Theorem eqrdav 2352

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

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 198	. . . . 5 |- ((ph /\ x e. C) -> (x e. A -> x e. B))
4	3	impancom 427	. . . 4 |- ((ph /\ x e. A) -> (x e. C -> x e. B))
5	1, 4	mpd 14	. . 3 |- ((ph /\ x e. A) -> x e. B)
6		eqrdav.2	. . . 4 |- ((ph /\ x e. B) -> x e. C)
7	2	exbiri 605	. . . . . 6 |- (ph -> (x e. C -> (x e. B -> x e. A)))
8	7	com23 72	. . . . 5 |- (ph -> (x e. B -> (x e. C -> x e. A)))
9	8	imp 418	. . . 4 |- ((ph /\ x e. B) -> (x e. C -> x e. A))
10	6, 9	mpd 14	. . 3 |- ((ph /\ x e. B) -> x e. A)
11	5, 10	impbida 805	. 2 |- (ph -> (x e. A <-> x e. B))
12	11	eqrdv 2351	1 |- (ph -> A = B)

Syntax hints:   -> wi 4   <-> wb 176   /\ wa 358   = wceq 1642   e. wcel 1710

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-an 360  df-cleq 2346

This theorem is referenced by: (None)