Theorem eqrd 2957

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

Hypotheses

Ref	Expression
eqrd.0	|- F/xph
eqrd.1	|- F/_xA
eqrd.2	|- F/_xB
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/xph
2		eqrd.1	. . 3 |- F/_xA
3		eqrd.2	. . 3 |- F/_xB
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 2944	. 2 |- (ph -> A C_ B)
7	4	biimprd 147	. . 3 |- (ph -> (x e. B -> x e. A))
8	1, 3, 2, 7	ssrd 2944	. 2 |- (ph -> B C_ A)
9	6, 8	eqssd 2956	1 |- (ph -> A = B)

Syntax hints:   -> wi 4   <-> wb 98   = wceq 1242   F/wnf 1346   e. wcel 1390   F/_wnfc 2162

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 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bndl 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-in 2918  df-ss 2925

This theorem is referenced by: (None)