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 ‘ ∅ )