Assume (←X,→X)={(←x,→x),...,(←xn,→xn)t},(←Y,→Y)={(←y1,→y1),...(←ym,→ym)},((←L,→L),(X←,X→)={(x←,x→),...,(x←n,x→n)t},(Y←,Y→)={(y←1,y→1),...(y←m,y→m)},((L←,L→), ≤) is a DF grid, and that 0 and 1 respectively represent the minimum and maximum elements of (←L,→L).(L←,L→).
Definition 6.7 If (←R,→R):(←X,→X)×(←Y,→Y)→(←L,→L),(R←,R→):(X←,X→)×(Y←,Y→)→(L←,L→), we say that (←R,→R)(R←,R→) is an (←L,→L)(L←,L→)-type DF matrix, written as
(←R,→R)=((←r11,→r11)(←r12,→r12)⋯(←r1m,→r1m)(←r21,→r21)(←r22,→r22)⋮(←r2m,→r2m)⋮⋮⋮⋮(←rn1,→rn1)(←rn2,→rn2)…(←rnm,→rnm)),(R←,R→)=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜(r←11,r→11)(r←21,r→21)⋮(r←n1,r→n1)(r←12,r→12)(r←22,r→22)⋮(r←n2,r→n2)⋯⋮⋮…(r←1m,r→1m)(r←2m,r→2m)⋮(r←nm,r→nm)⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟,
where (←rij,→rij)=(←R,→R)((←xi,→xi),(←yj,→yj))(i≤n, j≤m).(r←ij,r→ij)=(R←,R→)((x←i,x→i),(y←j,y→j))(i≤n, j≤m).
In particular, (←R,→R):(←X,→X)×(←Y,→Y)→[0,1]×[←,→](R←,R→):(X←,X→)×(Y←,Y→)→[0,1]×[←,→] is a DF matrix.
Obviously, an L-type DF matrix is an L-type DF relationship, and a DF matrix is a DF relationship.
Write DFL((←X,→X)×(←Y,→Y))={(←R,→R);(←R,→R):(←X,→X)×(←Y,→Y)→(←L,→L)}.DFL((X←,X→)×(Y←,Y→))={(R←,R→);(R←,R→):(X←,X→)×(Y←,Y→)→(L←,L→)}. If (←R,→R),(←S,→S)∈DFL((←X,→X)×(←Y,→Y)),(←R,→R)=(←rij,→rij),(←S,→S)=(←sij,→sij).(R←,R→),(S←,S→)∈DFL((X←,X→)×(Y←,Y→)),(R←,R→)=(r←ij,r→ij),(S←,S→)=(s←ij,s→ij). If
∀i≤n,j≤m, (←rij,→rij)≤(←sij,→sij),(←R,→R)∀i≤n,j≤m, (r←ij,r→ij)≤(s←ij,s→ij),(R←,R→) is contained in (←S,→S),(S←,S→), which we write as (←R,→R)⊂(←S,→S).(R←,R→)⊂(S←,S→).
Obviously, DFL(((←X,→X)×(←Y,→Y)),⊂)DFL(((X←,X→)×(Y←,Y→)),⊂) is a poset.
Assume (L,S,T,N) is a module system, and write
(←R,→R)∪(←S,→S)=((←S,→S)(←rij,→rij),(←sij,→sij));i≤n,j≤m(←R,→R)∪(←S,→S)=((←S,→S)(←rij,→rij),(←sij,→sij));i≤n,j≤m (←R,→R)c=(←N,→N)((→rij,→rij);l≤n,j≤m).(R←,R→)∪(S←,S→)=((S←,S→)(r←ij,r→ij),(sij←−,s→ij));i≤n,j≤m(R←,R→)∪(S←,S→)=((S←,S→)(r←ij,r→ij),(sij←−,s→ij));i≤n,j≤m (R,←−R→)c=(N←,N→)((r→ij,r→ij);l≤n,j≤m).
