Theorem eqrd 3115

Description: Deduce equality of classes from equivalence of membership. (Contributed by Thierry Arnoux, 21-Mar-2017.)

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.1	. . 3 ⊢ Ⅎ𝑥𝐴
3		eqrd.2	. . 3 ⊢ Ⅎ𝑥𝐵
4		eqrd.3	. . . 4 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
5	4	biimpd 143	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
6	1, 2, 3, 5	ssrd 3102	. 2 ⊢ (𝜑 → 𝐴 ⊆ 𝐵)
7	4	biimprd 157	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴))
8	1, 3, 2, 7	ssrd 3102	. 2 ⊢ (𝜑 → 𝐵 ⊆ 𝐴)
9	6, 8	eqssd 3114	1 ⊢ (𝜑 → 𝐴 = 𝐵)

Syntax hints:   → wi 4   ↔ wb 104   = wceq 1331  Ⅎwnf 1436   ∈ wcel 1480  Ⅎwnfc 2268

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-in 3077  df-ss 3084

This theorem is referenced by:  dfss4st  3309  imasnopn  12468