Theorem eqrd 3197

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 3184	. 2 ⊢ (𝜑 → 𝐴 ⊆ 𝐵)
7	4	biimprd 158	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴))
8	1, 3, 2, 7	ssrd 3184	. 2 ⊢ (𝜑 → 𝐵 ⊆ 𝐴)
9	6, 8	eqssd 3196	1 ⊢ (𝜑 → 𝐴 = 𝐵)

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1364  Ⅎwnf 1471   ∈ wcel 2164  Ⅎwnfc 2323

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 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-in 3159  df-ss 3166

This theorem is referenced by:  dfss4st  3392  imasnopn  14467