Theorem eqrd 3950

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 1568   = wceq 1570  Ⅎwnf 1816   ∈ wcel 2145  Ⅎwnfc 2907

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-11 2194  ax-12 2213  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-cleq 2752  df-clel 2835  df-nfc 2909

This theorem is used by:  eqri  3951  eqrrabd  4034  sniota  6524  fimarab  6953  dissnlocfin  23756  imasnopn  23917  imasncld  23918  imasncls  23919  blval2  24789  ofpreima  33139  algextdeglem6  34233  constrfin  34257  zarcls  34385  ordtconnlem1  34435  qqhval2  34493  reprdifc  35136  topdifinfindis  38101  icorempo  38106  isbasisrelowllem1  38110  isbasisrelowllem2  38111  sticksstones11  43023  areaquad  44058  rfcnpre1  45854  rfcnpre2  45866  preimagelt  47528  preimalegt  47529