Metamath Proof Explorer


Theorem eleq1d

Description: Deduction from equality to equivalence of membership. (Contributed by NM, 21-Jun-1993) Allow shortening of eleq1 . (Revised by Wolf Lammen, 20-Nov-2019)

Ref Expression
Hypothesis eleq1d.1
|- ( ph -> A = B )
Assertion eleq1d
|- ( ph -> ( A e. C <-> B e. C ) )

Proof

Step Hyp Ref Expression
1 eleq1d.1
 |-  ( ph -> A = B )
2 1 eqeq2d
 |-  ( ph -> ( x = A <-> x = B ) )
3 2 anbi1d
 |-  ( ph -> ( ( x = A /\ x e. C ) <-> ( x = B /\ x e. C ) ) )
4 3 exbidv
 |-  ( ph -> ( E. x ( x = A /\ x e. C ) <-> E. x ( x = B /\ x e. C ) ) )
5 dfclel
 |-  ( A e. C <-> E. x ( x = A /\ x e. C ) )
6 dfclel
 |-  ( B e. C <-> E. x ( x = B /\ x e. C ) )
7 4 5 6 3bitr4g
 |-  ( ph -> ( A e. C <-> B e. C ) )