Then, DFL(((←X,→X)×(←Y,→Y)),∩,∪,c)DFL(((X←,X→)×(Y←,Y→)),∩,∪,c) is also a module system, (←0,→0)=((←oij,→oij))=(0←,0→)=((o←ij,o→ij))=(←0,→0);i≤n,j≤m) and (←E,→E)=((←eij,→eij)=(←1,→1);i≤n,j≤m)(0←,0→);i≤n,j≤m) and (E←,E→)=((e←ij,e→ij)=(1←,1→);i≤n,j≤m) are the minimum element and maximum element of DFL((←X,→X)×(←Y,→Y)).DFL((X←,X→)×(Y←,Y→)).
Definition 6.8 Assume (←R,→R)(R←,R→) is an L-type DF matrix and (←R,→R)=((←rij,→rij);(R←,R→)=((r←ij,r→ij);i≤n, j≤m).i≤n, j≤m). If there exists an L-type DF matrix (←Q,→Q)=((←qij,→qij); j≤m,i≤n)(Q←,Q→)=((q←ij,q→ij); j≤m,i≤n) such that (←R,→R)∘(←Q,→Q)∘(←R,→R)=(←R,→R),we call (←R,→R)(R←,R→)∘(Q←,Q→)∘(R←,R→)=(R←,R→),we call (R←,R→) regular and say that (←Q,→Q)(Q←,Q→) is the general type-T matrix of
If there is some (←Q,→Q)∘(←R,→R)∘(←Q,→Q)=(←Q,→Q), then (←Q,→Q)(Q←,Q→)∘(R←,R→)∘(Q←,Q→)=(Q←,Q→), then (Q←,Q→) is the T inverse matrix of (←R,→R)(R←,R→)
Theorem 6.1 Assume (←Q,→Q)(Q←,Q→) is the general T inverse matrix of (←R,→R). Then, (←S,→S)=(R←,R→). Then, (S←,S→)=(←Q,→Q)∘(←R,→R)∘(←Q,→Q)(Q←,Q→)∘(R←,R→)∘(Q←,Q→) is the T inverse matrix of (←R,→R)(R←,R→) where T is a distribution grid.
Proof: According to Definition 6.8,
(←R,→R)∘(←S,→S)∘(←R,→R)=(←R,→R)∘(←Q,→Q)∘(←R,→R)∘(←Q,→Q)∘(←R,→R) =(←R,→R)∘(←Q,→Q)∘(←R,→R)∘(←R,→R).(R←,R→)∘(S←,S→)∘(R←,R→)=(R←,R→)∘(Q←,Q→)∘(R←,R→)∘(Q←,Q→)∘(R←,R→) =(R←,R→)∘(Q←,Q→)∘(R←,R→)∘(R←,R→).
Thus,
(←S,→S)∘(←R,→R)∘(←S,→S)=(←Q,→Q)∘(←R,→R)∘(←Q,→Q)∘(←R,→R)∘(←Q,→Q)∘(←R,→R)∘(←Q,→Q) =(←Q,→Q)∘(←R,→R)∘(←Q,→Q)=(←S,→S),(S←,S→)∘(R←,R→)∘(S←,S→)=(Q←,Q→)∘(R←,R→)∘(Q←,Q→)∘(R←,R→)∘(Q←,Q→)∘(R←,R→)∘(Q←,Q→) =(Q←,Q→)∘(R←,R→)∘(Q←,Q→)=(S←,S→),
which completes the proof.
Theorem 6.2 Assume (←R,→R)(R←,R→) is a DF matrix, and the sufficient and necessary condition for (←R,→R)(R←,R→) to be regular is
(←R,→R)∘(←X,→X)∘(←R,→R)=(←R,→R).(R←,R→)∘(X←,X→)∘(R←,R→)=(R←,R→).
