Theorem elrab2w 3653

Description: Membership in a restricted class abstraction. This is to elrab2 3652 what elab2gw 3635 is to elab2g 3637. (Contributed by SN, 2-Sep-2024.)

Hypotheses

Ref	Expression
elrab2w.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
elrab2w.2	⊢ (𝑦 = 𝐴 → (𝜓 ↔ 𝜒))
elrab2w.3	⊢ 𝐶 = {𝑥 ∈ 𝐵 ∣ 𝜑}

Assertion

Ref	Expression
elrab2w	⊢ (𝐴 ∈ 𝐶 ↔ (𝐴 ∈ 𝐵 ∧ 𝜒))

Distinct variable groups:   𝑦,𝐴   𝑥,𝐵,𝑦   𝜑,𝑦   𝜒,𝑦   𝜓,𝑥

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑦)   𝜒(𝑥)   𝐴(𝑥)   𝐶(𝑥, 𝑦)

Proof of Theorem elrab2w

Step	Hyp	Ref	Expression
1		elex 3474	. 2 ⊢ (𝐴 ∈ 𝐶 → 𝐴 ∈ V)
2		elex 3474	. . 3 ⊢ (𝐴 ∈ 𝐵 → 𝐴 ∈ V)
3	2	adantr 486	. 2 ⊢ ((𝐴 ∈ 𝐵 ∧ 𝜒) → 𝐴 ∈ V)
4		eleq1w 2845	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵))
5		elrab2w.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
6	4, 5	anbi12d 644	. . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 ∈ 𝐵 ∧ 𝜑) ↔ (𝑦 ∈ 𝐵 ∧ 𝜓)))
7		eleq1 2850	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
8		elrab2w.2	. . . 4 ⊢ (𝑦 = 𝐴 → (𝜓 ↔ 𝜒))
9	7, 8	anbi12d 644	. . 3 ⊢ (𝑦 = 𝐴 → ((𝑦 ∈ 𝐵 ∧ 𝜓) ↔ (𝐴 ∈ 𝐵 ∧ 𝜒)))
10		elrab2w.3	. . . 4 ⊢ 𝐶 = {𝑥 ∈ 𝐵 ∣ 𝜑}
11		df-rab 3415	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
12	10, 11	eqtri 2785	. . 3 ⊢ 𝐶 = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
13	6, 9, 12	elab2gw 3635	. 2 ⊢ (𝐴 ∈ V → (𝐴 ∈ 𝐶 ↔ (𝐴 ∈ 𝐵 ∧ 𝜒)))
14	1, 3, 13	pm5.21nii 381	1 ⊢ (𝐴 ∈ 𝐶 ↔ (𝐴 ∈ 𝐵 ∧ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145  {cab 2740  {crab 3414  Vcvv 3453

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3455

This theorem is used by:  uspgrlimlem2  48913