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Metamath Proof Explorer

Theorem chlejb1

Description: Hilbert lattice ordering in terms of join. (Contributed by NM, 30-Jun-2004) (New usage is discouraged.)

Ref Expression
Assertion chlejb1 ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Cℋ ) → ( 𝐴 ⊆ 𝐵 ↔ ( 𝐴 ∨ℋ 𝐵 ) = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 sseq1 ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → ( 𝐴 ⊆ 𝐵 ↔ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ 𝐵 ) )
2 oveq1 ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → ( 𝐴 ∨ℋ 𝐵 ) = ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∨ℋ 𝐵 ) )
3 2 eqeq1d ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → ( ( 𝐴 ∨ℋ 𝐵 ) = 𝐵 ↔ ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∨ℋ 𝐵 ) = 𝐵 ) )
4 1 3 bibi12d ⊢ ( 𝐴 = if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) → ( ( 𝐴 ⊆ 𝐵 ↔ ( 𝐴 ∨ℋ 𝐵 ) = 𝐵 ) ↔ ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ 𝐵 ↔ ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∨ℋ 𝐵 ) = 𝐵 ) ) )
5 sseq2 ⊢ ( 𝐵 = if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) → ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ 𝐵 ↔ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) ) )
6 oveq2 ⊢ ( 𝐵 = if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) → ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∨ℋ 𝐵 ) = ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∨ℋ if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) ) )
7 id ⊢ ( 𝐵 = if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) → 𝐵 = if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) )
8 6 7 eqeq12d ⊢ ( 𝐵 = if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) → ( ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∨ℋ 𝐵 ) = 𝐵 ↔ ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∨ℋ if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) ) = if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) ) )
9 5 8 bibi12d ⊢ ( 𝐵 = if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) → ( ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ 𝐵 ↔ ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∨ℋ 𝐵 ) = 𝐵 ) ↔ ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) ↔ ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∨ℋ if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) ) = if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) ) ) )
10 h0elch ⊢ 0ℋ ∈ Cℋ
11 10 elimel ⊢ if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∈ Cℋ
12 10 elimel ⊢ if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) ∈ Cℋ
13 11 12 chlejb1i ⊢ ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ⊆ if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) ↔ ( if ( 𝐴 ∈ Cℋ , 𝐴 , 0ℋ ) ∨ℋ if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) ) = if ( 𝐵 ∈ Cℋ , 𝐵 , 0ℋ ) )
14 4 9 13 dedth2h ⊢ ( ( 𝐴 ∈ Cℋ ∧ 𝐵 ∈ Cℋ ) → ( 𝐴 ⊆ 𝐵 ↔ ( 𝐴 ∨ℋ 𝐵 ) = 𝐵 ) )