Metamath Proof Explorer


Theorem eleq12d

Description: Deduction from equality to equivalence of membership. (Contributed by NM, 31-May-1994)

Ref Expression
Hypotheses eleq12d.1
|- ( ph -> A = B )
eleq12d.2
|- ( ph -> C = D )
Assertion eleq12d
|- ( ph -> ( A e. C <-> B e. D ) )

Proof

Step Hyp Ref Expression
1 eleq12d.1
 |-  ( ph -> A = B )
2 eleq12d.2
 |-  ( ph -> C = D )
3 2 eleq2d
 |-  ( ph -> ( A e. C <-> A e. D ) )
4 1 eleq1d
 |-  ( ph -> ( A e. D <-> B e. D ) )
5 3 4 bitrd
 |-  ( ph -> ( A e. C <-> B e. D ) )