Theorem eqrd 3782

Description: Deduce equality of classes from equivalence of membership. (Contributed by Thierry Arnoux, 21-Mar-2017.) (Proof shortened by BJ, 1-Dec-2021.)

Hypotheses

Ref	Expression
eqrd.0	⊢ Ⅎ𝑥𝜑
eqrd.1	⊢ Ⅎ𝑥𝐴
eqrd.2	⊢ Ⅎ𝑥𝐵
eqrd.3	⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))

Assertion

Ref	Expression
eqrd	⊢ (𝜑 → 𝐴 = 𝐵)

Proof of Theorem eqrd

Step	Hyp	Ref	Expression
1		eqrd.0	. . 3 ⊢ Ⅎ𝑥𝜑
2		eqrd.3	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
3	1, 2	alrimi 2246	. 2 ⊢ (𝜑 → ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
4		eqrd.1	. . 3 ⊢ Ⅎ𝑥𝐴
5		eqrd.2	. . 3 ⊢ Ⅎ𝑥𝐵
6	4, 5	cleqf 2933	. 2 ⊢ (𝐴 = 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
7	3, 6	sylibr 225	1 ⊢ (𝜑 → 𝐴 = 𝐵)

Syntax hints:   → wi 4   ↔ wb 197  ∀wal 1650   = wceq 1652  Ⅎwnf 1878   ∈ wcel 2155  Ⅎwnfc 2894

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1890  ax-4 1904  ax-5 2005  ax-6 2070  ax-7 2105  ax-9 2164  ax-10 2183  ax-11 2198  ax-12 2211  ax-13 2352  ax-ext 2743

This theorem depends on definitions:  df-bi 198  df-an 385  df-or 874  df-tru 1656  df-ex 1875  df-nf 1879  df-sb 2063  df-cleq 2758  df-clel 2761  df-nfc 2896

This theorem is referenced by:  sniota  6060  dissnlocfin  21628  imasnopn  21789  imasncld  21790  imasncls  21791  blval2  22662  eqri  29794  fimarab  29918  ofpreima  29938  ordtconnlem1  30440  qqhval2  30496  reprdifc  31179  topdifinfindis  33650  icorempt2  33655  isbasisrelowllem1  33659  isbasisrelowllem2  33660  areaquad  38504  rfcnpre1  39856  rfcnpre2  39868