Theorem 3.2. If
a
Eb, then we must get an IFN
c
, which satisfies
a
c
Eb.
Proof. Let
c
be an IFN that satisfies
c
Eb
a
, then by Theorem 3.1,
we have
a
c
Eb
a
a
¼ b
which completes the proof. h
Theorem 3.3. If
c
E
a
Eb, then
a
c
Eb
c
.
Proof. We can get ð0; 1ÞEðb
a
Þ, and by Theorem 3.1, we have
a
c
Eð
a
c
Þðb
a
Þ¼b
c
Theorem 3.4. If
a
Eb satisfies
l
b
– 1 and
m
b
– 0, then k
a
Ekb and
kb k
a
¼ kðb
a
Þ when k P 0.
Proof.
kb k
a
¼ 1 ð1
l
b
Þ
k
;
m
k
b
1 ð1
l
a
Þ
k
;
m
k
a
¼ 1
1
l
b
1
l
a
k
;
m
b
m
a
k
!
¼ 1 1
l
b
l
a
1
l
a
k
;
m
b
m
a
k
!
¼ kðb
a
Þ
Then ka kðb aÞ¼kb, and kaEkb is also proven. h
Theorem 3.5. If
l
a
– 1 and
m
a
– 0, then k
1
a
Ek
2
a
and
k
2
a
k
1
a
¼ðk
2
k
1
Þ
a
when 0 6 k
1
6 k
2
.
Proof.
k
2
a
k
1
a
¼ 1 ð1
l
Þ
k
2
;
m
k
2
1 ð1
l
Þ
k
1
;
m
k
1
¼ 1 ð1
l
Þ
k
2
k
1
;
m
k
2
k
1
¼ðk
2
k
1
Þ
a
That is, k
1
a ðk
2
k
1
Þa ¼ k
2
a,sok
1
aEk
2
a. h
Theorem 3.6. If
a
Eb, which satisfies
l
b
– 1 and
m
b
– 0, then
k
1
a
Ek
2
b when 0 6 k
1
6 k
2
.
Proof. By Theorems 3.5 and 3.6, we can know k
1
a
Ek
2
a
Ek
2
b and
k
1
a
Ek
1
bEk
2
b.Sok
1
a
Ek
2
b is proven. h
Theorem 3.7.
(1) If b
1
E
a
1
and b
2
E
a
2
, then b
1
b
2
E
a
1
a
2
.
(2) If bE
a
, then we can get an IFN
c
, which satisfies bE
a
c
.
(3) If
c
E
bE
a
, then
c
ø
a
E
c
øb.
(4) If bE
a
satisfies
l
b
– 0 and
m
b
– 1, then b
k
E
a
k
and
b
k
ø
a
k
¼ðbø
a
Þ
k
when k P 0.
(5) If
l
a
– 0 and
m
a
– 1, then
a
k
2
E
a
k
1
and
a
k
2
ø
a
k
1
¼
a
ðk
2
k
1
Þ
when 0 6 k
1
6 k
2
.
(6) If bE
a
satisfies
l
b
– 0 and
m
b
– 1, then b
k
2
E
a
k
1
when
0 6 k
1
6 k
2
.
Proof.
(1) Because b
1
E
a
1
and b
2
E
a
2
, then there are two IFNs
c
1
and
c
2
that satisfy
a
1
c
1
¼ b
1
, and
a
2
c
2
¼ b
2
. Thus, we have
ð
a
1
a
2
Þð
c
1
c
2
Þ¼ð
a
1
c
1
Þð
a
2
c
2
Þ¼b
1
b
2
i.e., b
1
b
2
E
a
1
a
2
.
(2) Let
c
be an IFN that satisfies bø
a
E
c
, then by (1), we have
b ¼ðbø
a
Þ
a
E
c
a
which completes the proof.
(3) If
c
E
bE
a
, then bø
a
E
ð1; 0Þ, by which we get
c
ø
a
¼ð
c
øbÞðbø
a
ÞE
ð
c
øbÞð1; 0Þ¼ð
c
øbÞ
(4) If bE
a
and k P 0, then
b
k
ø
a
k
¼
l
k
b
; 1 ð1
m
b
Þ
k
ø
l
k
a
; 1 ð1
m
a
Þ
k
¼
l
b
l
a
k
; 1
1
m
b
1
m
a
k
!
¼ðbø
a
Þ
k
That is, b
k
¼ a
k
ðbøaÞ
k
and b
k
E
a
k
.
(5) If 0 6 k
1
6 k
2
, then
a
k
2
ø
a
k
1
¼
l
k
2
a
; 1 ð1
m
a
Þ
k
2
ø
l
k
1
a
; 1 ð1
m
a
Þ
k
1
¼
l
k
2
k
1
a
; 1 ð1
m
a
Þ
k
2
k
1
¼
a
ðk
2
k
1
Þ
which means that a
k
2
¼ a
k
1
a
ðk
2
k
1
Þ
and a
k
2
E
a
k
1
.
(6) If 0 6 k
1
6 k
2
and bE
a
, then we know by (4) and (5) that
b
k
2
E
a
k
2
E
a
k
1
or b
k
2
E
b
k
1
E
a
k
1
.
So b
k
2
E
a
k
1
. This completes the proof of Theorem 3.7. h
We can express S
ð
a
;
u
Þ¼fbj
a
Eb and
u
ð
a
ÞE
u
ðbÞg by using
the new order, and S
ð
a
;
u
Þ can also be defined as S
ð
a
;
u
Þ
¼fbjbE
a
and
u
ðbÞE
u
ð
a
Þg.
Now we discuss some properties of the subtraction and division
of IFNs as follows:
Theorem 3.8. Let
a
1
;
a
2
and
a
3
be three IFNs, which satisfy
a
3
E
a
2
E
a
1
. Then there are the following conclusions:
(1) ð
a
1
a
2
Þð
a
2
a
3
Þ¼
a
1
a
3
.
(2) ð
a
1
a
3
Þð
a
2
a
3
Þ¼
a
1
a
2
.
(3) ð
a
1
a
2
Þð
a
2
a
3
Þ¼
a
1
a
3
.
Proof.
(1) By those operations between IFNs given in Section 2, we can
get
ð
a
1
a
2
Þð
a
2
a
3
Þ¼ð1ð1
l
1
Þð1
l
2
Þ;
m
1
m
2
Þð1ð1
l
2
Þ
ð1
l
3
Þ;
m
2
m
3
Þ
¼
ð1
l
2
Þð1
l
3
Þð1
l
1
Þð1
l
2
Þ
ð1
l
2
Þð1
l
3
Þ
;
m
1
m
2
m
2
m
3
¼ 1
1
l
1
1
l
3
;
m
1
m
3
¼
l
1
l
3
1
l
3
;
m
1
m
3
¼
a
1
a
3
(2) In a similar way to (1), we have
4 Q. Lei, Z. Xu / Knowledge-Based Systems 76 (2015) 1–16
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