Metamath Proof Explorer


Theorem shjval

Description: Value of join in SH . (Contributed by NM, 9-Aug-2000) (New usage is discouraged.)

Ref Expression
Assertion shjval ⊢ A ∈ S ℋ ∧ B ∈ S ℋ → A ∨ ℋ B = ⊥ ⁡ ⊥ ⁡ A ∪ B

Proof

Step Hyp Ref Expression
1 shss ⊢ A ∈ S ℋ → A ⊆ ℋ
2 shss ⊢ B ∈ S ℋ → B ⊆ ℋ
3 sshjval ⊢ A ⊆ ℋ ∧ B ⊆ ℋ → A ∨ ℋ B = ⊥ ⁡ ⊥ ⁡ A ∪ B
4 1 2 3 syl2an ⊢ A ∈ S ℋ ∧ B ∈ S ℋ → A ∨ ℋ B = ⊥ ⁡ ⊥ ⁡ A ∪ B