Metamath Proof Explorer


Theorem chj0

Description: Join with Hilbert lattice zero. (Contributed by NM, 22-Jun-2004) (New usage is discouraged.)

Ref Expression
Assertion chj0 ( 𝐴 ∈ Cℋ → ( 𝐴 ∨ℋ 0ℋ ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 oveq1 ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → ( 𝐴 ∨ℋ 0ℋ ) = ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∨ℋ 0ℋ ) )
2 id ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) )
3 1 2 eqeq12d ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → ( ( 𝐴 ∨ℋ 0ℋ ) = 𝐴 ↔ ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∨ℋ 0ℋ ) = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ) )
4 h0elch ⊢ 0ℋ ∈ Cℋ
5 4 elimel ⊢ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∈ Cℋ
6 5 chj0i ⊢ ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∨ℋ 0ℋ ) = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ )
7 3 6 dedth ⊢ ( 𝐴 ∈ Cℋ → ( 𝐴 ∨ℋ 0ℋ ) = 𝐴 )