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Metamath Proof Explorer

Theorem ereldm

Description: Equality of equivalence classes implies equivalence of domain membership. (Contributed by NM, 28-Jan-1996) (Revised by Mario Carneiro, 12-Aug-2015)

Ref Expression
Hypotheses ereldm.1 
|- ( ph -> R Er X )
ereldm.2 
|- ( ph -> [ A ] R = [ B ] R )
Assertion ereldm 
|- ( ph -> ( A e. X <-> B e. X ) )

Proof

Step Hyp Ref Expression
1 ereldm.1 
 |-  ( ph -> R Er X )
2 ereldm.2 
 |-  ( ph -> [ A ] R = [ B ] R )
3 2 neeq1d 
 |-  ( ph -> ( [ A ] R =/= (/) <-> [ B ] R =/= (/) ) )
4 ecdmn0 
 |-  ( A e. dom R <-> [ A ] R =/= (/) )
5 ecdmn0 
 |-  ( B e. dom R <-> [ B ] R =/= (/) )
6 3 4 5 3bitr4g 
 |-  ( ph -> ( A e. dom R <-> B e. dom R ) )
7 erdm 
 |-  ( R Er X -> dom R = X )
8 1 7 syl 
 |-  ( ph -> dom R = X )
9 8 eleq2d 
 |-  ( ph -> ( A e. dom R <-> A e. X ) )
10 8 eleq2d 
 |-  ( ph -> ( B e. dom R <-> B e. X ) )
11 6 9 10 3bitr3d 
 |-  ( ph -> ( A e. X <-> B e. X ) )