Scientiae Mathematicae Vol. 2, No. 3(1999), 373{384 373
TRICE AND TWO DELEGATES OPERATION
KIYOMITSU HORIUCHI
Received Octob er 1, 1999
Abstract. The concept of \trice " was intro duced in [5] and [6] for the purp ose of using
truth value of fuzzy sets. In this pap er, weintro duce the notion of trice and some of
its mathematical prop erties. Moreover, we shall showsometypical concrete examples
of trice including those having two delegates op erations.
1. Intro duction. A semilattice (S; ) is a set S with a single binary, idemp otent, com-
mutative and asso ciative op eration .
(1) a a = a (idempotent)
(2) a b = b a (commutativ e)
(3) a (b c)=(a b) c (associativ e)
The following Prop osition is well-known (see [1] p.10).
Prop osition 1. Let (S; ) b e a semilattice. Under the relation de ned by
a b () a b = b;
any semilattice (S; ) is a partially ordered set (S; ).
A lattice has two op erations. Let L b e a lattice with two op erations _ (join) and
^ (meet). Then, (L; _) and (L; ^) are semilattices. We can construct two ordered sets
(L; ) and (L; ), resp ectively,by using the Prop osition 1 (a b , a _ b = b and
_ ^ _
a b , a ^ b = b). The dual of (L; )is(L; ). That is, the order is nothing but
^ _ ^ _
the reverse of the order . The reason why _ and ^ intro duce the reverse order is that
^
lattices satisfy the absorption law:
(4) a _ (a ^ b)=a; a ^ (a _ b)= a (absor ption)
So, when a b (that is, a _ b = b), wehave a ^ b = a ^ (a _ b)=a by (4). Hence b a.
_ ^
Supp ose that there is an ob ject with a rop e on a line, and that we are able to pull the
rop e from the right (see Fig. 1). Then, wecanmove the ob ject to the right direction, but
not to the left direction. This situation is considered to b e irreversible.
Next, see Fig. 2. By pulling the rop e from either right or left directions, we can move
the ob ject to anypoint on the line. This situation is considered to b e reversible.
From Prop osition 1, a semilattice is a set having one order, i.e. one direction, likein
Fig. 1. A lattice is a set having two orders, i.e. two directions, likeinFig.2.Any ob ject
on the line can b e moved to another arbitrary p oint on the line by pulling the ob ject to the
1991 Mathematics Subject Classi cation. 06A12, 20N10, 20M10.
Key words and phrases. semilattice, trice, roundab out-absorption law, triangular situation, two dele- gates op eration.
374 K.HORIUCHI
-
-
Figure 1. one dimensional irreversible
-
-
Figure 2. one dimensional reversible
p ositive or negative directions.
6
>
HY
H
H
H
H
J
J
J
J
-
J
J^
Figure 3. two dimensional reversible
Supp ose that there is an ob ject not on a line but in a plane (two dimensional Euclidian
space). If wewanttomove the ob ject to an arbitrary p oint in the plane, it is sucientto
have three \suitable" directions to pull the ob ject, as shown in Fig. 3.
If the three directions are not \suitable", as shown in Fig. 4, we cannot move the original
ob ject to an arbitrarily chosen target p oint. Movements in three directions are not nec-
essarily restricted to the two dimensional plane (see Fig. 5 right). Consider that a lattice
corresp onds to the one dimensional reversible case (Fig. 2). What systems with three binary
op erations (semilattice) corresp ond to the two dimensional reversible cases (Fig. 3)?
2. Roundab out-Absorption Law. For A a nonempty set and n a p ositiveinteger, let
(A; ; ;:::; ) b e an algebra with n binary op erations, and (A; ) b e a semilattice for
1 2 n i
every i 2f1; 2; :::; ng. Then, (A; ; ;:::; )iscalledan-semilattice (See [7 ]).
1 2 n
In this pap er, we will deal mainly with triple-semilattice, (i.e. n =3).
We denote each order on A by
(5) a b () a b = b;
i i
resp ectively.
De nition 1. Let S b e the symmetric group on f1; 2; :::ng. An algebra (A; ; ; :::; )
n 1 2 n
has the n-roundab out-absorption law if it satis es the following n!identities:
(6) ((((a b) b) b)::: b)= b:
(1) (2) (3) (n)
TRICE 375
6
CO
> >
C
HY
H
f f H f C
H
H
Hj
? =
? -
Figure 4. two dimensional irreversible
6 6
f f
H H
H A H
H H
A H
H
Hj Hj
-
A
A
A
AU
Figure 5. three dimensional cases
for all a; b 2 A and for all 2 S .
n
We remark that \n-roundab out-absorption law" have di erent name \absorptivelaw."
