Theorem eqri 3958

Description: Infer equality of classes from equivalence of membership. (Contributed by Thierry Arnoux, 7-Oct-2017.)

Hypotheses

Ref	Expression
eqri.1	⊢ Ⅎ𝑥𝐴
eqri.2	⊢ Ⅎ𝑥𝐵
eqri.3	⊢ (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)

Assertion

Ref	Expression
eqri	⊢ 𝐴 = 𝐵

Proof of Theorem eqri

Step	Hyp	Ref	Expression
1		nftru 1837	. . 3 ⊢ Ⅎ𝑥⊤
2		eqri.1	. . 3 ⊢ Ⅎ𝑥𝐴
3		eqri.2	. . 3 ⊢ Ⅎ𝑥𝐵
4		eqri.3	. . . 4 ⊢ (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)
5	4	a1i 11	. . 3 ⊢ (⊤ → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
6	1, 2, 3, 5	eqrd 3957	. 2 ⊢ (⊤ → 𝐴 = 𝐵)
7	6	mptru 1577	1 ⊢ 𝐴 = 𝐵

This proof depends on syntax axioms:   ↔ wb 209   = wceq 1570  ⊤wtru 1571   ∈ wcel 2146  Ⅎwnfc 2912

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 2148  ax-9 2156  ax-11 2195  ax-12 2216  ax-ext 2737

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 2757  df-clel 2840  df-nfc 2914

This theorem is used by:  iunab  5018  iinab  5034  rnep  5919  difrab2  32915  esum2dlem  34546  eulerpartlemn  34836