Theorem clelab 2907

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 2192, see sbc5ALT 3774 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 2844	. 2 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} → ∃𝑦 𝑦 = 𝐴)
2		exsimpl 1889	. . 3 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) → ∃𝑥 𝑥 = 𝐴)
3		iseqsetv-cleq 2827	. . 3 ⊢ (∃𝑥 𝑥 = 𝐴 ↔ ∃𝑦 𝑦 = 𝐴)
4	2, 3	sylib 220	. 2 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) → ∃𝑦 𝑦 = 𝐴)
5		eleq1 2851	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜑}))
6		df-clab 2742	. . . . . 6 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
7		sb5 2311	. . . . . 6 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
8	6, 7	bitri 277	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
9		eqeq2 2775	. . . . . . 7 ⊢ (𝑦 = 𝐴 → (𝑥 = 𝑦 ↔ 𝑥 = 𝐴))
10	9	anbi1d 640	. . . . . 6 ⊢ (𝑦 = 𝐴 → ((𝑥 = 𝑦 ∧ 𝜑) ↔ (𝑥 = 𝐴 ∧ 𝜑)))
11	10	exbidv 1942	. . . . 5 ⊢ (𝑦 = 𝐴 → (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
12	8, 11	bitrid 285	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
13	5, 12	bitr3d 283	. . 3 ⊢ (𝑦 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
14	13	exlimiv 1951	. 2 ⊢ (∃𝑦 𝑦 = 𝐴 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
15	1, 4, 14	pm5.21nii 380	1 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))

Syntax hints:   ↔ wb 208   ∧ wa 399   = wceq 1561  ∃wex 1800  [wsb 2091   ∈ wcel 2143  {cab 2741

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1816  ax-4 1830  ax-5 1931  ax-6 1988  ax-7 2029  ax-8 2145  ax-9 2153  ax-10 2176  ax-12 2213  ax-ext 2735

This theorem depends on definitions:  df-bi 209  df-an 400  df-ex 1801  df-nf 1805  df-sb 2092  df-clab 2742  df-cleq 2755  df-clel 2838

This theorem is referenced by:  sbc5  3773  bj-csbsnlem  37393  frege55c  44499  spr0nelg  48073