Theorem eqrd 3210

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 144	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
6	1, 2, 3, 5	ssrd 3197	. 2 ⊢ (𝜑 → 𝐴 ⊆ 𝐵)
7	4	biimprd 158	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴))
8	1, 3, 2, 7	ssrd 3197	. 2 ⊢ (𝜑 → 𝐵 ⊆ 𝐴)
9	6, 8	eqssd 3209	1 ⊢ (𝜑 → 𝐴 = 𝐵)

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1372  Ⅎwnf 1482   ∈ wcel 2175  Ⅎwnfc 2334

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1469  ax-7 1470  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-10 1527  ax-11 1528  ax-i12 1529  ax-bndl 1531  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557  ax-ext 2186

This theorem depends on definitions:  df-bi 117  df-nf 1483  df-sb 1785  df-clab 2191  df-cleq 2197  df-clel 2200  df-nfc 2336  df-in 3171  df-ss 3178

This theorem is referenced by:  dfss4st  3405  imasnopn  14713