Metamath Proof Explorer


Theorem chjo

Description: The join of a closed subspace and its orthocomplement is all of Hilbert space. (Contributed by NM, 31-Oct-2005) (New usage is discouraged.)

Ref Expression
Assertion chjo ( 𝐴C → ( 𝐴 ( ⊥ ‘ 𝐴 ) ) = ℋ )

Proof

Step Hyp Ref Expression
1 id ( 𝐴 = if ( 𝐴C , 𝐴 , ℋ ) → 𝐴 = if ( 𝐴C , 𝐴 , ℋ ) )
2 fveq2 ( 𝐴 = if ( 𝐴C , 𝐴 , ℋ ) → ( ⊥ ‘ 𝐴 ) = ( ⊥ ‘ if ( 𝐴C , 𝐴 , ℋ ) ) )
3 1 2 oveq12d ( 𝐴 = if ( 𝐴C , 𝐴 , ℋ ) → ( 𝐴 ( ⊥ ‘ 𝐴 ) ) = ( if ( 𝐴C , 𝐴 , ℋ ) ∨ ( ⊥ ‘ if ( 𝐴C , 𝐴 , ℋ ) ) ) )
4 3 eqeq1d ( 𝐴 = if ( 𝐴C , 𝐴 , ℋ ) → ( ( 𝐴 ( ⊥ ‘ 𝐴 ) ) = ℋ ↔ ( if ( 𝐴C , 𝐴 , ℋ ) ∨ ( ⊥ ‘ if ( 𝐴C , 𝐴 , ℋ ) ) ) = ℋ ) )
5 ifchhv if ( 𝐴C , 𝐴 , ℋ ) ∈ C
6 5 chjoi ( if ( 𝐴C , 𝐴 , ℋ ) ∨ ( ⊥ ‘ if ( 𝐴C , 𝐴 , ℋ ) ) ) = ℋ
7 4 6 dedth ( 𝐴C → ( 𝐴 ( ⊥ ‘ 𝐴 ) ) = ℋ )