And arlgebaic researches yielded interesting results. (See [7]). However, weadoptn-
roundab out-absorption law from the image of Prop osition 2.
Of cause, the 2-roundab out-absorption law is the absorption law. An algebra (A; ; )
1 2
which satis es the 2-roundab out-absorption law is a lattice. The op erations and are
1 2
denoted by _ and ^.
De nition 2. An algebra (A; ; ; )which satis es the 3-roundab out-absorption
1 2 3
law issaidtobeatrice. To simplify explanation, we often omit \3" of \3-roundab out-
absorption law." The op erations , and will b e denoted by % , - and # .
1 2 3 1 2
3
The \trice" is a notion to corresp ond to \lattice."
Let T b e a set. Weintro duce three orders into T , under the condition that all two elements
of T have a least upp er b ound for each order. Then, we can construct the set into a
triple-semilattice.
Example 1. Let T b e a set which consists of six p oints. Weintro duce three orders in the
set by arrows of Fig. 6. Supp ose that arrowhead is larger than the other end (e.g. d e).
2
Then, T is a trice. That is, T has the 3-roundab out-absorption law.
1 2 3
c - c - c c c c c c c
J J
J] J]
J^ J^
- J J c c c c c c
e d
J
J]
J^
J c c c c
Figure 6. Example 1
376 K.HORIUCHI
Example 2. Denote by R the set of all real numbers. Weintro duce three orders in the
2
two dimensional Euclidian space R as follows :
p
x y () x y and 0 y x 3(y x )
1 1 1 2 2 1 1
p
x y () x y and 0 y x 3(x y )
2 1 1 2 2 1 1
p
3 jy x j: x y () x y
1 1 3 2 2
2 2
for x =(x ;x );y=(y ;y ) 2 R . This (R ; % ; - ; # )isatrice. Needless to say,
1 2 1 2 1 2
3
2
wecanmake another trice on R .
De nition 3. Let (T; % ; - ; # ) b e a triple-semilattice. A sub-triple-semilatti ce
1 2
3
of T is a subset S of T such that
(7) a; b 2 S impl y a % b; a - b; a # b 2 S:
1 2
3
If T is a trice, then a sub-triple-semilattice S is a trice. Wesay that S is a subtrice of T .
Let T b e a trice. The emptysubsetof T is a subtrice. For any a 2 T ,onepoint set fag
is a subtrice of T .Inabove Example 1, the set fc, d, eg isasubtriceofT .Wecanembed
Example 1 in Example 2, as a subtrice.
We can easily check that a lattice L has following prop erties:
8a; b 2 L 9c 2 Ls:t:a c b
_ ^
8a; b 2 L 9c 2 L s:t: a c b:
^ _
Prop osition 2. Let T b e a trice. For every a; b 2 T , there exist c; d 2 T suchthat
a c d b.
1 2 3
Pro of Obviously, a a % b (a % b) - b. From ((a % b) - b) # b = b
1 1 2 1 2 1 2
3
(the roundab out-absorption law), (a % b) - b b. Let c = a % b and
1 2 3 1
d =(a % b) - b. Then, wehave a c d b. This completes the pro of.
1 2 1 2 3
One can prove the next Prop osition in a similary way.
Prop osition 3. Let an algebra (A; ; ;:::; )have the n-roundab out-absorption law.
1 2 n
For every a; b 2 A and for evry 2 S , there exists c ;:::;c 2 A such that
n 1 n 1
(8) a c c ::: c b:
1 2 n 1
(1) (2) (3) (n 1) (n)
De nition 4. Wesay that (A; ; ;:::; )is attainable if (8) is true for every a; b 2 A.
1 2 n
The \attainablity" of lattice corresp onds to the notion of reversiblity in the case of one
dimension (Fig. 2). We can consider that the \attainablity" of trice corresp onds to the case
of Fig. 3, that is, the two dimensional reversible cases.
1 2 3
c c c - - c c c c c c
* HY
H
J
J]
?
b c H c c
J^
* HY
c c J H c c c c
c d
J
J]
?
