Theorem eqrd 3202

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 144	. . 3 |- ( ph -> ( x e. A -> x e. B ) )
6	1, 2, 3, 5	ssrd 3189	. 2 |- ( ph -> A C_ B )
7	4	biimprd 158	. . 3 |- ( ph -> ( x e. B -> x e. A ) )
8	1, 3, 2, 7	ssrd 3189	. 2 |- ( ph -> B C_ A )
9	6, 8	eqssd 3201	1 |- ( ph -> A = B )

Syntax hints:   -> wi 4   <-> wb 105   = wceq 1364   F/wnf 1474   e. wcel 2167   F/_wnfc 2326

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-in 3163  df-ss 3170

This theorem is referenced by:  dfss4st  3397  imasnopn  14619