Theorem clelab 2239

Description: Membership of a class variable in a class abstraction. (Contributed by NM, 23-Dec-1993.)

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		df-clab 2102	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
2	1	anbi2i 450	. . 3 ⊢ ((𝑦 = 𝐴 ∧ 𝑦 ∈ {𝑥 ∣ 𝜑}) ↔ (𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑))
3	2	exbii 1567	. 2 ⊢ (∃𝑦(𝑦 = 𝐴 ∧ 𝑦 ∈ {𝑥 ∣ 𝜑}) ↔ ∃𝑦(𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑))
4		df-clel 2111	. 2 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑦(𝑦 = 𝐴 ∧ 𝑦 ∈ {𝑥 ∣ 𝜑}))
5		nfv 1491	. . 3 ⊢ Ⅎ𝑦(𝑥 = 𝐴 ∧ 𝜑)
6		nfv 1491	. . . 4 ⊢ Ⅎ𝑥 𝑦 = 𝐴
7		nfs1v 1890	. . . 4 ⊢ Ⅎ𝑥[𝑦 / 𝑥]𝜑
8	6, 7	nfan 1527	. . 3 ⊢ Ⅎ𝑥(𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑)
9		eqeq1 2121	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝐴 ↔ 𝑦 = 𝐴))
10		sbequ12 1727	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ [𝑦 / 𝑥]𝜑))
11	9, 10	anbi12d 462	. . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 = 𝐴 ∧ 𝜑) ↔ (𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑)))
12	5, 8, 11	cbvex 1712	. 2 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ ∃𝑦(𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑))
13	3, 4, 12	3bitr4i 211	1 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))

Syntax hints:   ∧ wa 103   ↔ wb 104   = wceq 1314  ∃wex 1451   ∈ wcel 1463  [wsb 1718  {cab 2101

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1406  ax-7 1407  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-8 1465  ax-11 1467  ax-4 1470  ax-17 1489  ax-i9 1493  ax-ial 1497  ax-ext 2097

This theorem depends on definitions:  df-bi 116  df-nf 1420  df-sb 1719  df-clab 2102  df-cleq 2108  df-clel 2111

This theorem is referenced by:  elrabi  2806