
()*
=�
!
"
$##
= 1 0 0
0 1 0 2, 1 or 0
(4)
,#
(or four solutions).
Then, applying the practical gadget of TABLES of the correspondent
“Discreet Function of Discreet Function” of the ordinary Algebra, we have:
(161) ( ) ( ) ( )Ax
Cx
Bx
− → =1
2 23 2 22 2 23
NOTE:
It is suitable to observe that as it is an Inverse Operation, whose respective
Direct Operation does not have the mathematical commutative property of the
given terms. Therefore, this Second Inverse Operation is not UNIQUE either.
Then, there are several solutions and one of them is the same, which was
considered in the 1st EXAMPLE of the Direct Boolean Arithmetical Function of
Boolean Arithmetical function Operation or BAF(BAF) Operation of Section 7.1.
Therefore, we have:
GIVEN DATA RESULTS
( )Ax
−1
2 23( )C
x2 22 ( )B
x2 23
col. k ak#
−1 col. k#
ck col. k bk#
7 Φ2 3 0 7 Φ2
6 2 2 2 6 25 3 1 1 5 04 0 0 3 4 33 Φ2 3 Φ2
2 Φ2 2 Φ2
1 1 1 10 Φ2 0 Φ2
238
This problem offers a result of “44 = 256” full solutions, which in the
numerical vertical compact representation of a system of BAF is the following,
(162) B x NTZ
ZY Y Y1 6 = B2
1
22 2 2 2
223 1 2 3 23 2 0 3 1
(4) (4) (4) (4)
=%&'K
()*K
=�
!
"
$##Φ Φ Φ Φ#
. ; ,
or, as binary Boolean Matrix form,
(163) B x NTZ
ZY Y Y1 6 = B2
1
2
4 4 4 4
22
3 1 2 3 23
* * * *
1 0 1 0
0 0 1 1 =
%&'K
()*K
=
�
!
"
$
######
Φ Φ Φ ΦΦ Φ Φ Φ
( ) ( ) ( ) ( )
#
. ;
(with “44 = 256” full solutions)
NOTE:
In that global numerical expression of all 256 full solutions (i.e. solutions
without any time-restriction) of expression (163), the abscissa components marked
with (*) correspond to the abscissa components of the solution, that is, it is the
same as the data in the 1st EXAMPLE of the Direct BAF(BAF) Operation of Section
7.1. In fact, this data was the following numerical expression (5):
(5) NT }3Y2Y1Y;3.{2#2
2213] 1203[2Z1Z
NT32x2)B( )}y(z{ =
== , or
(164) B x NTZ
ZY Y Y1 6 = B2
1
2
22
3 1 2 3
0 1 1
1 0 0 1 23
* * * *
1 0 1 1 0
0 0 1 1 =
%&'K
()*K
=
�
!
"
$
######
. ; , where the
bit columns not marked with (*) are one of the “44 = 256” possible solutions showed
in (163) numerical expression.
239
12th EXAMPLE:
The Systems of Boolean Arithmetical Functions (BAFs) are given:
( ) . ;#
165 1 23 2
1
2
3
2 1 22 3A NT
Y
Y
Y
X Xx1 6 ; @=%&K
'K
()K
*K=
%&K
'K 6 7 3
and
( ) . ;#
166 11 25 2
1
2
3
4
5
2 1 22 5C NT
Z
Z
Z
Z
Z
X Xx1 6 ; @=
%
&
KKK
'
KKK
(
)
KKK
*
KKK
=
%
&
KKK
'
KKK
15 15 29 , determine the following
system of Boolean Arithmetical Functions (BAFs),
( ) . ;#
167 35 2
1
2
3
4
5
7 6 5 4 3 2 1 02 1 2 33 5B NT
Z
Z
Z
Z
Z
b b b b b b b b Y Y Yx1 6 ; @=
%
&
KKK
'
KKK
(
)
KKK
*
KKK
=
%
&
KKK
'
KKK
, so that we have:
( )168 3 2 5 2 5 22 3 2 A1 6 1 6 1 6x x xB C→ =
NOTE: This example corresponds to the Second Inverse Consequent BAF(BAF)
Operation, referring to the 2nd EXAMPLE considered in Section 7.1.
