Theorem eqrd 4003

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 2213	. 2 ⊢ (𝜑 → ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
4		eqrd.1	. . 3 ⊢ Ⅎ𝑥𝐴
5		eqrd.2	. . 3 ⊢ Ⅎ𝑥𝐵
6	4, 5	cleqf 2934	. 2 ⊢ (𝐴 = 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
7	3, 6	sylibr 234	1 ⊢ (𝜑 → 𝐴 = 𝐵)

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1538   = wceq 1540  Ⅎwnf 1783   ∈ wcel 2108  Ⅎwnfc 2890

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-11 2157  ax-12 2177  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-ex 1780  df-nf 1784  df-cleq 2729  df-clel 2816  df-nfc 2892

This theorem is referenced by:  eqri  4004  eqrrabd  4086  sniota  6552  fimarab  6983  dissnlocfin  23537  imasnopn  23698  imasncld  23699  imasncls  23700  blval2  24575  ofpreima  32675  algextdeglem6  33763  constrfin  33787  zarcls  33873  ordtconnlem1  33923  qqhval2  33983  reprdifc  34642  topdifinfindis  37347  icorempo  37352  isbasisrelowllem1  37356  isbasisrelowllem2  37357  sticksstones11  42157  areaquad  43228  rfcnpre1  45024  rfcnpre2  45036  preimagelt  46714  preimalegt  46715