Metamath Proof Explorer


Theorem indifbi

Description: Two ways to express equality relative to a class A . (Contributed by Thierry Arnoux, 23-Jun-2024)

Ref Expression
Assertion indifbi
|- ( ( A i^i B ) = ( A i^i C ) <-> ( A \ B ) = ( A \ C ) )

Proof

Step Hyp Ref Expression
1 inss1
 |-  ( A i^i B ) C_ A
2 inss1
 |-  ( A i^i C ) C_ A
3 rcompleq
 |-  ( ( ( A i^i B ) C_ A /\ ( A i^i C ) C_ A ) -> ( ( A i^i B ) = ( A i^i C ) <-> ( A \ ( A i^i B ) ) = ( A \ ( A i^i C ) ) ) )
4 1 2 3 mp2an
 |-  ( ( A i^i B ) = ( A i^i C ) <-> ( A \ ( A i^i B ) ) = ( A \ ( A i^i C ) ) )
5 difin
 |-  ( A \ ( A i^i B ) ) = ( A \ B )
6 difin
 |-  ( A \ ( A i^i C ) ) = ( A \ C )
7 5 6 eqeq12i
 |-  ( ( A \ ( A i^i B ) ) = ( A \ ( A i^i C ) ) <-> ( A \ B ) = ( A \ C ) )
8 4 7 bitri
 |-  ( ( A i^i B ) = ( A i^i C ) <-> ( A \ B ) = ( A \ C ) )