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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 ∈ 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 ) ) ) )