Metamath Proof Explorer


Theorem iscss2

Description: It is sufficient to prove that the double orthocomplement is a subset of the target set to show that the set is a closed subspace. (Contributed by Mario Carneiro, 13-Oct-2015)

Ref Expression
Hypotheses cssss.v ⊢ 𝑉 = ( Base ‘ 𝑊 )
cssss.c ⊢ 𝐶 = ( ClSubSp ‘ 𝑊 )
ocvcss.o ⊢ ⊥ = ( ocv ‘ 𝑊 )
Assertion iscss2 ( ( 𝑊 ∈ PreHil ∧ 𝑆 ⊆ 𝑉 ) → ( 𝑆 ∈ 𝐶 ↔ ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ⊆ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 cssss.v ⊢ 𝑉 = ( Base ‘ 𝑊 )
2 cssss.c ⊢ 𝐶 = ( ClSubSp ‘ 𝑊 )
3 ocvcss.o ⊢ ⊥ = ( ocv ‘ 𝑊 )
4 3 2 iscss ⊢ ( 𝑊 ∈ PreHil → ( 𝑆 ∈ 𝐶 ↔ 𝑆 = ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ) )
5 4 adantr ⊢ ( ( 𝑊 ∈ PreHil ∧ 𝑆 ⊆ 𝑉 ) → ( 𝑆 ∈ 𝐶 ↔ 𝑆 = ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ) )
6 1 3 ocvocv ⊢ ( ( 𝑊 ∈ PreHil ∧ 𝑆 ⊆ 𝑉 ) → 𝑆 ⊆ ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) )
7 eqss ⊢ ( 𝑆 = ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ↔ ( 𝑆 ⊆ ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ∧ ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ⊆ 𝑆 ) )
8 7 baib ⊢ ( 𝑆 ⊆ ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) → ( 𝑆 = ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ↔ ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ⊆ 𝑆 ) )
9 6 8 syl ⊢ ( ( 𝑊 ∈ PreHil ∧ 𝑆 ⊆ 𝑉 ) → ( 𝑆 = ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ↔ ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ⊆ 𝑆 ) )
10 5 9 bitrd ⊢ ( ( 𝑊 ∈ PreHil ∧ 𝑆 ⊆ 𝑉 ) → ( 𝑆 ∈ 𝐶 ↔ ( ⊥ ‘ ( ⊥ ‘ 𝑆 ) ) ⊆ 𝑆 ) )