Metamath Proof Explorer


Theorem mdsym

Description: M-symmetry of the Hilbert lattice. Lemma 5 of Maeda p. 168. (Contributed by NM, 6-Jul-2004) (New usage is discouraged.)

Ref Expression
Assertion mdsym ( ( 𝐴C𝐵C ) → ( 𝐴 𝑀 𝐵𝐵 𝑀 𝐴 ) )

Proof

Step Hyp Ref Expression
1 breq1 ( 𝐴 = if ( 𝐴C , 𝐴 , ℋ ) → ( 𝐴 𝑀 𝐵 ↔ if ( 𝐴C , 𝐴 , ℋ ) 𝑀 𝐵 ) )
2 breq2 ( 𝐴 = if ( 𝐴C , 𝐴 , ℋ ) → ( 𝐵 𝑀 𝐴𝐵 𝑀 if ( 𝐴C , 𝐴 , ℋ ) ) )
3 1 2 bibi12d ( 𝐴 = if ( 𝐴C , 𝐴 , ℋ ) → ( ( 𝐴 𝑀 𝐵𝐵 𝑀 𝐴 ) ↔ ( if ( 𝐴C , 𝐴 , ℋ ) 𝑀 𝐵𝐵 𝑀 if ( 𝐴C , 𝐴 , ℋ ) ) ) )
4 breq2 ( 𝐵 = if ( 𝐵C , 𝐵 , ℋ ) → ( if ( 𝐴C , 𝐴 , ℋ ) 𝑀 𝐵 ↔ if ( 𝐴C , 𝐴 , ℋ ) 𝑀 if ( 𝐵C , 𝐵 , ℋ ) ) )
5 breq1 ( 𝐵 = if ( 𝐵C , 𝐵 , ℋ ) → ( 𝐵 𝑀 if ( 𝐴C , 𝐴 , ℋ ) ↔ if ( 𝐵C , 𝐵 , ℋ ) 𝑀 if ( 𝐴C , 𝐴 , ℋ ) ) )
6 4 5 bibi12d ( 𝐵 = if ( 𝐵C , 𝐵 , ℋ ) → ( ( if ( 𝐴C , 𝐴 , ℋ ) 𝑀 𝐵𝐵 𝑀 if ( 𝐴C , 𝐴 , ℋ ) ) ↔ ( if ( 𝐴C , 𝐴 , ℋ ) 𝑀 if ( 𝐵C , 𝐵 , ℋ ) ↔ if ( 𝐵C , 𝐵 , ℋ ) 𝑀 if ( 𝐴C , 𝐴 , ℋ ) ) ) )
7 ifchhv if ( 𝐴C , 𝐴 , ℋ ) ∈ C
8 ifchhv if ( 𝐵C , 𝐵 , ℋ ) ∈ C
9 7 8 mdsymi ( if ( 𝐴C , 𝐴 , ℋ ) 𝑀 if ( 𝐵C , 𝐵 , ℋ ) ↔ if ( 𝐵C , 𝐵 , ℋ ) 𝑀 if ( 𝐴C , 𝐴 , ℋ ) )
10 3 6 9 dedth2h ( ( 𝐴C𝐵C ) → ( 𝐴 𝑀 𝐵𝐵 𝑀 𝐴 ) )