Theorem eqrrabd 31741

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 1918	. 2 ⊢ Ⅎ𝑥𝜑
2		nfcv 2904	. 2 ⊢ Ⅎ𝑥𝐵
3		nfrab1 3452	. 2 ⊢ Ⅎ𝑥{𝑥 ∈ 𝐴 ∣ 𝜓}
4		eqrrabd.1	. . . . . 6 ⊢ (𝜑 → 𝐵 ⊆ 𝐴)
5	4	sseld 3982	. . . . 5 ⊢ (𝜑 → (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴))
6	5	pm4.71rd 564	. . . 4 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)))
7		eqrrabd.2	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝑥 ∈ 𝐵 ↔ 𝜓))
8	7	pm5.32da 580	. . . 4 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝜓)))
9	6, 8	bitrd 279	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∧ 𝜓)))
10		rabid 3453	. . 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 4002	1 ⊢ (𝜑 → 𝐵 = {𝑥 ∈ 𝐴 ∣ 𝜓})

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 397   = wceq 1542   ∈ wcel 2107  {crab 3433   ⊆ wss 3949

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-10 2138  ax-11 2155  ax-12 2172  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-tru 1545  df-ex 1783  df-nf 1787  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-nfc 2886  df-rab 3434  df-v 3477  df-in 3956  df-ss 3966

This theorem is referenced by: (None)