Theorem eqrd 3928

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 2249	. 2 ⊢ (𝜑 → ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
4		eqrd.1	. . 3 ⊢ Ⅎ𝑥𝐴
5		eqrd.2	. . 3 ⊢ Ⅎ𝑥𝐵
6	4, 5	cleqf 2950	. 2 ⊢ (𝐴 = 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
7	3, 6	sylibr 237	1 ⊢ (𝜑 → 𝐴 = 𝐵)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567   = wceq 1569  Ⅎwnf 1815   ∈ wcel 2144  Ⅎwnfc 2907

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-cleq 2752  df-clel 2835  df-nfc 2909

This theorem is used by:  eqri  3929  eqrrabd  4010  sniota  6534  fimarab  6964  dissnlocfin  23857  imasnopn  24018  imasncld  24019  imasncls  24020  blval2  24890  ofpreima  33310  algextdeglem6  34404  constrfin  34428  zarcls  34556  ordtconnlem1  34606  qqhval2  34664  reprdifc  35306  topdifinfindis  38406  icorempo  38411  isbasisrelowllem1  38415  isbasisrelowllem2  38416  sticksstones11  43343  areaquad  44355  rfcnpre1  46172  rfcnpre2  46184  preimagelt  47845  preimalegt  47846