In terms of Boolean Geometry language, we have the following Fig. 17:
240
It is opportune to remember that the Systems of BAFs, in the compact
vertical numerical notation, could be also represented by the TABLES of the
practical gadget. This practical gadget corresponds to the “Discreet Function of
Discreet Function” of the ordinary Algebra.
Then, applying expression (151) we have:
1692 23 5 22 5 23
1 1 11 6 NT
-1 ( ) ( ) ( )Ax
NT Cx
NT Bx
− − −%&'()*
→ %&'()*
= %&'()*
Therefore, firstly we must determine the Inverse Analytical System of the
given Antecedent System of BAFs, that is:
A1 63 22x ( )Ax
−1
2 23
col.k
ak#
col.k
ak#
−1
3 1 7 12 6 6 21 7 5 Φ2
0 3 4 Φ2
3 02 Φ2
1 30 Φ2
Therefore, we have:
(170) ( )
#
. ;Ax
NTX
XY Y Y− =
%&'K
()*K
=1 2 2 2 2
2 231
21
22 3 1 2 3 2 0 3
Φ Φ Φ Φ = B , where
(171) ΦΦΦ
2
2
21
13 2=
%&'
()*
=�
!
"
$##
= 1 0 0
0 1 0 2, 1 or 0
(4)
,#
(or four solutions).
Then, applying the practical gadget of TABLES of the correspondent
“Discreet Function of Discreet Function” of the ordinary Algebra, we have:
(172) ( ) ( ) ( )Ax
Cx
Bx
− → =1
2 23 5 22 5 23
241
NOTE:
It is suitable to observe that as it is an Inverse Operation, whose respective
Direct Operation does not have the mathematical commutative property of the
given terms. Therefore, this Second Inverse Operation is not UNIQUE either.
Then, there are several solutions and one of them is the same, which was
considered in the 2nd EXAMPLE of the Direct Boolean Arithmetical Function of
Boolean Arithmetical Function Operation or BAF(BAF) Operation of Section 7.1.
Therefore, we have:
GIVEN DATA RESULTS
( )Ax
−1
2 23( )C
x5 22 ( )B
x5 23
col. k ak#
−1 col. k#
ck col. K bk#
7 1 3 11 7 156 2 2 15 6 155 Φ2 1 15 5 Φ5
4 Φ2 0 29 4 Φ5
3 0 3 292 Φ2 2 Φ5
1 3 1 110 Φ2 0 Φ5
In this TABLE, “)5”, means the following “25 = 32” solutions:
(173) Φ
ΦΦΦΦΦ
5231 5=
%
&
KKK
'
KKK
(
)
KKK
*
KKK
= ,#
30, 29, ... , 2, 1, 0
This problem offers a result of “(25)4 = 220” full solutions, which in the
numerical vertical compact representation of a system of BAF is the following,
242
( ) . ;
#
174 15 15 35 2
1
2
3
4
5
21 2 33 5
B NT
Z
Z
Z
Z
Z
Y Y Yx1 6 ; @=
%
&
KKK
'
KKK
(
)
KKK
*
KKK
=
%
&
KKK
'
KKK
Φ Φ Φ Φ5 5 5 5 29 11 ,
or,
with “(25)4 = 220” full solutions”,
NOTE:
In that global numerical expression of all “(25)4 = 220” full solutions (i.e.
solutions without any time-restriction) of expression (175), the abscissa
components marked with (*) correspond to the abscissa components of the
solution, that is, it is the same as the data in the 2nd EXAMPLE of the Direct
BAF(BAF) Operation of Section 7.1. In fact, this data was the following numerical
expression:
( ) . ;
#
176 18 01 25 11 35 2
1
2
3
4
5
21 2 33
5B NT
Z
Z
Z
Z
Z
Y Y Yx1 6 ; @=
%
&
KKK
'
KKK
(
)
KKK
*
KKK
= �!
