Metamath Proof Explorer


Theorem lssn0

Description: A subspace is not empty. (Contributed by NM, 12-Jan-2014) (Revised by Mario Carneiro, 8-Jan-2015)

Ref Expression
Hypothesis lssn0.s
|- S = ( LSubSp ` W )
Assertion lssn0
|- ( U e. S -> U =/= (/) )

Proof

Step Hyp Ref Expression
1 lssn0.s
 |-  S = ( LSubSp ` W )
2 eqid
 |-  ( Scalar ` W ) = ( Scalar ` W )
3 eqid
 |-  ( Base ` ( Scalar ` W ) ) = ( Base ` ( Scalar ` W ) )
4 eqid
 |-  ( Base ` W ) = ( Base ` W )
5 eqid
 |-  ( +g ` W ) = ( +g ` W )
6 eqid
 |-  ( .s ` W ) = ( .s ` W )
7 2 3 4 5 6 1 islss
 |-  ( U e. S <-> ( U C_ ( Base ` W ) /\ U =/= (/) /\ A. x e. ( Base ` ( Scalar ` W ) ) A. a e. U A. b e. U ( ( x ( .s ` W ) a ) ( +g ` W ) b ) e. U ) )
8 7 simp2bi
 |-  ( U e. S -> U =/= (/) )