Theorem clelab 2207

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 2070	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
2	1	anbi2i 445	. . 3 ⊢ ((𝑦 = 𝐴 ∧ 𝑦 ∈ {𝑥 ∣ 𝜑}) ↔ (𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑))
3	2	exbii 1537	. 2 ⊢ (∃𝑦(𝑦 = 𝐴 ∧ 𝑦 ∈ {𝑥 ∣ 𝜑}) ↔ ∃𝑦(𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑))
4		df-clel 2079	. 2 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑦(𝑦 = 𝐴 ∧ 𝑦 ∈ {𝑥 ∣ 𝜑}))
5		nfv 1462	. . 3 ⊢ Ⅎ𝑦(𝑥 = 𝐴 ∧ 𝜑)
6		nfv 1462	. . . 4 ⊢ Ⅎ𝑥 𝑦 = 𝐴
7		nfs1v 1858	. . . 4 ⊢ Ⅎ𝑥[𝑦 / 𝑥]𝜑
8	6, 7	nfan 1498	. . 3 ⊢ Ⅎ𝑥(𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑)
9		eqeq1 2089	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝐴 ↔ 𝑦 = 𝐴))
10		sbequ12 1696	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ [𝑦 / 𝑥]𝜑))
11	9, 10	anbi12d 457	. . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 = 𝐴 ∧ 𝜑) ↔ (𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑)))
12	5, 8, 11	cbvex 1681	. 2 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ ∃𝑦(𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑))
13	3, 4, 12	3bitr4i 210	1 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))

Syntax hints:   ∧ wa 102   ↔ wb 103   = wceq 1285  ∃wex 1422   ∈ wcel 1434  [wsb 1687  {cab 2069

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-11 1438  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-ext 2065

This theorem depends on definitions:  df-bi 115  df-nf 1391  df-sb 1688  df-clab 2070  df-cleq 2076  df-clel 2079

This theorem is referenced by:  elrabi  2755