"$#
%
&
KKK
'
KKK
15 15 29 11 * * * *
or,
( ) . ;
* * * *
177
0 1 0
1 0 1
0 0 0 0
1 1 0 0 1
1 1 0 1 1 1 1 1
35 2
1
2
3
4
52
1 2 33B NT
Z
Z
Z
Z
Z
Y Y Yx1 6 ; @=
%
&
KKK
'
KKK
(
)
KKK
*
KKK
=
�
!
"
$
#######
%
&
KKK
'
KKKK
0 1 0 1 1
1 0 1 1 1
1 1 1 0
1 0 1
,
( ) . ;
( ) ( ) ( ) ( )
* * * *
175
0
1
1
1 1 1 1
35 2
1
2
3
4
5
2 2 2 2
2
1 2 33
5 5 5 5
B NT
Z
Z
Z
Z
Z
Y Y Yx1 6 ; @=
%
&
KKK
'
KKK
(
)
KKK
*
KKK
=
�
!
"
$
########
%
&
KKKK
'
KKKK
0 1 0
1 1 1
1 1 1 0
1 0 1
Φ Φ Φ ΦΦ Φ Φ ΦΦ Φ Φ ΦΦ Φ Φ ΦΦ Φ Φ Φ
243
where the bit columns not marked with (*) are one of the “(25)4 = 220” possible
solutions showed in (175) numerical expression.
13th EXAMPLE:
The Systems of Boolean Arithmetical Functions (BAFs) are given:
( ) . ;#
178 0 32 21
22 1 2 33 2A NT
Y
YX X Xx1 6 ; @=
%&'
()*
=%&'
1 2 0 2 0 0 1
and
( ) . ;#
179 6 33 2
1
2
3
2 1 2 33 3C NT
Z
Z
Z
X X Xx1 6 ; @=%&K
'K
()K
*K=
%&K
'K 2 7 5 1 3 0 4 , determine the
following system of Boolean Arithmetical Functions (BAFs),
( ) . ;#
180 53x2
1
2
3
31 30 29 28 3 2 1 02 1 2 1 2 35 3B NT
Z
Z
Z
b b b b b b b b Y Y X X X1 6 ; @=%&K
'K
()K
*K=
%&K
'K ...
, so that we have:
( )1815 2 2 2 3 23 5 3 A2 7 1 6 1 6x x xB C→ = , where, with the Complementary System
referring to the proper variables, that is:
( )
#
. ;1821
2
3
07 23 3 1 2 3 06 05 04 03 02 01 00NT
X
X
X
X X X
%&KK
'KK
()KK
*KK
=
%
&KK
'KK
= B ,
we have:
( ) . ; ,183
0010
35 2
1
2
1
2
3 2
1 2 33
1000
0100 0001
1111 0000
1100 1100
1010 1010
orA NT
Y
Y
X
X
X
X X Xx2 7 ; @=
%
&
KKK
'
KKK
(
)
KKK
*
KKK
=
�
!
"
$
######
244
( ) . ; ,183 07 35 2
1
2
1
2
3
2 1 2 33 14 21 04 19 02 01 08 orA NT
Y
Y
X
X
X
X X Xx2 7 ; @=
%
&
KKK
'
KKK
(
)
KKK
*
KKK
=
Then, according to the definition of the Second Inverse Consequent
BAF(BAF) Operation, we have:
1843 25 3 23 3 25
11 6
( ) ( ) ( )Ax
Cx
Bx
− → = , and therefore:
1853 25 3 23 3 25
1 1 11 6 NT
-1 ( ) ( ) ( )Ax
NT Cx
NT Bx
− − −%&'()*
→ %&'()*
= %&'()*
NOTE: This example corresponds to the Second Inverse Consequent BAF(BAF)
Operation, referring to the 3rd EXAMPLE considered in Section 7.2.
