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 2250	. 2 ⊢ (𝜑 → ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
4		eqrd.1	. . 3 ⊢ Ⅎ𝑥𝐴
5		eqrd.2	. . 3 ⊢ Ⅎ𝑥𝐵
6	4, 5	cleqf 2951	. 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 2908

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 2733

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 2753  df-clel 2836  df-nfc 2910

This theorem is used by:  eqri  3951  eqrrabd  4034  sniota  6529  fimarab  6959  dissnlocfin  23848  imasnopn  24009  imasncld  24010  imasncls  24011  blval2  24881  ofpreima  33259  algextdeglem6  34354  constrfin  34378  zarcls  34506  ordtconnlem1  34556  qqhval2  34614  reprdifc  35256  topdifinfindis  38269  icorempo  38274  isbasisrelowllem1  38278  isbasisrelowllem2  38279  sticksstones11  43206  areaquad  44217  rfcnpre1  46035  rfcnpre2  46047  preimagelt  47708  preimalegt  47709