(8)Dualization law
((←A,→A)∪(←B,→B))C=(←A,→A)C∪(←B,→B)C((←A,→A)∩(←B,→B))C=(←A,→A)C∪(←B,→B)C
DF set does not satisfy the complementary law. The fundamental reason is that there is no clear boundary of DF set, that is, the boundary of the dynamic fuzzy.
In the transformation of DF set and common set, an important concept is the horizontal section set (←λ,→λ).
Definition 8.4 Assume(←A,→A)∈DF(U), (←λ,→λ)∈[0,1]×[←,→],
(1)←A←A={|←u|←u∈U,←A(←u)≥←λ}or ←A←λ={→u|→u∈U,→A(→u)≥→λ},(0≤(←λ,→λ)<1),we note that (←A←λ,→A→λ)is the horizontal section set(←λ,→λ)of(←A,→A).(2)←A←A={|←u|←u∈U,←A(←u)≥←λ}or ←A←λ={→u|→u∈U,→A(→u)≥→λ},(0≤(←λ,→λ)<1),we note that (←A←λ,→A→λ)is the weak section set(←λ,→λ)of(←A,→A).
According to definition, for ∀(←u,→u)∈U, if (←A(←u),←A(←u))≥(←λ,→λ), then(←u,→u)∈
Property 8.1 Assume (←A,→A),(←B,→B)∈DF(U),
then
((←A,→A)∪(←B,→B))(←A,→A)=(←A,→A)(←A,→A)∪(←B,→B)(←A,→A)((←A,→A)∩(←B,→B))(←A,→A)=(←A,→A)(←A,→A)∩(←B,→B)(←A,→A).
Obviously, for the finite number of DF set in DF(U), these conclusions still hold.
Namely:
(N∪i=1(←Aj,→Ai))(←λ,→λ)=n∪i=1(←Ai←λ,←Ai→λ)(N∩i=1(←Aj,→Ai))(←λ,→λ)=n∩i=1(←Ai←λ,←Ai→λ).
Property 8.2 If (←A←t,→A→t)(←t,→t)∈T⊆DF(U),
∪(←t,→t)∈T(←A←t,→A→t)(←x,→x)=(∪(←t,→t)∈T(←A←t,→A→t))(←x,→x)∩(←t,→t)∈T(←A←t,→A→t)(←x,→x)=(∩(←t,→t)∈T(←A←t,→A→t))(←x,→x).
Property 8.3 Assume (←λ1,→λ1),(←λ2,→λ2)∈[0,1]×[←,→];(←A,→A)∈DF(U).If(←λ2,→λ2)≤
(←A,→A)(←λ2,→λ2)⊃(←A,→A)(←λ1,→λ1).
Property 8.4 Assume
(←A,→A)(∨t∈T(←λt,→λt))=∪t∈T(←A←λt,→A→λt).
According to the concept of horizontal section set the on the last section, we can get the following theorem:
Theorem 8.2 The horizontal section set (←α,→α)
Theorem 8.3 If {(←A,→A)(←t,→t);(←t,→t)∈T}⊂DF(←X,→X),
Proof: If
(1). The rest of the situation is similar to the certificate. [Prove up].
Theorem 8.4 Assume (←A,→A)∈ DF(←X,→X), {(←a←t,→a→t);(←t,→t)∈T}⊂[(←0,→0),(←1,→1)].
where
(←α,→α)=∨(←t,→t)∈T(←α←t,→α→t),(←β,→β)=∨(←t,→t)∈T(←α←t,→α→t).
Proof: According to
(←A→α,→A→α)={(←x,→x);(→A,←A)(←x,→x)≥∨(←t,→t)∈T(←α←t,→α→t)} =∩(←t,→t)∈T(←x,→x);(←A,→A)(←x,→x)≥(←α←t,→α→t) =∩(←t,→t)∈T(←A←α←t,→A→α→t).
Then we know
[Prove up].
(2) It is Similar to permit, so we omitted.
Theorem 8.5 For any (←A,→A)∈DF(←X,→X),
(1)(←A→α,→A→α)=∩(←λ,→λ)<(→α,←α)(←A←λ,→A→λ)(2)(←A→α,→A→α)=∩(←λ,→λ)>(→α,←α)(←A←λ,→A→λ)
Definition 8.5 Assume (←α,→α)∈[(←0,→0),(←1,→1)],(←A,→A)∈DF(←X,→X),
Theorem 8.6 (DF set decomposition theorem) For any
(→A,→A)=∪(→α,←α)∈(←0,→0),(˙1,→1)(←α,→α)(←A,→A)(←α,→α)(→A,→A)=∪(→α,←α)∈(←0,→0),(˙1,→1)(←α,→α)(←A,→A)(←α,→α).
Proof: Since
(←A,→A)(→α,←α)(←x,→x)={(←1,→1)(←x,→x)∈(←A←α,→A→α)(←0,→0)(←x,→x)∈(←A←α,→A→α),
then we have
(∪(←α,→α∈[(←0,→0),(→1,←1)])(←α,→α)(←A,→A)(←α,→α))(←x,→x) =sup(←α,→α)⋅((←A←α,→A→α)(←x,→x)) =sup(→x,→x)∈(←A,→A)(←α,→α)(←α,→α) =sup(←α,→α)∈(←A→x,→A→x)(←α,→α) =(←A,→A)(→x,←x)).