In terms of Boolean Geometry language, we have the following Fig. 18:
It is opportune to remember that the Systems of BAFs, in the compact
vertical numerical notation, could be also represented by the TABLES of the
practical gadget. This practical gadget corresponds to the “Discreet Function of
Discreet Function” of the ordinary Algebra.
Therefore, to apply expression (183) we must firstly determine the Inverse
Analytical System of the given Antecedent System of BAFs, that is:
F ig. 18
245
A2 75 23x ( )Ax
−1
3 25
col.k
ak#
col.k
ak#
−1 col.k
ak#
−1
7 07 31 Φ3 15 Φ3
6 14 30 Φ3 14 65 21 29 Φ3 13 Φ3
4 04 28 Φ3 12 Φ3
3 19 27 Φ3 11 Φ3
2 02 26 Φ3 10 Φ3
1 01 25 Φ3 09 Φ3
0 08 24 Φ3 08 0
23 Φ3 07 7
22 Φ3 06 Φ3
21 5 05 Φ3
20 Φ3 04 4
19 3 03 Φ3
18 Φ3 02 2
17 Φ3 01 1
16 Φ3 00 8
Therefore, we have:
( ) ( ) [
] . ;#
1843 25
1
2
3
5
1 3 3 3 3
3 3 32 1 2 1 2 33
5 ( ) 3 ( ) 6 ( ) 0 7
( ) 4 ( ) 2 1 ( )
10 1 4 5
2 1 1
Ax
NT
X
X
X
Y Y X X X
− =
%&K
'K
()K
*K
= Φ Φ Φ Φ
Φ Φ Φ
4 9
; @
, where
(185) ΦΦΦΦ
3
2
2
1
1 7 3 3=%&K
'K
()K
*K=
�
!
"
$
#####
= 1 1 1 0 0 0 0
1 0 0 1 1 0 0
1 0 1 0 1 0 1 0
6, 5, 4, 2, 1 or 0
(8)
, ,#
(or, eight solutions).
246
Then, applying the practical gadget of TABLES of the correspondent
“Discreet Function of Discreet Function” of the ordinary Algebra to the numerical
expression (182), we have the following TABLES:
NOTE:
It is suitable to observe that as it is an Inverse Operation, whose respective
Direct Operation do not have the mathematical commutative property of the given
terms. Therefore, this Second Inverse Operation is not UNIQUE either.
Then, there are several solutions and one of them is the same, which was
considered in the 3rd EXAMPLE of the Direct Boolean Arithmetical Function of
Boolean Arithmetical function Operation or BAF(BAF) Operation of Section 7.1.
Therefore, we have:
247
GIVEN DATA RESULTS
( )Ax
−1
3 25( )C
x3 23 ( )B
x3 25
col. k ak#
−1 Col. k#
ck col. k bk#
31 Φ3 7 6 31 Φ3
30 Φ3 6 2 30 Φ3
29 Φ3 5 7 29 Φ3
28 Φ3 4 5 28 Φ3
27 Φ3 3 1 27 Φ3
26 Φ3 2 3 26 Φ3
25 Φ3 1 0 25 Φ3
24 Φ3 0 4 24 Φ3
23 Φ3 23 Φ3
22 Φ3 22 Φ3
21 5 21 720 Φ3 20 Φ3
19 3 19 118 Φ3 18 Φ3
17 Φ3 17 Φ3
16 Φ3 16 Φ3
15 Φ3 15 Φ3
14 6 14 213 Φ3 13 Φ3
12 Φ3 12 Φ3
11 Φ3 11 Φ3
10 Φ3 10 Φ3
09 Φ3 09 Φ3
08 0 08 407 7 07 606 Φ3 06 Φ3
05 Φ3 05 Φ3
04 4 04 503 Φ3 03 Φ3
02 2 02 301 1 01 000 Φ3 00 Φ3
248
In this TABLE, “)3”, means the following “23 = 8” solutions:
(186) ΦΦΦΦ
3
8)
2
1111
7 3=%&K
'K
()K
*K=
�
!
