Theorem eqrrabd 4010

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 1946	. 2 ⊢ Ⅎ𝑥𝜑
2		nfcv 2922	. 2 ⊢ Ⅎ𝑥𝐵
3		nfrab1 3431	. 2 ⊢ Ⅎ𝑥{𝑥 ∈ 𝐴 ∣ 𝜓}
4		eqrrabd.1	. . . . . 6 ⊢ (𝜑 → 𝐵 ⊆ 𝐴)
5	4	sseld 3908	. . . . 5 ⊢ (𝜑 → (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴))
6	5	pm4.71rd 572	. . . 4 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)))
7		eqrrabd.2	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝑥 ∈ 𝐵 ↔ 𝜓))
8	7	pm5.32da 590	. . . 4 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝜓)))
9	6, 8	bitrd 282	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ (𝑥 ∈ 𝐴 ∧ 𝜓)))
10		rabid 3432	. . 3 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜓} ↔ (𝑥 ∈ 𝐴 ∧ 𝜓))
11	9, 10	bitr4di 292	. 2 ⊢ (𝜑 → (𝑥 ∈ 𝐵 ↔ 𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜓}))
12	1, 2, 3, 11	eqrd 3928	1 ⊢ (𝜑 → 𝐵 = {𝑥 ∈ 𝐴 ∣ 𝜓})

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569   ∈ wcel 2144  {crab 3412   ⊆ wss 3896

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-rab 3413  df-ss 3897

This theorem is used by:  tmachlem-agreeprod  48054  usgrexmpl2nb0  49236  usgrexmpl2nb1  49237  usgrexmpl2nb2  49238  usgrexmpl2nb3  49239  usgrexmpl2nb4  49240  usgrexmpl2nb5  49241