Theorem eqrd 3219

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 3206	. 2 |- ( ph -> A C_ B )
7	4	biimprd 158	. . 3 |- ( ph -> ( x e. B -> x e. A ) )
8	1, 3, 2, 7	ssrd 3206	. 2 |- ( ph -> B C_ A )
9	6, 8	eqssd 3218	1 |- ( ph -> A = B )

Syntax hints:   -> wi 4   <-> wb 105   = wceq 1373   F/wnf 1484   e. wcel 2178   F/_wnfc 2337

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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2189

This theorem depends on definitions:  df-bi 117  df-nf 1485  df-sb 1787  df-clab 2194  df-cleq 2200  df-clel 2203  df-nfc 2339  df-in 3180  df-ss 3187

This theorem is referenced by:  dfss4st  3414  imasnopn  14886