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 ( ( 𝐴C𝐵C ) → ( 𝐴𝐵 ↔ ( 𝐴 𝐵 ) = 𝐵 ) )

Proof

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