J^
c J c c
a
Figure 7. Example 3
TRICE 377
Example 3. Let T b e a set which consists of seven p oints. Weintro duce three orders
in the set byarrows of Fig. 7. Then, T has not the roundab out-absorption law. But T is
attainable.
From the p oint of algebraic view, the concept of trices are available. Now, we will take
one example of many. The following prop osition is well knwon. (See [8]).
Prop osition 4. If V is a varietywhich admits a k-ary p olynomial m that satis es the
near unanimity identities, then V is congruence distributive.
1
Prop osition 5. The variety of trices is congruence-distributive.
Pro of Consider the following 4-ary term:
M (x; y; z; t) = ((x % y ) - (y % z ) - (x % z )) #
1 2 1 2 1
3
((x % y ) - (y % t) - (x % t)) #
1 2 1 2 1
3
((x % z ) - (z % t) - (x % t)) #
1 2 1 2 1
3
((y % z ) - (y % t) - (z % t)):
1 2 1 2 1
We have that:
M (a; a; a; b) = ((a % a) - (a % a) - (a % a)) #
1 2 1 2 1
3
((a % a) - (a % b) - (a % b)) #
1 2 1 2 1
3
((a % a) - (a % b) - (a % b)) #
1 2 1 2 1
3
((a % a) - (a % b) - (a % b))
1 2 1 2 1
= a # (a - (a % b)) # (a - (a % b)) #
2 1 2 1
3 3 3
(a - (a % b))
2 1
= a # (a - (a % b))
2 1
3
= a:
Similarly, M (a; a; b; a)=a, M (a; b; a; a)=a and M (b; a; a; a)=a.
From Prop osition 4, variety of trices is congruence distributive.
Similary, one can prove that a variety of algebras (A; ;:::; ), whichhave n-roundab out-
1 n
absorption law, is congruence-distributive.
3. Triangular Situation. In this section, we discuss the concept to distinguish b etween
trice and lattice.
Let (T; % ; - ; # ) b e a trice and a; b; c 2 T (a 6= b 6= c 6= a).
1 2
3
De nition 5. Wesay that an ordered triple (a; b; c)isina triangular situation if
(a; b; c)have the following prop erties:
(9) a # b = c and a - c = b and b % c = a:
2 1
3
Strictly sp eaking, a corresp ond to % , b corresp ond to - and c corresp ond to # .
1 2
3
Therefore, if necessary,wemust say that (a; b; c)isina % - # -triangular situation.
1 2
3
If there is a triple in the triangular situation on T ,wesay that T has a triangle.
1
After I gave a presentation at the conference on lattices and universal algebra in Szeged 1998, I received
an email message from Professor A. Tepavcevic of UniversityofNovi Sad saying that she proved Prop osition 5.
378 K.HORIUCHI
If we add another op eration to a lattice, we can construct a trice. However, this trice
has no triangle. Hence, triangular situation is an imp ortant concept in the theory of trices.
Let T b e a trice. If (a; b; c) is in a triangular situation in T , then the set fa; b; cg is a
subtrice and
(c - a) % c = b % c = a 6= c
2 1 1
(a # b) - a = c - a = b 6= a
2 2
3
(b % c) # b = a # b = c 6= b:
1
3 3
Hence, two op erations in T do not satisfy the absorption law. That is, eachof(T; % ; - ),
1 2
(T; - ; # ) and (T; - ; # ) is not a lattice. If we desire to consider the case that trices
2 2
3 3
corresp onding to the two dimensional reversible cases (Fig. 3), we should supp ose that T
is not a lattice. Of course, there is an example in which \(T; % ; - ; # ) is a trice
1 2
3
and (T; % ; - ) is a lattice". Let I b e the unit interval in R, then (I; _; ^) is a lattice.
1 2
Denote _ by % and ^ by - . Supp ose # is a \mo de-typ e op erator" in [4]. Then
1 2
3
(I; % ; - ; # ) is a trice.
1 2
3
The triangular situation is easy to understand. Let n 4. When an algebra (A; ; ; :::; )
1 2 n
have the n-roundab out-absorption law, it is dicult to represent a concept corresp onding
with triangular situation on trice. Thus, the "triangular situation" would b e a key word for
trices.
From the algebraic p oint of view, distributivity is imp ortant and essential. However, if
T has a triangle, then T is not distributive. Each of the Example 1, 2 and 3 has a triangle.
