$\left(\overleftarrow{R},\overrightarrow{R}\right)=\left[\begin{array}{cccc}\left({\text{<mover accent="true"> r <mo>←</mo> </mover>}}_{11},{\text{<mover accent="true"> r <mo>→</mo> </mover>}}_{11}\right)& \left({\text{<mover accent="true"> r <mo>←</mo> </mover>}}_{12},{\text{<mover accent="true"> r <mo>→</mo> </mover>}}_{12}\right)& \dots & \left({\text{<mover accent="true"> r <mo>←</mo> </mover>}}_{1\text{m}},{\text{<mover accent="true"> r <mo>→</mo> </mover>}}_{1\text{m}}\right)\\ \left({\text{<mover accent="true"> r <mo>←</mo> </mover>}}_{21},{\text{<mover accent="true"> r <mo>→</mo> </mover>}}_{21}\right)& \left({\text{<mover accent="true"> r <mo>←</mo> </mover>}}_{22},{\text{<mover accent="true"> r <mo>→</mo> </mover>}}_{22}\right)& \dots & \left({\text{<mover accent="true"> r <mo>←</mo> </mover>}}_{2\text{m}},{\text{<mover accent="true"> r <mo>→</mo> </mover>}}_{2\text{m}}\right)\\ \dots & \dots & \dots & \dots \\ \left({\text{<mover accent="true"> r <mo>←</mo> </mover>}}_{\text{n}1},{\text{<mover accent="true"> r <mo>→</mo> </mover>}}_{\text{n}1}\right)& \left({\text{<mover accent="true"> r <mo>←</mo> </mover>}}_{\text{n2}},{\text{<mover accent="true"> r <mo>→</mo> </mover>}}_{\text{n2}}\right)& \dots & \left({\text{<mover accent="true"> r <mo>←</mo> </mover>}}_{\text{nm}},{\text{<mover accent="true"> r <mo>→</mo> </mover>}}_{\text{nm}}\right)\end{array}\right],$

where, $\left({\stackrel{\leftharpoonup }{r}}_{\text{ij}},{\stackrel{\rightharpoonup }{r}}_{\text{ij}}\right)=\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\left(\left({\stackrel{\leftharpoonup }{x}}_i,\stackrel{\rightharpoonup }{x_i}\right),\left({\stackrel{\leftharpoonup }{y}}_j,{\stackrel{\rightharpoonup }{y}}_j\right)\right)\left(i\le n,j\le m\right).$

Obviously, $\left(\stackrel{\leftharpoonup }{L},\stackrel{\rightharpoonup }{L}\right)$-type DF matrix is $\left(\stackrel{\leftharpoonup }{L},\stackrel{\rightharpoonup }{L}\right)$-type DF relations, and DF matrix is DF relations.

Definition 8.13 Assume $\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)$ is type DF matrix, and $\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)={\left({\stackrel{\leftharpoonup }{r}}_{\text{ij}},\stackrel{\rightharpoonup }{r_{\text{ij}}}\right)}_{n\times m}(i\le$$n,j\le m),\text{if}\text{it exits}\left(\stackrel{\leftharpoonup }{L},\stackrel{\rightharpoonup }{L}\right)$-type DF matrix $\left(\text{<mover accent="true"> Q <mo>↼</mo> </mover> ,}\stackrel{\rightharpoonup }{\text{Q}}\right)={\left({\text{<mover accent="true"> q <mo>↼</mo> </mover>}}_{\text{ji}},\stackrel{\rightharpoonup }{{\text{q}}_{\text{ji}}}\right)}_{m\times n}\left(j\le m,i\le n\right),$ let satisfy $\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\circ \left(\text{<mover accent="true"> Q <mo>↼</mo> </mover> ,}\stackrel{\rightharpoonup }{\text{Q}}\right)\circ \left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)=\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right),\text{then}\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)$ is regular, $\left(\text{<mover accent="true"> Q <mo>↼</mo> </mover> ,}\stackrel{\rightharpoonup }{\text{Q}}\right)$ is the generalized $\left(\stackrel{\leftharpoonup }{L},\stackrel{\rightharpoonup }{L}\right)$-type DF matrix of $\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right).$

If we further have $\left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)\circ \left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\circ \left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)=\left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right),$ is the T inverse matrix of $\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right).$

Theorem 8.10 Assume $\left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)$ is the generalized $\left(\stackrel{\leftharpoonup }{L},\stackrel{\rightharpoonup }{L}\right)$-type DF inverse matrix of $\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right),\text{then}\left(\stackrel{\leftharpoonup }{S},\stackrel{\rightharpoonup }{S}\right)=\left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)\circ \left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\circ \left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)$ is T inverse matrix of $\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right),$ where L is distributive lattice.

