Metamath Proof Explorer


Theorem 00lss

Description: The empty structure has no subspaces (for use with fvco4i ). (Contributed by Stefan O'Rear, 31-Mar-2015)

Ref Expression
Assertion 00lss = LSubSp

Proof

Step Hyp Ref Expression
1 noel ¬ a
2 base0 = Base
3 eqid LSubSp = LSubSp
4 2 3 lssss a LSubSp a
5 ss0 a a =
6 4 5 syl a LSubSp a =
7 3 lssn0 a LSubSp a
8 7 neneqd a LSubSp ¬ a =
9 6 8 pm2.65i ¬ a LSubSp
10 1 9 2false a a LSubSp
11 10 eqriv = LSubSp