This shows that wemust deal with \non-distributive" algebra, and the aim of this pap er is
to investigate characteristic p op erties of non-algebraic algebra.
Example 4. Let T b e a set of four p oints. Weintro duce three orders in the set by
1 2 3
c c c c c c
b a
H H
* HY
H
H H d
Hj Hj
c c H c
6 6
?
c c c
c
Figure 8. Example 4
arrows of Fig. 8. This T is a trice, but T do es not have a triangle. Let S b e the set
fa; d; bg2 T . Then, S is a subtrice of T , but again S do es not have a triangle. Note that
(S , % , - ) is a lattice.
1 2
There are some prop erties related to the triangular situation. We de ne \the triangle
constructivelaw" and \the triangle natural law".
De nition 6. Let T b e a triple-semilattice. Wesay that T has the triangle construc-
tivelaw if T has the following prop erties:
(10) (d % e) # (d - e)= d # e
1 2
3 3
(11) (d % e) - (d # e)=d - e
1 2 2
3
(12) (d # e) % (d - e)=d % e
1 2 1
3
for all d; e 2 T .
TRICE 379
Supp ose that a triple-semilattice T satis es the identities ((10) (12)) for d; e 2 T . Let
a = d % e, b = d - e, c = d # e. If a 6= b 6= c 6= a, then (a; b; c) is in a triangular
1 2
3
situation. That is whywe named the identities \the triangle constructivelaw".
De nition 7. Let T b e a triple-semilattice. Wesay that T has the triangle natural
law if T has the following prop erties:
(13) if x % y = zandx- z = y; then y # z = x
1 2
3
(14) if x # y = z and x - z = y; then y % z = x
2 1
3
(15) if x % y = zandx# z = y; then y - z = x
1 2
3
for all x; y ; z 2 T .
Example 1 and Example 2 have b oth the triangle constructivelaw and the triangle
natural law. Example 3 has neither the triangle constructivelaw nor the triangle natural
law.
Now, wegive other examples. Example 5 has the triangle constructivelaw, but it has
neither the roundab out-absorption law nor the triangle natural law. Example 6 has the
roundab out-absorption law, but it has neither the triangle constructivelaw nor the triangle
natural law. Example 7 has the triangle natural law, but it has neither the roundab out-
absorption law nor the triangle constructivelaw.
Example 5. Let T be a vepoints set fa; b; c; d; eg. See Fig. 9. Weintro duce three
1 3 2
c c - c c c c
b a
* HY
H
J
@I
c J] d c H c
@
J
J
c c @ e c
J
J
J^
?
c J c
c
c
Figure 9. Example 5
orders in T by:
(16) b a; c a; d a; e a
1 1 1 1
(17) a b; c b; d b; e b
2 2 2 2
(18) a c; b c; d c; e c:
3 3 3 3
From ((b % e) - e) # e = c,thisT do es not satisfy the roundab out-absorption law.
1 2
3
From e % b = a , e - a = b and b # a = c, T do es not have the triangle natural law.
1 2
3
But T has the triangle constructivelaw.
Example 6. Let T b e a set of four p oints. Weintro duce three orders in the set by
arrows of Fig. 10. We can easily check T has the roundab out-absorption law, and the
(a; b; d) is a in a triangular situation on T . But, from (c % b) # (c - b)=a # b = d
1 2
3 3
and c # b = c, T do es not have the triangle constructivelaw. From c % b = a, a - c = b
1 2
3
and b # a = d 6= c, T do es not have the triangle natural law, either. 3
380 K.HORIUCHI
1 2 3
c c c c c c -
b a
H
HY *
H
H
Hj
c Hc c d
6 6
?
c c c
c
Figure 10. Example 6
Example 7. Let T be a set fx ;y;z j n 2 N [f0gg.Weintro duce three orders in
n n n
T by:
(19) z x ; y x ; x x ; y x ; z x
0 1 0 0 1 0 k 1 k +1 k 1 k +1 k 1 k +1
(20) x y ; z y ; x y ; y y ; z y
0 2 0 0 2 0 k 2 k +1 k 2 k +1 k 2 k +1
(21) y z ; x z ; x z ; y z ; z z
0 3 0 0 3 0 k 3 k +1 k 3 k +1 k 3 k +1
for k 2 N [f0g. In this triple-semilattice, the only (x ;y;z) is in a triangular situation.
0 0 0
In case of another combination, the assumption of statements (13) (14) (15) do es not hold.
