Theorem eqrdav 2621

Description: Deduce equality of classes from an equivalence of membership that depends on the membership variable. (Contributed by NM, 7-Nov-2008.) (Proof shortened by Wolf Lammen, 19-Nov-2019.)

Hypotheses

Ref	Expression
eqrdav.1	|- ( ( ph /\ x e. A ) -> x e. C )
eqrdav.2	|- ( ( ph /\ x e. B ) -> x e. C )
eqrdav.3	|- ( ( ph /\ x e. C ) -> ( x e. A <-> x e. B ) )

Assertion

Ref	Expression
eqrdav	|- ( ph -> A = B )

Distinct variable groups:   x, A   x, B   ph, x

Allowed substitution hint:   C( x)

Proof of Theorem eqrdav

Step	Hyp	Ref	Expression
1		eqrdav.1	. . . 4 |- ( ( ph /\ x e. A ) -> x e. C )
2		eqrdav.3	. . . . . 6 |- ( ( ph /\ x e. C ) -> ( x e. A <-> x e. B ) )
3	2	biimpd 219	. . . . 5 |- ( ( ph /\ x e. C ) -> ( x e. A -> x e. B ) )
4	3	impancom 456	. . . 4 |- ( ( ph /\ x e. A ) -> ( x e. C -> x e. B ) )
5	1, 4	mpd 15	. . 3 |- ( ( ph /\ x e. A ) -> x e. B )
6		eqrdav.2	. . . 4 |- ( ( ph /\ x e. B ) -> x e. C )
7	2	biimprd 238	. . . . 5 |- ( ( ph /\ x e. C ) -> ( x e. B -> x e. A ) )
8	7	impancom 456	. . . 4 |- ( ( ph /\ x e. B ) -> ( x e. C -> x e. A ) )
9	6, 8	mpd 15	. . 3 |- ( ( ph /\ x e. B ) -> x e. A )
10	5, 9	impbida 877	. 2 |- ( ph -> ( x e. A <-> x e. B ) )
11	10	eqrdv 2620	1 |- ( ph -> A = B )

Syntax hints:   -> wi 4   <-> wb 196   /\ wa 384   = wceq 1483   e. wcel 1990

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-an 386  df-ex 1705  df-cleq 2615

This theorem is referenced by:  boxcutc  7951  supminf  11775  f1omvdconj  17866  fmucndlem  22095  ballotlemsima  30577  supminfxr  39694