Theorem eqrrabd 4068

Description: Deduce equality with a restricted abstraction. (Contributed by Thierry Arnoux, 11-Apr-2024.)

Hypotheses

Ref	Expression
eqrrabd.1	⊢ (𝜑 → 𝐵 ⊆ 𝐴)
eqrrabd.2	⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝑥 ∈ 𝐵 ↔ 𝜓))

Assertion

Ref	Expression
eqrrabd	⊢ (𝜑 → 𝐵 = {𝑥 ∈ 𝐴 ∣ 𝜓})

Distinct variable groups:   𝑥,𝐵   𝜑,𝑥

Allowed substitution hints:   𝜓(𝑥)   𝐴(𝑥)

Proof of Theorem eqrrabd

Step	Hyp	Ref	Expression
1		nfv 1913	. 2 ⊢ Ⅎ𝑥𝜑
2		nfcv 2897	. 2 ⊢ Ⅎ𝑥𝐵
3		nfrab1 3441	. 2 ⊢ Ⅎ𝑥{𝑥 ∈ 𝐴 ∣ 𝜓}
4		eqrrabd.1	. . . . . 6 ⊢ (𝜑 → 𝐵 ⊆ 𝐴)
5	4	sseld 3964	. . . . 5 ⊢ (𝜑 → (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴))
6	5	pm4.71rd 562	. . . 4 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)))
7		eqrrabd.2	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝑥 ∈ 𝐵 ↔ 𝜓))
8	7	pm5.32da 579	. . . 4 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝜓)))
9	6, 8	bitrd 279	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∧ 𝜓)))
10		rabid 3442	. . 3 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜓} ↔ (𝑥 ∈ 𝐴 ∧ 𝜓))
11	9, 10	bitr4di 289	. 2 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ 𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜓}))
12	1, 2, 3, 11	eqrd 3985	1 ⊢ (𝜑 → 𝐵 = {𝑥 ∈ 𝐴 ∣ 𝜓})

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1539   ∈ wcel 2107  {crab 3420   ⊆ wss 3933

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-10 2140  ax-11 2156  ax-12 2176  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1542  df-ex 1779  df-nf 1783  df-sb 2064  df-clab 2713  df-cleq 2726  df-clel 2808  df-nfc 2884  df-rab 3421  df-ss 3950

This theorem is referenced by:  usgrexmpl2nb0  47936  usgrexmpl2nb1  47937  usgrexmpl2nb2  47938  usgrexmpl2nb3  47939  usgrexmpl2nb4  47940  usgrexmpl2nb5  47941