Theorem clelab 2178

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 2043	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
2	1	anbi2i 438	. . 3 ⊢ ((𝑦 = 𝐴 ∧ 𝑦 ∈ {𝑥 ∣ 𝜑}) ↔ (𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑))
3	2	exbii 1512	. 2 ⊢ (∃𝑦(𝑦 = 𝐴 ∧ 𝑦 ∈ {𝑥 ∣ 𝜑}) ↔ ∃𝑦(𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑))
4		df-clel 2052	. 2 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑦(𝑦 = 𝐴 ∧ 𝑦 ∈ {𝑥 ∣ 𝜑}))
5		nfv 1437	. . 3 ⊢ Ⅎ𝑦(𝑥 = 𝐴 ∧ 𝜑)
6		nfv 1437	. . . 4 ⊢ Ⅎ𝑥 𝑦 = 𝐴
7		nfs1v 1831	. . . 4 ⊢ Ⅎ𝑥[𝑦 / 𝑥]𝜑
8	6, 7	nfan 1473	. . 3 ⊢ Ⅎ𝑥(𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑)
9		eqeq1 2062	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝐴 ↔ 𝑦 = 𝐴))
10		sbequ12 1670	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ [𝑦 / 𝑥]𝜑))
11	9, 10	anbi12d 450	. . 3 ⊢ (𝑥 = 𝑦 → ((𝑥 = 𝐴 ∧ 𝜑) ↔ (𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑)))
12	5, 8, 11	cbvex 1655	. 2 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ ∃𝑦(𝑦 = 𝐴 ∧ [𝑦 / 𝑥]𝜑))
13	3, 4, 12	3bitr4i 205	1 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))

Syntax hints:   ∧ wa 101   ↔ wb 102   = wceq 1259  ∃wex 1397   ∈ wcel 1409  [wsb 1661  {cab 2042

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-11 1413  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-ext 2038

This theorem depends on definitions:  df-bi 114  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052

This theorem is referenced by:  elrabi  2717