Metamath Proof Explorer


Theorem eqelb

Description: Substitution of equal classes into elementhood relation. (Contributed by Peter Mazsa, 17-Jul-2019)

Ref Expression
Assertion eqelb
|- ( ( A = B /\ A e. C ) <-> ( A = B /\ B e. C ) )

Proof

Step Hyp Ref Expression
1 simpl
 |-  ( ( B = A /\ A e. C ) -> B = A )
2 eqeltr
 |-  ( ( B = A /\ A e. C ) -> B e. C )
3 1 2 jca
 |-  ( ( B = A /\ A e. C ) -> ( B = A /\ B e. C ) )
4 eqcom
 |-  ( B = A <-> A = B )
5 4 anbi1i
 |-  ( ( B = A /\ A e. C ) <-> ( A = B /\ A e. C ) )
6 4 anbi1i
 |-  ( ( B = A /\ B e. C ) <-> ( A = B /\ B e. C ) )
7 3 5 6 3imtr3i
 |-  ( ( A = B /\ A e. C ) -> ( A = B /\ B e. C ) )
8 simpl
 |-  ( ( A = B /\ B e. C ) -> A = B )
9 eqeltr
 |-  ( ( A = B /\ B e. C ) -> A e. C )
10 8 9 jca
 |-  ( ( A = B /\ B e. C ) -> ( A = B /\ A e. C ) )
11 7 10 impbii
 |-  ( ( A = B /\ A e. C ) <-> ( A = B /\ B e. C ) )