Proof: According to Definition 8.13, we can prove $\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\circ \left(\stackrel{\leftharpoonup }{S},\stackrel{\rightharpoonup }{S}\right)\circ \left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)=$$\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\circ \left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)\circ \left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\circ \left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)\circ \left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)=\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\circ \left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)\circ \left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)=\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right).$ Corollaries, $\left(\stackrel{\leftharpoonup }{S},\stackrel{\rightharpoonup }{S}\right)\circ \left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\circ \left(\stackrel{\leftharpoonup }{S},\stackrel{\rightharpoonup }{S}\right)=\left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)\circ \left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\circ \left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)\circ \left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\circ \left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)\circ \left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\circ$$\left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)=\left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)\circ \left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\circ \left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)=\left(\stackrel{\leftharpoonup }{S},\stackrel{\rightharpoonup }{S}\right)$

[Prove up].

Theorem 8.11 Assume that $\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)$ is DF matrix, then the necessary and sufficient condition of $\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)$ is regular is $\left(\overleftarrow{R},\overrightarrow{R}\right)\circ \left(\overleftarrow{X},\overrightarrow{X}\right)\circ \left(\overleftarrow{R},\overrightarrow{R}\right)=\left(\overleftarrow{R},\overrightarrow{R}\right),\text{where,}\left(\overleftarrow{X},\overrightarrow{X}\right)$ is the maximum generalized DF inverse matrix of $\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right).$ Namely, if $\left(\overleftarrow{R},\overrightarrow{R}\right)\circ \left(\overleftarrow{Q},\overrightarrow{Q}\right)\circ$$\left(\overleftarrow{R},\overrightarrow{R}\right)=\left(\overleftarrow{R},\overrightarrow{R}\right),$ then we must have $\left(\overleftarrow{Q},\overrightarrow{Q}\right)\subset \left(\overleftarrow{X},\overrightarrow{X}\right).$

Proof: Sufficiency is obvious, and the necessity is presented as follows. Assume $\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)$ is regular, then it exits $\left(\overleftarrow{Q},\overrightarrow{Q}\right);$ let satisfy $\left(\overleftarrow{R},\overrightarrow{R}\right)\circ \left(\overleftarrow{Q},\overrightarrow{Q}\right)\circ \left(\overleftarrow{R},\overrightarrow{R}\right)=\left(\overleftarrow{R},\overrightarrow{R}\right).$

Therefore,

$\underset{k=1}{\overset{n}{\vee }}\left(\underset{j=1}{\overset{m}{\vee }}\left(\left({\text{<mover accent="true"> r <mo>←</mo> </mover>}}_{ie},{\text{<mover accent="true"> r <mo>←</mo> </mover>}}_{ie}\right)\Lambda \left({\text{<mover accent="true"> q <mo>←</mo> </mover>}}_{ik},{\text{<mover accent="true"> q <mo>←</mo> </mover>}}_{ik}\right)\right)\Lambda \left({\overleftarrow{r}}_{ik},{\overrightarrow{r}}_{kj}\right)\right)=\left({\overleftarrow{r}}_{ij},{\overrightarrow{r}}_{ij}\right)\left(i\le n,j\le m\right).$

Namely,

Thus, for $\forall k\le n,e\le m,\text{we}\text{have}\left(\left({\overleftarrow{r}}_{ie},{\overrightarrow{r}}_{ie}\right)\wedge \left({\overleftarrow{\text{q}}}_{{}_{ek},}{\overrightarrow{\text{q}}}_{ek}\right)\wedge \left({\overleftarrow{r}}_{kj,}{\overrightarrow{r}}_{kj}\right)\right)\le r_{ij}$$\left(i\le n,j\le m\right).$ We hence have

$\begin{array}{l}\left(\left({\overleftarrow{r}}_{ie},{\overleftarrow{r}}_{ie}\right)\Lambda \left({\overleftarrow{r}}_{kj},{\overleftarrow{r}}_{kj}\right)\right)\le \left({\overleftarrow{r}}_{ij},{\overleftarrow{r}}_{ij}\right)\Rightarrow 0\le \left({\overleftarrow{q}}_{ek},{\overrightarrow{q}}_{ek}\right)\le 1;\left(\left({\overleftarrow{r}}_{ie},{\overleftarrow{r}}_{ie}\right)\Lambda \left({\overleftarrow{r}}_{kj},{\overleftarrow{r}}_{kj}\right)\right)>\\ \left({\overleftarrow{r}}_{ij},{\overleftarrow{r}}_{ij}\right)\Rightarrow 0\le \left({\overleftarrow{q}}_{ek},{\overrightarrow{q}}_{ek}\right)\le \left({\overleftarrow{r}}_{ij},{\overleftarrow{r}}_{ij}\right).\end{array}$