Therefore, T has the triangle natural law. Let a = x ;b= x ,then((a % b) - b) # b =
2 1 1 2
3
((x % x ) - x ) # x =(x - x ) # x = y # x = z 6= b. Hence, T do es not
2 1 1 2 1 1 2 2 1 1 3 1 4
3 3 3
have the roundab out-absorption law. From a % b = x , a - b = y and a # b = z , T
1 2 2 3 3
3
do es not have the triangle constructivelaw.
Prop osition 6. Every trice T has the following prop erty:
(22) if x % y = z and x - z = y; then y # z x:
1 2 3
3
Pro of From x % y = z and x - z = y , x - (x % y )= y . By the roundab out-
1 2 2 1
absorption law, x # (x - (x % y )) = x. Hence, wehave x # y = x. This implies y x.
2 1 3
3 3
Similarly,wehave x # z = x, that is, z x. Therefore, we see that y # z x.
3 3
3 3
Prop osition 7. Let T b e a trice. If T has the triangle constructivelaw, then T has the
triangle natural law.
Pro of By the Prop osition 6, if x % y = z and x - z = y ,theny # z x for
1 2 3
3
x; y ; z 2 T . Let d be y # z . And supp ose that d 6= x.From x y , x - y = y . Hence
2 2
3
(x % y ) # (x - y )= z # y = d. But, from y y # z = d x, x # y = x.
1 2 3 3
3 3 3 3
This implies T do es not have the triangle constructivelaw. It is in contradiction with the
assumption. Therefore, d = x. Hence, T has the triangle natural law.
Question 1. Do es every trice which has the triangle natural law have the triangle
constructivelaw?
We answer this by Example 9 in the next section.
Let L b e a lattice and a; b 2 L. If a b, then the set fa; bg is a sublattice of L.
Moreover, the interval [a; b] is a sublattice of L. The interval can b e considered the set
fx 2 Ljx a; x bg. Similarly,letT b e a trice and a; b; c 2 T (a 6= b 6= c 6= a). If
^ _
(a; b; c) is in a triangular situation on T , then the set fa; b; cg is a subtrice of T .
Now, we consider the set fx 2 T jx a; x b; x cg.We will denote it by[a; b; c],
1 2 3
and call it hyp er-interval. It seems the concept corresp onding with the interval in a lattice.
TRICE 381
Example 8. Let T b e a set of seven p oints fa; b; c; d; e; f ; g g.Weintro duce three orders
in T by:
(23) b a; c a; d a; e a; a g; f g
1 1 1 1 1 1
(24) a b; c b; d b; e b; f b; g b
2 2 2 2 2 2
(25) a c; b c; d f; e f; f c; g c:
3 3 3 3 3 3
This T is a triple-semilattice. The (a; b; c) is in a triangular situation on T . And the set
[a; b; c]is fa; b; c; d; eg. From d # e = f 2= [a; b; c], the hyp er-interval [a; b; c] is not a
3
sub-triple-semilattice of T .
Prop osition 8. Let T b e a triple-semilattice. If T has the triangle constructivelaw,
the hyp er-interval [a; b; c]isasubtriceofT for every a; b; c 2 T .
Pro of Toprove this prop osition, it suces to show that x % y 2 [a; b; c] (that is,
1
x % y a, x % y b and x % y c) for x; y 2 [a; b; c].
1 1 1 2 1 3
From x a and y a, it is clear that x % y a. By the triangle constructivelaw,
1 1 1 1
(x % y ) - (x # y )=x - y . Hence x % y x - y . From x b and y b,
1 2 2 1 2 2 2 2
3
x - y b. Therefore x % y b. Similarly,we can prove x % y c. This
2 2 1 2 1 3
completes the pro of.
Question 2. Let T b e a trice and (a; b; c) b e in a triangular situation on T .Isittrue
that the hyp er-interval [a; b; c]isasubtriceofT ?