Now, (←X,→X)(X←,X→) is the biggest general DF inverse matrix. When (←R,→R)∘(←Q,→Q)∘(←R,→R)=(R←,R→)∘(Q←,Q→)∘(R←,R→)=(←R,→R), then (←Q,→Q)⊂(←X,→X).(R←,R→), then (Q←,Q→)⊂(X←,X→).
Proof: The sufficient condition is obviously true, so we need to prove the necessary condition.
Assume (←R,→R)(R←,R→) is regular and that there exists some (←Q,→Q)(Q←,Q→) satisfying (←R,→R)∘(R←,R→)∘(←Q,→Q)∘(←R,→R)=(←R,→R).(Q←,Q→)∘(R←,R→)=(R←,R→). Thus, there is
n∨k=1(m∨j=1((←rie,→rie)∧(←qik,→qik))∧(←rkj,→rkj)=(←rij,→rij)(i≤n,j≤m),∨k=1n(∨j=1m((r←ie,r→ie)∧(q←ik,q→ik))∧(r←kj,r→kj)=(r←ij,r→ij)(i≤n,j≤m),
that is,
n∨k=1(m∨e=1((←rie,→rie)∧(←qek,→qek))∧(←rkj,→rkj)=(←rij,→rij)(i≤n,j≤m).∨k=1n(∨e=1m((r←ie,r→ie)∧(q←ek,q→ek))∧(r←kj,r→kj)=(r←ij,r→ij)(i≤n,j≤m).
Thus, for ∀k≤n, e≤m,∀k≤n, e≤m, there is ((←rie,→rie)∧(←qik,←qik))∧(←rkj,→rkj)≤(←rij,→rij)(i≤n,((r←ie,r→ie)∧(q←ik,q←ik))∧(r←kj,r→kj)≤(r←ij,r→ij)(i≤n, j ≧ m).
We can conclude the following:
1. ((←rie,→rie)∧(←rkj,→rkj))≤(←rij,→rij)⇒0≤(←qek,→qel)≤1;2. ((←rie,→rie)∧(←rkj,→rkj))>(←rij,→rij)⇒0≤(←qek,→qel)≤(←rij,→rij)1. ((r←ie,r→ie)∧(r←kj,r→kj))≤(r←ij,r→ij)⇒0≤(q←ek,q→el)≤1;2. ((r←ie,r→ie)∧(r←kj,r→kj))>(r←ij,r→ij)⇒0≤(q←ek,q→el)≤(r←ij,r→ij)
Hence, 0≤(←qek,→qek)≤{(←rst,→rst); (←rst,→rst)<((←rst,→rst)∧(←rnt,→rnt))}=0≤(q←ek,q→ek)≤{(r←st,r→st); (r←st,r→st)<((r←st,r→st)∧(r←nt,r→nt))}=(←xek,→xek);(x←ek,x→ek); that is, (←Q,→Q)⊂(←X,→X) and (←R,→R)∘(←Q,→Q)∘(←R,→R)⊂(←R,→R)∘(←X,→X)∘(Q←,Q→)⊂(X←,X→) and (R←,R→)∘(Q←,Q→)∘(R←,R→)⊂(R←,R→)∘(X←,X→)∘(←R,→R)(R←,R→)
Assuming that (←U,→U)=(←R,→R)∘(←X,→X)∘(←R,→R)(U←,U→)=(R←,R→)∘(X←,X→)∘(R←,R→) and
uij=n∨k=1n∨e=1((←rie,→rie)∧(←xek,→xek)∧(←rke,→rke)) (i≤n,j≤m),uij=∨k=1n∨e=1n((r←ie,r→ie)∧(x←ek,x→ek)∧(r←ke,r→ke)) (i≤n,j≤m),
Thus, we can prove that uij ≤ rij(i ≤ n, ≤ j ≤ m) is true.
In fact, when ((←rie,→rie)∧(←rie,→rie))≤(←rij,→rij)((r←ie,r→ie)∧(r←ie,r→ie))≤(r←ij,r→ij) is true, we have