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Metamath Proof Explorer

Theorem eqrrabd

Description: Deduce equality with a restricted abstraction. (Contributed by Thierry Arnoux, 11-Apr-2024)

Ref Expression
Hypotheses eqrrabd.1 
|- ( ph -> B C_ A )
eqrrabd.2 
|- ( ( ph /\ x e. A ) -> ( x e. B <-> ps ) )
Assertion eqrrabd 
|- ( ph -> B = { x e. A | ps } )

Proof

Step Hyp Ref Expression
1 eqrrabd.1 
 |-  ( ph -> B C_ A )
2 eqrrabd.2 
 |-  ( ( ph /\ x e. A ) -> ( x e. B <-> ps ) )
3 nfv 
 |-  F/ x ph
4 nfcv 
 |-  F/_ x B
5 nfrab1 
 |-  F/_ x { x e. A | ps }
6 1 sseld 
 |-  ( ph -> ( x e. B -> x e. A ) )
7 6 pm4.71rd 
 |-  ( ph -> ( x e. B <-> ( x e. A /\ x e. B ) ) )
8 2 pm5.32da 
 |-  ( ph -> ( ( x e. A /\ x e. B ) <-> ( x e. A /\ ps ) ) )
9 7 8 bitrd 
 |-  ( ph -> ( x e. B <-> ( x e. A /\ ps ) ) )
10 rabid 
 |-  ( x e. { x e. A | ps } <-> ( x e. A /\ ps ) )
11 9 10 bitr4di 
 |-  ( ph -> ( x e. B <-> x e. { x e. A | ps } ) )
12 3 4 5 11 eqrd 
 |-  ( ph -> B = { x e. A | ps } )