take notice of

{A 1，

A2，

...,

. AK}

be

A

A division must meet two conditions:

1)∪Ai

=

a；

2)Ai∩Aj

=

Φ

(i≠j).

1) obviously. The following proofs 2):

If there is me,

J, manufacturing

Ai∩Aj

≠

φ, that is, there is

a

Included in the

Ai∩Aj

Yes, so for any

B∈Ai because

A∈Ai, there should be ∈R, but

A∈Aj, knowing that there should also be b∈Aj, therefore

artificial intelligence

Included in the

Aj, it contradicts the topic. Get a license.