Theorem eqrd 2963

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 132	. . 3 |- ( ph -> ( x e. A -> x e. B ) )
6	1, 2, 3, 5	ssrd 2950	. 2 |- ( ph -> A C_ B )
7	4	biimprd 147	. . 3 |- ( ph -> ( x e. B -> x e. A ) )
8	1, 3, 2, 7	ssrd 2950	. 2 |- ( ph -> B C_ A )
9	6, 8	eqssd 2962	1 |- ( ph -> A = B )

Syntax hints:   -> wi 4   <-> wb 98   = wceq 1243   F/wnf 1349   e. wcel 1393   F/_wnfc 2165

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-in 2924  df-ss 2931

This theorem is referenced by: (None)