Theorem eqri 3987

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 1805	. . 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 3986	. 2 ⊢ (⊤ → 𝐴 = 𝐵)
7	6	mptru 1544	1 ⊢ 𝐴 = 𝐵

Syntax hints:   ↔ wb 208   = wceq 1537  ⊤wtru 1538   ∈ wcel 2114  Ⅎwnfc 2961

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-11 2161  ax-12 2177  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1540  df-ex 1781  df-nf 1785  df-cleq 2814  df-clel 2893  df-nfc 2963

This theorem is referenced by:  rnep  5797  difrab2  30261  esum2dlem  31351  eulerpartlemn  31639