"
$
#####
(
,#
0000
1100 1100
1010 1010
= 6, 5, 4, 3, 2, 1 and 0
2
This problem offers a result of “(23)24 = 232” full solutions, which in the
numerical vertical compact representation of a system in the BAF is the following,
( ) [
] . ;
*
#
187 4
5
3 2
1
2
3
3 10 3 1 3 4 3 5 3 2
3 1 3 1
2 1 2 1 2 3
5
3
B NT
Z
Z
Z
Y Y X X X
x1 6 4 9 4 9 4 9 4 9 4 9
4 9 4 9 ; @
=%&K
'K
()K
*K=
%&K
'K 7 1 2 6
5 3 0
* * * *
* * *
Φ Φ Φ Φ Φ
Φ Φ
,
NOTE:
In that global numerical expression of all “(23)24 = 272” full solutions (i.e.
solutions without any time-restriction) of expression (187), the abscissa
components marked with (*) corresponds to the abscissa components of the
solution, that is, it is the same as the data in the 3rd EXAMPLE of the Direct
BAF(BAF) Operation of Section 7.2. In fact, this data was the following numerical
expression:
( ) ( ) [
]
#
.{ ; },
233
1
2
3
5476
23 5 1 2 1 2 3
x 25 1032 547 6 1032 3210
7654 6745 2301
* * *
* * * * *
B NT
Z
Z
Z
Y Y X X X
=
%&K
'K
()K
*K
=
where the bit columns not marked with (*) are one of the “(25)4 = 220” possible
solutions showed in (187) numerical expression.
249
14th EXAMPLE:
The Systems of Boolean Arithmetical Functions (BAFs) are given:
( ) . ;188 1 21 2 2 1 22A NT Y X Xx1 6 ; @ ; @= = = 0 0 1
and
( ) . ;#
189 2103 42 21
22 1 2 3 44 2C NT
Z
ZX X X Xx1 6 ; @=
%&'
()*
=%&'
2213 2213 2103 , determine
the following system of Boolean Arithmetical Functions (BAFs),
( ) . ;#
190 32 21
27 6 5 4 3 2 1 02 3 43 2B NT
Z
Zb b b b b b b b YX Xx1 6 ; @=
%&'
()*
=%&'
, so that we have:
( )1913 2 2 2 2 24 3 4 A4 9 1 6 1 6x x xB C→ =
Then, according to the definition of the Second Inverse Consequent
BAF(BAF) Operation, we have:
1924 23 2 24 2 23
11 6
( ) ( ) ( )Ax
Cx
Bx
− → = , and therefore:
1934 23 2 24 2 23
1 1 11 6 NT
-1 ( ) ( ) ( )Ax
NT Cx
NT Bx
− − −%&'()*
→ %&'()*
= %&'()*
NOTE: This example corresponds to the Second Inverse Consequent BAF(BAF)
Operation, relative to the 4th EXAMPLE considered in Section 7.2.
In terms of Boolean Geometry language, we have the following Fig. 19:
250
The data given by numerical expression (188), must be prepared to
become “ Ax
4 93 24 ” of expression (191), through the following procedure:
We consider the Complementary System referred to the variables {X3 X4},
extended to the BAFi of ordinality “ω4 3 4 1 2= X X X X; @ ”, that is:
(194) NTX
XX X X X3
4 2
3 4 1 2
11114
%&'
()*
=�!
"$#
1111 0000 0000
1111 0000 1111 0000. ;; @ .
