Metamath Proof Explorer


Theorem dmecd

Description: Equality of the coset of B and the coset of C implies equivalence of domain elementhood (equivalence is not necessary as opposed to ereldm ). (Contributed by Peter Mazsa, 9-Oct-2018)

Ref Expression
Hypotheses dmecd.1
|- ( ph -> dom R = A )
dmecd.2
|- ( ph -> [ B ] R = [ C ] R )
Assertion dmecd
|- ( ph -> ( B e. A <-> C e. A ) )

Proof

Step Hyp Ref Expression
1 dmecd.1
 |-  ( ph -> dom R = A )
2 dmecd.2
 |-  ( ph -> [ B ] R = [ C ] R )
3 2 neeq1d
 |-  ( ph -> ( [ B ] R =/= (/) <-> [ C ] R =/= (/) ) )
4 ecdmn0
 |-  ( B e. dom R <-> [ B ] R =/= (/) )
5 ecdmn0
 |-  ( C e. dom R <-> [ C ] R =/= (/) )
6 3 4 5 3bitr4g
 |-  ( ph -> ( B e. dom R <-> C e. dom R ) )
7 1 eleq2d
 |-  ( ph -> ( B e. dom R <-> B e. A ) )
8 1 eleq2d
 |-  ( ph -> ( C e. dom R <-> C e. A ) )
9 6 7 8 3bitr3d
 |-  ( ph -> ( B e. A <-> C e. A ) )