Theorem clelab 2905

Description: Membership of a class variable in a class abstraction. (Contributed by NM, 23-Dec-1993.) (Proof shortened by Wolf Lammen, 16-Nov-2019.) Avoid ax-11 2190, see sbc5ALT 3772 for more details. (Revised by SN, 2-Sep-2024.)

Assertion

Ref	Expression
clelab	⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem clelab

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		elissetv 2842	. 2 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} → ∃𝑦 𝑦 = 𝐴)
2		exsimpl 1896	. . 3 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) → ∃𝑥 𝑥 = 𝐴)
3		iseqsetv-cleq 2825	. . 3 ⊢ (∃𝑥 𝑥 = 𝐴 ↔ ∃𝑦 𝑦 = 𝐴)
4	2, 3	sylib 221	. 2 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) → ∃𝑦 𝑦 = 𝐴)
5		eleq1 2849	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
6		df-clab 2740	. . . . . 6 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
7		sb5 2309	. . . . . 6 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
8	6, 7	bitri 278	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
9		eqeq2 2773	. . . . . . 7 ⊢ (𝑦 = 𝐴 → (𝑥 = 𝑦 ↔ 𝑥 = 𝐴))
10	9	anbi1d 642	. . . . . 6 ⊢ (𝑦 = 𝐴 → ((𝑥 = 𝑦 ∧ 𝜑) ↔ (𝑥 = 𝐴 ∧ 𝜑)))
11	10	exbidv 1949	. . . . 5 ⊢ (𝑦 = 𝐴 → (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
12	8, 11	bitrid 286	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
13	5, 12	bitr3d 284	. . 3 ⊢ (𝑦 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
14	13	exlimiv 1958	. 2 ⊢ (∃𝑦 𝑦 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
15	1, 4, 14	pm5.21nii 381	1 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))

Syntax hints:   ↔ wb 209   ∧ wa 400   = wceq 1568  ∃wex 1807  [wsb 2094   ∈ wcel 2141  {cab 2739

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1823  ax-4 1837  ax-5 1938  ax-6 1995  ax-7 2036  ax-8 2143  ax-9 2151  ax-10 2174  ax-12 2211  ax-ext 2733

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1808  df-nf 1812  df-sb 2095  df-clab 2740  df-cleq 2753  df-clel 2836

This theorem is referenced by:  sbc5  3771  bj-csbsnlem  37482  frege55c  44592  spr0nelg  48170