Then, we have:
(195) A NT
Y
X
X
X X X Xx
4 9 ; @3 2 3
4 2
3 4 1 24
1001
1111 4=%&K
'K
()K
*K=
�
!
"
$
###
1001 1001 1001
1111 0000 0000
1111 0000 1111 0000
. ; ,
or, in the numerical vertical compact representation,
(196) A NT
Y
X
X
X X X Xx
4 9 ; @3 2
3
4
2 3 4 1 2437337 4=
%&K
'K
()K
*K= 6226 5115 4004
#
. ;
Applying the properties given in CHAPTER 3 of changes in the
permutation of ordinality to the BAFi “ω4 1 2 3 4= X X X X; @ ”, we find:
(197) A NT
Y
X
X
X X X Xx
4 9 ; @3 2 3
4
2 1 2 3 44 37654 4=%&K
'K
()K
*K= 3210 3210 7654
#
. ;
Therefore, to apply expression (191) we must firstly determine the Inverse
Analytical System of the given Antecedent System of BAFs, suitably prepared by
expression (197), that is:
251
A4 93 24x
A -14 94 23x
Col.k
ak#
Col.k
ak#
−1
15 7 7 15 0314 6 6 14 0213 5 5 13 0112 4 4 12 0011 3 3 11 0710 2 2 10 0609 1 1 09 0508 0 0 08 0407 306 205 104 003 702 601 500 4
Then, with “28 = 256” full solutions, we have:
( ) [
]
#
. ;
( )198
1
2
315 03 02 01 00
07 06 05 0423 3 3 4
4 2
4
2
3 A = NT 14 13 12
11 10 09 08
-1(2) (2) (2)
(2) (2) (2) (2)
4 9
= B
x
X
X
X
X
YX X
%
&KK
'KK
(
)KK
*KK
=
Then, we have:
252
GIVEN DATA RESULTS
Ax
−1
4 234 9 C x1 62 24 B x1 62 23
Col. k ak#
−1 Col. k ck#
Col. k bk#
7 15 03 15 2 7 26 14 02 14 1 6 15 13 01 13 0 5 04 12 00 12 3 4 33 11 07 11 2 3 22 10 06 10 2 2 21 09 05 09 1 1 10 08 04 08 3 0 3
07 206 205 104 303 202 101 000 3
This example has the following unique solution, which is the same as the
one considered as data given by numerical expression (37), in the 4th EXAMPLE of
the Direct BAF(BAF) Operation of the Section 7.2, that is:
( )
#
. ;199 1
22 22 3 3 2 NT 1 0 3 2 2 1 3
Z
ZYX X
%&'K
()*K
= = B ,
15th EXAMPLE:
The Systems of Boolean Arithmetical Functions (BAFs) are given:
( ) . ;200 1 21 2 2 1 22 0 0 1A NT Y X Xx1 6 ; @ ; @= = =
and
253
( ) . ;#
201 5476 76 43 2
1
2
3
2 1 2 3 44 3C NT
Z
Z
Z
X X X Xx1 6 ; @=%&K
'K
()K
*K=
%&K
'K54 6745 1032 , determine
the following system of Boolean Arithmetical Functions (BAFs),
( ) . ;#
202 52 2
1
2
3
31 30 29 28 3 2 1 02 1 2 3 43 3 ... B NT
Z
Z
Z
b b b b b b b b YX X X Xx1 6 ; @=%&K
'K
()K
*K=
%&K
'K, so that
we have:
( )2035 2 3 2 3 24
5 4 A��
�� → =
x x xB C1 6 1 6
Then, according to the definition of the Second Inverse Consequent
BAF(BAF) Operation, we have:
2044 25 3 24 3 25
11 6
( ) ( ) ( )Ax
Cx
Bx
− → = , and therefore:
2054 25 3 24 3 25
1 1 11 6 NT
-1 ( ) ( ) ( )Ax
NT Cx
NT Bx
− − −%&'()*
→ %&'()*
= %&'()*
NOTE: This example corresponds to the Second Inverse Consequent BAF(BAF)
Operation, referring to the 5th EXAMPLE considered in Section 7.2.
