Theorem eqrd 3115

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 143	. . 3 |- ( ph -> ( x e. A -> x e. B ) )
6	1, 2, 3, 5	ssrd 3102	. 2 |- ( ph -> A C_ B )
7	4	biimprd 157	. . 3 |- ( ph -> ( x e. B -> x e. A ) )
8	1, 3, 2, 7	ssrd 3102	. 2 |- ( ph -> B C_ A )
9	6, 8	eqssd 3114	1 |- ( ph -> A = B )

Syntax hints:   -> wi 4   <-> wb 104   = wceq 1331   F/wnf 1436   e. wcel 1480   F/_wnfc 2268

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-in 3077  df-ss 3084

This theorem is referenced by:  dfss4st  3309  imasnopn  12468