Theorem elabd2 3623

Description: Membership in a class abstraction, using implicit substitution. Deduction version of elab 3632. (Contributed by GG, 12-Oct-2024.) (Revised by BJ, 16-Oct-2024.)

Hypotheses

Ref	Expression
elabd2.ex	⊢ (𝜑 → 𝐴 ∈ 𝑉)
elabd2.eq	⊢ (𝜑 → 𝐵 = {𝑥 ∣ 𝜓})
elabd2.is	⊢ ((𝜑 ∧ 𝑥 = 𝐴) → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
elabd2	⊢ (𝜑 → (𝐴 ∈ 𝐵 ↔ 𝜒))

Distinct variable groups:   𝜑,𝑥   𝜒,𝑥   𝑥,𝐴

Allowed substitution hints:   𝜓(𝑥)   𝐵(𝑥)   𝑉(𝑥)

Proof of Theorem elabd2

Step	Hyp	Ref	Expression
1		elabd2.ex	. 2 ⊢ (𝜑 → 𝐴 ∈ 𝑉)
2		elabd2.eq	. . . . 5 ⊢ (𝜑 → 𝐵 = {𝑥 ∣ 𝜓})
3	2	eleq2d 2846	. . . 4 ⊢ (𝜑 → (𝐴 ∈ 𝐵 ↔ 𝐴 ∈ {𝑥 ∣ 𝜓}))
4		elab6g 3622	. . . 4 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝑥 = 𝐴 → 𝜓)))
5	3, 4	sylan9bb 519	. . 3 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑉) → (𝐴 ∈ 𝐵 ↔ ∀𝑥(𝑥 = 𝐴 → 𝜓)))
6		elisset 2842	. . . 4 ⊢ (𝐴 ∈ 𝑉 → ∃𝑥 𝑥 = 𝐴)
7		elabd2.is	. . . . . . . 8 ⊢ ((𝜑 ∧ 𝑥 = 𝐴) → (𝜓 ↔ 𝜒))
8	7	pm5.74da 816	. . . . . . 7 ⊢ (𝜑 → ((𝑥 = 𝐴 → 𝜓) ↔ (𝑥 = 𝐴 → 𝜒)))
9	8	albidv 1953	. . . . . 6 ⊢ (𝜑 → (∀𝑥(𝑥 = 𝐴 → 𝜓) ↔ ∀𝑥(𝑥 = 𝐴 → 𝜒)))
10		19.23v 1975	. . . . . 6 ⊢ (∀𝑥(𝑥 = 𝐴 → 𝜒) ↔ (∃𝑥 𝑥 = 𝐴 → 𝜒))
11	9, 10	bitrdi 290	. . . . 5 ⊢ (𝜑 → (∀𝑥(𝑥 = 𝐴 → 𝜓) ↔ (∃𝑥 𝑥 = 𝐴 → 𝜒)))
12		pm5.5 364	. . . . 5 ⊢ (∃𝑥 𝑥 = 𝐴 → ((∃𝑥 𝑥 = 𝐴 → 𝜒) ↔ 𝜒))
13	11, 12	sylan9bb 519	. . . 4 ⊢ ((𝜑 ∧ ∃𝑥 𝑥 = 𝐴) → (∀𝑥(𝑥 = 𝐴 → 𝜓) ↔ 𝜒))
14	6, 13	sylan2 605	. . 3 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑉) → (∀𝑥(𝑥 = 𝐴 → 𝜓) ↔ 𝜒))
15	5, 14	bitrd 282	. 2 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑉) → (𝐴 ∈ 𝐵 ↔ 𝜒))
16	1, 15	mpdan 700	1 ⊢ (𝜑 → (𝐴 ∈ 𝐵 ↔ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568   = wceq 1570  ∃wex 1812   ∈ wcel 2145  {cab 2738

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 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835

This theorem is used by:  elabd3  3624  elimasng1  6077