Theorem eqri 3975

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 1804	. . 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 3974	. 2 ⊢ (⊤ → 𝐴 = 𝐵)
7	6	mptru 1547	1 ⊢ 𝐴 = 𝐵

Syntax hints:   ↔ wb 206   = wceq 1540  ⊤wtru 1541   ∈ wcel 2109  Ⅎwnfc 2878

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 2008  ax-8 2111  ax-9 2119  ax-11 2158  ax-12 2178  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-nf 1784  df-cleq 2722  df-clel 2804  df-nfc 2880

This theorem is referenced by:  rnep  5898  difrab2  32434  esum2dlem  34090  eulerpartlemn  34380