Then, $0\le \left({\overleftarrow{q}}_{ek,}{\overrightarrow{q}}_{ek}\right)\le \wedge \left\{\left({\overleftarrow{\text{r}}}_{st},{\overrightarrow{\text{r}}}_{st}\right),\left({\overleftarrow{\text{r}}}_{st},{\overrightarrow{\text{r}}}_{st}\right)<\left(\left({\text{<mover accent="true"> r <mo>↼</mo> </mover>}}_{st},\stackrel{\rightharpoonup }{{\text{r}}_{st}}\right)\wedge \left({\text{<mover accent="true"> r <mo>↼</mo> </mover>}}_{nt},\stackrel{\rightharpoonup }{{\text{r}}_{nt}}\right)\right)\right\}=$$\left({\stackrel{\leftharpoonup }{x}}_{ek},{\stackrel{\rightharpoonup }{x}}_{ek}\right),\text{Namely,}\left(\stackrel{\leftharpoonup }{Q},\stackrel{\rightharpoonup }{Q}\right)\subset \left(\stackrel{\leftharpoonup }{X},\stackrel{\rightharpoonup }{X}\right)\mathrm{and}\left(\stackrel{\leftharpoonup }{R},\stackrel{\rightharpoonup }{R}\right)\circ \left(\overleftarrow{Q},\overrightarrow{Q}\right)\circ \left(\overleftarrow{R},\overrightarrow{R}\right)\subset \left(\overleftarrow{R},\overrightarrow{R}\right)\circ \left(\overleftarrow{X},\overrightarrow{X}\right)\circ$$\left(\overleftarrow{R},\overrightarrow{R}\right).$

Assume $\left(\overleftarrow{U},\overrightarrow{U}\right)=\left(\overleftarrow{R},\overrightarrow{R}\right)\circ \left(\overleftarrow{X},\overrightarrow{X}\right)\circ \left(\overleftarrow{R},\overrightarrow{R}\right),$ then

$u_{ij}=\underset{k=1}{\overset{n}{\vee }}\underset{e=1}{\overset{m}{\vee }}\left(\left({\overleftarrow{r}}_{ie},{\overleftarrow{r}}_{ie}\right)\Lambda \left({\overrightarrow{x}}_{ek},{\overleftarrow{x}}_{ek}\right)\Lambda \left({\overrightarrow{x}}_{ke},{\overleftarrow{x}}_{ke}\right)\right)\left(i\le n,j\le m\right).$

We can prove uij ≤ n, ≤ j ≤ m).

In fact, when $u_{\text{ij}}\le r_{\text{ij}}\left(i\le n,j\le m\right).$ we have $\left(\left({\overleftarrow{\text{r}}}_{ie},{\overrightarrow{\text{r}}}_{ie}\right)\wedge \left({\overleftarrow{\text{r}}}_{ei},\overrightarrow{{\text{r}}_{\text{ej}}}\right)\right)\le \left({\overleftarrow{\text{r}}}_{\text{ij}},{\overrightarrow{\text{r}}}_{\text{ij}}\right),$$(\left({\overleftarrow{\text{r}}}_{ie},{\overrightarrow{\text{r}}}_{ie}\right)\wedge \left({\overleftarrow{x}}_{ik},{\overrightarrow{x}}_{ik}\right)\wedge$ Therefore, $\left({\overleftarrow{u}}_{ij},{\overrightarrow{u}}_{ij}\right)\le \left({\overleftarrow{\text{r}}}_{\text{ij}},\overrightarrow{{\text{r}}_{\text{ij}}}\right)\left(i\le n,i\le m\right),$ namely, $\left(\overleftarrow{R},\overrightarrow{R}\right)\circ \left(\overleftarrow{X},\overrightarrow{X}\right)\circ \left(\overleftarrow{R},\overrightarrow{R}\right)\subset \left(\overleftarrow{R},\overrightarrow{R}\right)=$$\left(\overleftarrow{R},\overrightarrow{R}\right)\circ \left(\overleftarrow{Q},\overrightarrow{Q}\right)\circ \left(\overleftarrow{R},\overrightarrow{R}\right);\text{hence,}\left(\overleftarrow{R},\overrightarrow{R}\right)\circ \left(\overleftarrow{X},\overrightarrow{X}\right)\circ \left(\overleftarrow{R},\overrightarrow{R}\right)=\left(\overleftarrow{R},\overrightarrow{R}\right).$

If $\left(\overleftarrow{Q^{\prime }},\overrightarrow{Q^{\prime }}\right)$ satisfies $\left(\overleftarrow{R},\overrightarrow{R}\right)\circ \left(\overleftarrow{Q^{\prime }},\overrightarrow{Q^{\prime }}\right)\circ \left(\overleftarrow{R},\overrightarrow{R}\right)=\left(\overleftarrow{R},\overrightarrow{R}\right)\mathrm{and}\left(\overleftarrow{R},\overrightarrow{R}\right)\subset \left(\overleftarrow{X},\overrightarrow{X}\right),$ then $\left(\overleftarrow{X},\overrightarrow{X}\right)$ is the maximum generalized DF inverse matrix of $\left(\overleftarrow{R},\overrightarrow{R}\right)\circ \left(\overleftarrow{Q},\overrightarrow{Q}\right)\circ \left(\overleftarrow{R},\overrightarrow{R}\right)=$$\left(\overleftarrow{R},\overrightarrow{R}\right).$

[Prove up].