Metamath Proof Explorer


Theorem sshjcl

Description: Closure of join for subsets of Hilbert space. (Contributed by NM, 1-Nov-2000) (New usage is discouraged.)

Ref Expression
Assertion sshjcl ( ( 𝐴 ⊆ ℋ ∧ 𝐵 ⊆ ℋ ) → ( 𝐴 𝐵 ) ∈ C )

Proof

Step Hyp Ref Expression
1 sshjval ( ( 𝐴 ⊆ ℋ ∧ 𝐵 ⊆ ℋ ) → ( 𝐴 𝐵 ) = ( ⊥ ‘ ( ⊥ ‘ ( 𝐴𝐵 ) ) ) )
2 unss ( ( 𝐴 ⊆ ℋ ∧ 𝐵 ⊆ ℋ ) ↔ ( 𝐴𝐵 ) ⊆ ℋ )
3 ocss ( ( 𝐴𝐵 ) ⊆ ℋ → ( ⊥ ‘ ( 𝐴𝐵 ) ) ⊆ ℋ )
4 occl ( ( ⊥ ‘ ( 𝐴𝐵 ) ) ⊆ ℋ → ( ⊥ ‘ ( ⊥ ‘ ( 𝐴𝐵 ) ) ) ∈ C )
5 3 4 syl ( ( 𝐴𝐵 ) ⊆ ℋ → ( ⊥ ‘ ( ⊥ ‘ ( 𝐴𝐵 ) ) ) ∈ C )
6 2 5 sylbi ( ( 𝐴 ⊆ ℋ ∧ 𝐵 ⊆ ℋ ) → ( ⊥ ‘ ( ⊥ ‘ ( 𝐴𝐵 ) ) ) ∈ C )
7 1 6 eqeltrd ( ( 𝐴 ⊆ ℋ ∧ 𝐵 ⊆ ℋ ) → ( 𝐴 𝐵 ) ∈ C )