Theorem elabd2 3569

Description: Membership in a class abstraction, using implicit substitution. Deduction version of elab 3576. (Contributed by Gino Giotto, 12-Oct-2024.)

Hypotheses

Ref	Expression
elabd2.1	⊢ (𝜑 → 𝐴 ∈ 𝑉)
elabd2.2	⊢ ((𝜑 ∧ 𝑥 = 𝐴) → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
elabd2	⊢ (𝜑 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒))

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

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

Proof of Theorem elabd2

Step	Hyp	Ref	Expression
1		elabd2.1	. 2 ⊢ (𝜑 → 𝐴 ∈ 𝑉)
2		elab6g 3568	. . . 4 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝑥 = 𝐴 → 𝜓)))
3	2	adantl 485	. . 3 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑉) → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝑥 = 𝐴 → 𝜓)))
4		elisset 2812	. . . 4 ⊢ (𝐴 ∈ 𝑉 → ∃𝑥 𝑥 = 𝐴)
5		elabd2.2	. . . . . . . 8 ⊢ ((𝜑 ∧ 𝑥 = 𝐴) → (𝜓 ↔ 𝜒))
6	5	pm5.74da 804	. . . . . . 7 ⊢ (𝜑 → ((𝑥 = 𝐴 → 𝜓) ↔ (𝑥 = 𝐴 → 𝜒)))
7	6	albidv 1928	. . . . . 6 ⊢ (𝜑 → (∀𝑥(𝑥 = 𝐴 → 𝜓) ↔ ∀𝑥(𝑥 = 𝐴 → 𝜒)))
8		19.23v 1950	. . . . . 6 ⊢ (∀𝑥(𝑥 = 𝐴 → 𝜒) ↔ (∃𝑥 𝑥 = 𝐴 → 𝜒))
9	7, 8	bitrdi 290	. . . . 5 ⊢ (𝜑 → (∀𝑥(𝑥 = 𝐴 → 𝜓) ↔ (∃𝑥 𝑥 = 𝐴 → 𝜒)))
10		pm5.5 365	. . . . 5 ⊢ (∃𝑥 𝑥 = 𝐴 → ((∃𝑥 𝑥 = 𝐴 → 𝜒) ↔ 𝜒))
11	9, 10	sylan9bb 513	. . . 4 ⊢ ((𝜑 ∧ ∃𝑥 𝑥 = 𝐴) → (∀𝑥(𝑥 = 𝐴 → 𝜓) ↔ 𝜒))
12	4, 11	sylan2 596	. . 3 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑉) → (∀𝑥(𝑥 = 𝐴 → 𝜓) ↔ 𝜒))
13	3, 12	bitrd 282	. 2 ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑉) → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒))
14	1, 13	mpdan 687	1 ⊢ (𝜑 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝜒))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399  ∀wal 1541   = wceq 1543  ∃wex 1787   ∈ wcel 2112  {cab 2714

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2018  ax-8 2114  ax-9 2122  ax-ext 2708

This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1546  df-ex 1788  df-sb 2073  df-clab 2715  df-cleq 2728  df-clel 2809

This theorem is referenced by:  sbcied  3728