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 ∨ ℋ B ∈ C ℋ

Proof

Step Hyp Ref Expression
1 sshjval ⊢ A ⊆ ℋ ∧ B ⊆ ℋ → A ∨ ℋ B = ⊥ ⁡ ⊥ ⁡ A ∪ B
2 unss ⊢ A ⊆ ℋ ∧ B ⊆ ℋ ↔ A ∪ B ⊆ ℋ
3 ocss ⊢ A ∪ B ⊆ ℋ → ⊥ ⁡ A ∪ B ⊆ ℋ
4 occl ⊢ ⊥ ⁡ A ∪ B ⊆ ℋ → ⊥ ⁡ ⊥ ⁡ A ∪ B ∈ C ℋ
5 3 4 syl ⊢ A ∪ B ⊆ ℋ → ⊥ ⁡ ⊥ ⁡ A ∪ B ∈ C ℋ
6 2 5 sylbi ⊢ A ⊆ ℋ ∧ B ⊆ ℋ → ⊥ ⁡ ⊥ ⁡ A ∪ B ∈ C ℋ
7 1 6 eqeltrd ⊢ A ⊆ ℋ ∧ B ⊆ ℋ → A ∨ ℋ B ∈ C ℋ