Image Operations

Priority-assignment image operation

The Priority-assignment image operation, denoted by the symbol '^─', operates on variables and on the priority-assignment of a prioritor. For example, a^─f means that this operation is to operate on the priority-assignment of the “a” prioritor not on the prioritor itself.

Definition 10.1 Priority-assignment Image operation:
The priority-assignment image operation, denoted by “^─f””, is defined as a^─f= (priority-assignment of a)^─f

Binary image operation

The binary image operation, denoted by the symbol '══', operates on variables and on the prioritor itself not on its priority-assignment.

Definition 10.2 The binary image operation 
The binary-image operation, denoted by a^══f, is defined as a^══f =f(function Table of a).

The binary image operation and priority-assignment image operation are different and not comparable when they operate on a prioritor because the first results in a unary operator while the second results in a binary operator.  The relation between the two operation is given by (AaB)^══f=Aa^══f B=(Aa B)^─f

Example-10.1: In the quaternary system, let f=4S1302 and a=Q_J=4S3012= 4S3333:3210:3110:3000. Using the priority-assignment image operation a^─f is 4S3012^─4S1302=4S1203.  Note that “f” operated on the priority-assignment “4S3012” not on the prioritor “4S3333:3210:3110:3000”.   Using the binary image operation a^══f =(4S3333:3210:3110:3000)^═4S1302= 4S1111:1302:1002:1222.  Note that both results are completely different and incomparable.

2.3 Uniform Image-Scaling Theorem

            When we take the image of a binary operation using the NOT or MV-NOT operator in Boolean and Post algebras, we use DeMorgan's laws to break out the image operation. The DeMorgan's laws work only for the NOT and MV-NOT operators. What about if we take the image of a binary operation, say MIN, by using a one-to-one unary operator, say f=4S3012, other than the MV-NOT operator (see Example-2.3.2)? Post algebra does not provide the means in this case to break the image operation. AOP solves this problem by replacing DeMorgan's laws by a new theorem (Theorem-2.3.1) called the "uniform image-scaling (UIS) theorem" which is a special case from the Image-Scaling Theorem.  

Examples

Example-2.3.2 shows how to break up the image of the MIN binary operator under the unary operator f=4S3012. Other similar results are shown in Example-2.3.3 where “¾” by default stands for the NOT (2S01) operator in the binary system and MV-NOT in MVL systems.

In the quaternary system, if we let f=4S1302 and a=Q_J=4S3012=4S3333:3210:3110:3000 then (A 4S3012 B)^{─ 4S1302}=A^─4S1302 4S3012^─4S1302 B^{─ 4S1302}; 
(A 4S3012 B)^─4S1302=A^─4S1302 4S1203 B^─4S1302.  According to the UIS theorem, the image on the left side in “(A a B)^¾f” is taken on the final result not on the priority-assignment of a, or it is taken on the function table of a as shown in “A a^══f B”. That is (A a B)^══4S1302 =(A a^══4S1302 B) =AbB where b=(Q_J=4S3333:3210:3110:3000)^══4S1302= 4S1111:1302:1002:1222. On the other hand, the image operation on the right side is taken on the priority-assignment of a. That is 4S3012 ^─4S1302=4S1203=Q₉. Assume A=2 and B=3, then (2 4S3012 3)^{─ 4S1302}=2^{─ 4S1302} 4S1203 3^{─ 4S1302}; 3^{─ 4S1302}=3 4S1203 1; 1=1.

Using Table-1 and Table-1 , the image of the binary operation AaB for a=Q₁=MIN=4S0123 and f=4S3012 is (A a B)^─f = A^─f a^─f B^─f= A^─fb B^─f where b=a^─f = a^─f=4S0123^─4SS3012=4S2103=Q_F. See Table-1 , which shows the function table of this example.

Since AND^─f=2S01^─2S01=2S10=OR and OR^─=2S10^─2S01=2S01=AND when f=2S01, then we obtain (A AND B)¯= A¯ OR B¯ and (A OR B)¯= A¯AND B¯ , which is DeMorgan’s law in Boolean Algebra.  Since MIN^─=MAX and MAX^─=MIN when f=D, then we obtain (A MAX B)¯= A¯MIN B¯ and (A MIN B)¯= A¯ MAX B¯, which DeMorgan’s law in Post Algebra.

Example-2.3.5 On Uniform Image-Scaling Theorem: In the quaternary system, let f=4S0123 and a=4S0123.

(A 4S0123 B)^─4S0123= A^─4S0123 4S0123^─4S0123 B^─4S0123

(A 4S0123 B)^─4S0123= A^─4S0123 4S3210 B^─4S0123

(A MIN B)^─4S0123= A^─4S0123 MAX B^─4S0123

(A MIN B)¯ = A¯ MAX B¯  using default notation.

where MIN=Q₁ and MAX=Q_O in Table-1

Example-2.3.6 On Uniform Image-Scaling Theorem: In the ternary system, let f=3S012 and a=3S012.

(A 3S012 B)^─3S012 =A^─3S012 3S012^─3S012 B^─3S012

(A 3S012 B)^─3S012=A^─3S012 3S210 B^─3S012

(A MIN B)^─3S012 =A^─3S012 MAX  B^─3S012

(A MIN B)¯ =A¯ MAX  B¯  using default notation.

where MIN=T₁ and MAX=T₆ in Table-1