In terms of Boolean Geometry language, we have the following Fig. 20:
254
The data given by numerical expression (200), must be prepared to
become the term “ Ax
��
��5 24
” of the expression (203), through the following
procedure:
We consider the Complementary System referred to the variables {X3 X4},
extended to the BAFi of ordinality “ω4 3 4 1 2= X X X X; @ ”, with “24 = 16” solutions,
that is:
(206) NT
X
X
X
X
X X X X
3
4
1
2 2
3 4 1 2
1111
4
%
&KK
'KK
(
)KK
*KK
=
�
!
"
$
####
1111 0000 0000
1111 0000 1111 0000
1100 1100 1100 1100
1010 1010 1010 1010
. ;; @ ,
or, in the numerical vertical compact representation,
(207) NT
X
X
X
X
X X X X
3
4
1
2
2 3 4 1 215 44
%
&KK
'KK
(
)KK
*KK
= 14 13 12 ... 03 02 01 00#
. ;; @
Then, we have:
(208) A NT
Y
X
X
X
X
X X X Xx
��
�� =
%
&
KKK
'
KKK
(
)
KKK
*
KKK
=
�
!
"
$
######
5 2
3
4
1
2 2
3 4 1 24
1001
1111
4
1001 1001 1001
1111 0000 0000
1111 0000 1111 0000
1100 1100 1100 1100
1010 1010 1010 1010
. ;; @ ,
or, in the numerical vertical compact representation,
(209)
A NT
Y
X
X
X
X
X X X Xx
��
�� =
%
&
KKK
'
KKK
(
)
KKK
*
KKK
=5 2
3
4
1
2
2 3 4 1 24531 4 14 13 28 27 10 09 24 23 06 05 20 19 02 01 16
#
. ;; @
Applying the properties given in Chapter 5 of changes in the permutation of
ordinality to the BAFi “ω4 1 2 3 4= X X X X; @ ”, we find:
255
(210)
A NT
Y
X
X
X
X
X X X Xx
��
�� =
%
&
KKK
'
KKK
(
)
KKK
*
KKK
=5 2
3
4
1
2
2 1 2 3 44531 4 27 23 19 14 10 06 02 13 09 05 01 28 24 20 16
#
. ;; @
Therefore, to apply expression (204) we must firstly determine the Inverse
Analytical System of the given Antecedent System of BAFs, suitably prepared by
expression (210), that is:
Ax
��
��5 24
( )Ax
−14 25
col.k
ak#
col.k
ak#
−1 col.k
ak#
−1
15 31 31 15 15 Φ4
14 27 30 Φ4 14 11
13 23 29 Φ4 13 07
12 19 28 03 12 Φ4
11 14 27 14 11 Φ4
10 10 26 Φ4 10 10
09 06 25 Φ4 09 06
08 02 24 02 08 Φ4
07 13 23 13 07 Φ4
06 09 22 Φ4 06 09
05 05 21 Φ4 05 05
04 01 20 01 04 Φ4
03 28 19 12 03 Φ4
02 24 18 Φ4 02 08
01 20 17 Φ4 01 04
00 16 16 00 00 Φ4
Then, with “ 2 24 16 644 9 = ” full solutions and “ Φ4 ” is given by (207), we have:
256
( ) [
]
#
. ;
211
1
2
315 01 12
24 5 3 4
4
2 2 2 2
1 2 2 2 11 2
A =NT 03 14 02 13
00 11 07 10 06 09 05 08 04
-1
4x2
4 4 4 4
4 4 4 4 4
5
��
��
%
&KK
'KK
(
)KK
*KK
=
X
X
X
X
YX X X X
