Theorem clelab 2474

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 2340	. . . 4 ⊢ (y ∈ {x ∣ φ} ↔ [y / x]φ)
2	1	anbi2i 675	. . 3 ⊢ ((y = A ∧ y ∈ {x ∣ φ}) ↔ (y = A ∧ [y / x]φ))
3	2	exbii 1582	. 2 ⊢ (∃y(y = A ∧ y ∈ {x ∣ φ}) ↔ ∃y(y = A ∧ [y / x]φ))
4		df-clel 2349	. 2 ⊢ (A ∈ {x ∣ φ} ↔ ∃y(y = A ∧ y ∈ {x ∣ φ}))
5		nfv 1619	. . 3 ⊢ Ⅎy(x = A ∧ φ)
6		nfv 1619	. . . 4 ⊢ Ⅎx y = A
7		nfs1v 2106	. . . 4 ⊢ Ⅎx[y / x]φ
8	6, 7	nfan 1824	. . 3 ⊢ Ⅎx(y = A ∧ [y / x]φ)
9		eqeq1 2359	. . . 4 ⊢ (x = y → (x = A ↔ y = A))
10		sbequ12 1919	. . . 4 ⊢ (x = y → (φ ↔ [y / x]φ))
11	9, 10	anbi12d 691	. . 3 ⊢ (x = y → ((x = A ∧ φ) ↔ (y = A ∧ [y / x]φ)))
12	5, 8, 11	cbvex 1985	. 2 ⊢ (∃x(x = A ∧ φ) ↔ ∃y(y = A ∧ [y / x]φ))
13	3, 4, 12	3bitr4i 268	1 ⊢ (A ∈ {x ∣ φ} ↔ ∃x(x = A ∧ φ))

This proof depends on syntax axioms:   ↔ wb 176   ∧ wa 358  ∃wex 1541   = wceq 1642  [wsb 1648   ∈ wcel 1710  {cab 2339

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349

This theorem is used by: (None)