4. Two Delegates Op eration.
4.1. On linearly ordered sets. Supp ose that X is a linearly ordered set and jX j2.
Let X (2) b e the set of all two p oints subset of X ,thatis,
X (2) = ffx ;x gj x ;x 2 X s:t: x 1 2 1 2 1 2 Supp ose that fa ;a g; fb ;b g2X (2). Let c be a _ b .Thisc is the largest member of 1 2 1 2 2 2 2 2 fa ;a ;b ;b g. And take 1 2 1 2 8 a _ b if a < 2 1 2 2 a _ b if b c = 1 2 2 2 1 : a _ b if a = b 1 1 2 2 Then c is the second largest member of fa ;a ;b ;b g. 1 1 2 1 2 We de ne the binary op eration t on X (2) by fa ;a gtfb ;b g = fc ;c g: 1 2 1 2 1 2 Clearly,(X (2); t) is a semilattice. For example, let X b e a set of students, and we want to select two students as deledats to attend a convention. Some memb ers of committee recommend a and a . The other memb ers of committee recommend b and b . Then, we 1 2 1 2 may select c and c from a , a , b and b according to this metho d. 1 2 1 2 1 2 Next, we de ne the dual op eration of t. Let d be the a ^ b . This d b e the smallest 1 1 1 1 member of fa ;a ;b ;b g. And take 1 2 1 2 8 a ^ b if a < 2 1 1 1 a ^ b if b d = 1 2 1 1 2 : a ^ b if a = b 2 2 1 1 382 K.HORIUCHI Then, d b e the second smallest member of fa ;a ;b ;b g.We de ne the binary op eration 2 1 2 1 2 u on X (2) by fa ;a gufb ;b g = fd ;d g: 1 2 1 2 1 2 This (X (2); u) is also a semilattice. But, (X (2); t; u) is not a lattice! And the op erations t and u do not satisfy the distributivelaw. Wehavetosay \this (X (2); t; u)isnogood algebra"? Now, we de ne another op eration on X (2) by fa ;a gfb ;b g = fd ;c g: 1 2 1 2 1 2 This is the op eration whichchose the smallest memb er and the largest memb er among fa ;a ;b ;b g. Clearly,(X (2); ) is a semilattice. 1 2 1 2 Prop osition 9. The algebra (X (2); t; u; ) is a trice. That is, it satis es the roundab out- absorption law. Pro of It suces to observe the case of four p oints, that is, next Example. Example 9. Let X b e the four p oints linearly ordered set f1; 2; 3; 4g.We denote the two p oints subset f1; 2g, f1; 3g, f1; 4g, f2; 3g, f2; 4g and f3; 4g by a, b, c, d, e and f . Then, the set X (2) is fa; b; c; d; e; f g (see Fig. 11). This (X (2); t; u; ) is a trice. We denote t, u and by % , - and # . From (a % d) - d = a, the algebra (X (2); % ; - )is 1 2 1 2 1 2 3 not a lattice. From (b % e) - (d % e)= c and (b - d) % e = e, the op erations % 1 2 1 2 1 1 and - do not satisfy the distributive. From (e - d) % e = e and e - (d % e)= b, 2 2 1 2 1 that is, the op erations % and - do not satisfy Birkho system. 1 2 1 2 3 a b c c c c - - c c c c c c H H J J J] J] H J^ J^ Hj c c c J c J c c d e J J] J^ c J c c f Figure 11. Example 9 The X (2) of Example 9 has the triangle natural law. But, from (d # e) % (d - e)=e 1 2 3 and d % e = c,theX (2) do es not have the triangle constructivelaw. This is an answer of 1 Question 1. Note that we can consider replacing "two delegates" with "three delegates". Then, we add two op erations, we can construct an algebra with 4-roundab out-absorption law. 4.2. On partially ordered sets. Next, we will try to expand our op erations t and u on a partially ordered set. Supp ose that P is a partially ordered set. And let P (2) b e the set ffx ;x gj x ;x 2 P s:t: x 1 2 1 2 1 2 and the notation akb to denote a and b are incomparable. We can not expand op erations on general partially ordered sets. To de nition of op era- tion t,we add next two conditions on P : (I ) P is a _-semilattice, _ TRICE 383 (I I )if hki for h; i 2 P , then there exists a unique j 2 P such that h _ i j . _ Let fa ;a g, fb ;b g2 P (2). And let j 2 P such that a _ b j . 1 2 1 2 2 2 Now, set 8 fa _ b ;jg if a kb > 2 2 2 2 > > > fa ;jg if a >b and a _ b = a > 2 2 2 1 2 2 > > > fb ;jg if a < 2 2 2 2 1 2 fa _ b ;a g if a >b and a _ b 1 2 2 2 2 1 2 2 1 2 > > fa _ b ;b g if a > 2 1 2 2 2 2 1 2 > > > fa ;jg if a = b and a _ b = a > 2 2 2 1 1 2 > :