Φ Φ Φ Φ
Φ Φ Φ Φ Φ
4 9 4 9 4 9 4 9
4 9 4 9 4 9 4 9 4 9 = B
Then, applying the definition given by expression (204), we have:
2044 25 3 24 3 25
11 6
( ) ( ) ( )Ax
Cx
Bx
− → =
GIVEN DATA RESULTS
( )Ax
−14 25
( )Cx3 24
( )Bx3 25
Col. k ak#
−1 Col. k ak#
−1 Col. k ck#
Col. k bk#
Col. k bk#
31 15 15 Φ4 15 5 31 5 15 Φ3
30 Φ4 14 11 14 4 30 Φ3 14 7
29 Φ4 13 07 13 7 29 Φ3 13 628 03 12 Φ4 12 6 28 1 12 Φ3
27 14 11 Φ4 11 7 27 4 11 Φ3
26 Φ4 10 10 10 6 26 Φ3 10 6
25 Φ4 09 06 09 5 25 Φ3 09 7
24 02 08 Φ4 08 4 24 0 08 Φ3
23 13 07 Φ4 07 6 23 7 07 Φ3
22 Φ4 06 09 06 7 22 Φ3 06 5
21 Φ4 05 05 05 4 21 Φ3 05 4
20 01 04 Φ4 04 5 20 3 04 Φ3
19 12 03 Φ4 03 1 19 6 03 Φ3
18 Φ4 02 08 02 0 18 Φ3 02 4
17 Φ4 01 04 01 3 17 Φ3 01 5
16 00 00 Φ4 00 2 16 2 00 Φ3
where “ Φ4 ” is given by expression (207) and “Φ3” is given by the following
numerical expression:
257
(212) NT
Z
Z
Z
1
2
3
7
%&K
'K
()K
*K= , 6, 5, 4 ,3, 2, 1 or 0
This problem offers a result of “(23)16 = 248” full solutions, which in the
numerical vertical compact representation of a system of BAF is the following,
( ) [
. ;
*
#
213 5
5
3 2
1
2
3
3 2 3 2 3 2 3 2 3 1
3 2 3 2 3 32 3 4 1 2
5
3
B NT
Z
Z
Z
YX X X X
x1 6 4 9 4 9 4 9 4 9 4 9
4 9 4 9 4 9 4 9 ; @
=%&K
'K
()K
*K=
%&K
'K 1 4 0 7 3 6 2
7 6 6 7 5 4 4 5 ]
* * * * * * *
* * * * * *
2
* *
1
Φ Φ Φ Φ Φ
Φ Φ Φ Φ
As we have seen in Property (27) in Chapter 3, Section 3.3, we can
determine the new numerical expression in a BAFi, whose cardinality/ordinality is
“ 5 1 2 3 4;YX X X X; @ ” and we find:
( ) ) ) ) .
;
* * * *#
214 5 0 7 6
5
2
3
8 4 4
2
1 2 3 4
3 B = NT
Z
4 7 6 ( 1 3 2 ( 6 5 4 7 4 5 (
.
3x2
1
* * *
3
* * *
3
* * * * * *
351 6
; @
Z
Z
YX X X X
%&K
'K
()K
*K= �
! "$#
Φ Φ Φ
NOTE: As we have seen, this example corresponds to the Second Inverse
Consequent BAF(BAF) Operation, relative to the 5th EXAMPLE considered in
Section 7.2 and the value expected for this system of BAF is given by the
numerical expression:
( )
#
.{ ; },B NT
Z
Z
Z
YX X X X3
1
2
223
5 1 2 3 4 x 25 5476 1032 5476 1032 3210 7654 6745 2301 * * * * * * * * * * * * * * * *
=%&K
'K
()K
*K= �
! "$#
It is clear that expression has, apart from the sixteen common number
marked by (*), other numbers, that is one of those “(23)16 = 248” full solutions of
global numerical expression (214).