Theorem clelab 1187

Description: Membership of a class variable in a class abstraction.

Assertion

Ref	Expression
clelab	⊢ (A ∈ {x∣φ} ↔ ∃x(x = A ∧ φ))

Distinct variable group(s):   x,A

Proof of Theorem clelab

Step	Hyp	Ref	Expression
1		df-clab 1093	. . . 4 ⊢ (y ∈ {x∣φ} ↔ [y / x]φ)
2	1	anbi2i 367	. . 3 ⊢ ((y = A ∧ y ∈ {x∣φ}) ↔ (y = A ∧ [y / x]φ))
3	2	biex 733	. 2 ⊢ (∃y(y = A ∧ y ∈ {x∣φ}) ↔ ∃y(y = A ∧ [y / x]φ))
4		df-clel 1099	. 2 ⊢ (A ∈ {x∣φ} ↔ ∃y(y = A ∧ y ∈ {x∣φ}))
5		ax-17 925	. . 3 ⊢ ((x = A ∧ φ) → ∀y(x = A ∧ φ))
6		ax-17 925	. . . 4 ⊢ (y = A → ∀x y = A)
7		hbs1 986	. . . 4 ⊢ ([y / x]φ → ∀x[y / x]φ)
8	6, 7	hban 704	. . 3 ⊢ ((y = A ∧ [y / x]φ) → ∀x(y = A ∧ [y / x]φ))
9		cleq1 1107	. . . 4 ⊢ (x = y → (x = A ↔ y = A))
10		sbequ12 865	. . . 4 ⊢ (x = y → (φ ↔ [y / x]φ))
11	9, 10	anbi12d 476	. . 3 ⊢ (x = y → ((x = A ∧ φ) ↔ (y = A ∧ [y / x]φ)))
12	5, 8, 11	cbvex 849	. 2 ⊢ (∃x(x = A ∧ φ) ↔ ∃y(y = A ∧ [y / x]φ))
13	3, 4, 12	3bitr4 158	1 ⊢ (A ∈ {x∣φ} ↔ ∃x(x = A ∧ φ))

Syntax hints:   ↔ wb 127   ∧ wa 196  ∃wex 678   = weq 797  [wsb 852  {cab 1090   = wceq 1091   ∈ wcel 1092

This theorem is referenced by:  opabid 2099

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099

metamath.org