Theorem eqrd 3018

Description: Deduce equality of classes from equivalence of membership. (Contributed by Thierry Arnoux, 21-Mar-2017.)

Hypotheses

Ref	Expression
eqrd.0	|- F/ x ph
eqrd.1	|- F/_ x A
eqrd.2	|- F/_ x B
eqrd.3	|- ( ph -> ( x e. A <-> x e. B ) )

Assertion

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

Proof of Theorem eqrd

Step	Hyp	Ref	Expression
1		eqrd.0	. . 3 |- F/ x ph
2		eqrd.1	. . 3 |- F/_ x A
3		eqrd.2	. . 3 |- F/_ x B
4		eqrd.3	. . . 4 |- ( ph -> ( x e. A <-> x e. B ) )
5	4	biimpd 142	. . 3 |- ( ph -> ( x e. A -> x e. B ) )
6	1, 2, 3, 5	ssrd 3005	. 2 |- ( ph -> A C_ B )
7	4	biimprd 156	. . 3 |- ( ph -> ( x e. B -> x e. A ) )
8	1, 3, 2, 7	ssrd 3005	. 2 |- ( ph -> B C_ A )
9	6, 8	eqssd 3017	1 |- ( ph -> A = B )

Syntax hints:   -> wi 4   <-> wb 103   = wceq 1285   F/wnf 1390   e. wcel 1434   F/_wnfc 2207

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2064

This theorem depends on definitions:  df-bi 115  df-nf 1391  df-sb 1687  df-clab 2069  df-cleq 2075  df-clel 2078  df-nfc 2209  df-in 2980  df-ss 2987

This theorem is referenced by: (None)