Metamath Proof Explorer

Theorem 0psubN

Description: The empty set is a projective subspace. Remark below Definition 15.1 of MaedaMaeda p. 61. (Contributed by NM, 13-Oct-2011) (New usage is discouraged.)

Ref Expression
Hypothesis 0psub.s 
|- S = ( PSubSp ` K )
Assertion 0psubN 
|- ( K e. V -> (/) e. S )

Proof

Step Hyp Ref Expression
1 0psub.s 
 |-  S = ( PSubSp ` K )
2 0ss 
 |-  (/) C_ ( Atoms ` K )
3 ral0 
 |-  A. p e. (/) A. q e. (/) A. r e. ( Atoms ` K ) ( r ( le ` K ) ( p ( join ` K ) q ) -> r e. (/) )
4 2 3 pm3.2i 
 |-  ( (/) C_ ( Atoms ` K ) /\ A. p e. (/) A. q e. (/) A. r e. ( Atoms ` K ) ( r ( le ` K ) ( p ( join ` K ) q ) -> r e. (/) ) )
5 eqid 
 |-  ( le ` K ) = ( le ` K )
6 eqid 
 |-  ( join ` K ) = ( join ` K )
7 eqid 
 |-  ( Atoms ` K ) = ( Atoms ` K )
8 5 6 7 1 ispsubsp 
 |-  ( K e. V -> ( (/) e. S <-> ( (/) C_ ( Atoms ` K ) /\ A. p e. (/) A. q e. (/) A. r e. ( Atoms ` K ) ( r ( le ` K ) ( p ( join ` K ) q ) -> r e. (/) ) ) ) )
9 4 8 mpbiri 
 |-  ( K e. V -> (/) e. S )