Theorem shjval

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

	Ref	Expression
Assertion	shjval	⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( 𝐴 ∨ℋ 𝐵 ) = ( ⊥ ‘ ( ⊥ ‘ ( 𝐴 ∪ 𝐵 ) ) ) )

Proof

Step	Hyp	Ref	Expression
1		shss	⊢ ( 𝐴 ∈ Sℋ → 𝐴 ⊆ ℋ )
2		shss	⊢ ( 𝐵 ∈ Sℋ → 𝐵 ⊆ ℋ )
3		sshjval	⊢ ( ( 𝐴 ⊆ ℋ ∧ 𝐵 ⊆ ℋ ) → ( 𝐴 ∨ℋ 𝐵 ) = ( ⊥ ‘ ( ⊥ ‘ ( 𝐴 ∪ 𝐵 ) ) ) )
4	1 2 3	syl2an	⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( 𝐴 ∨ℋ 𝐵 ) = ( ⊥ ‘ ( ⊥ ‘ ( 𝐴 ∪ 𝐵 ) ) ) )