Theorem cleqabr 1175

Description: Equality of a class variable and a class abstraction.

Assertion

Ref	Expression
cleqabr	⊢ ({x∣φ} = A ↔ ∀x(φ ↔ x ∈ A))

Distinct variable group(s):   x,A

Proof of Theorem cleqabr

Step	Hyp	Ref	Expression
1		cleqab 1174	. 2 ⊢ (A = {x∣φ} ↔ ∀x(x ∈ A ↔ φ))
2		cleqcom 1103	. 2 ⊢ ({x∣φ} = A ↔ A = {x∣φ})
3		bicom 398	. . 3 ⊢ ((φ ↔ x ∈ A) ↔ (x ∈ A ↔ φ))
4	3	bial 695	. 2 ⊢ (∀x(φ ↔ x ∈ A) ↔ ∀x(x ∈ A ↔ φ))
5	1, 2, 4	3bitr4 158	1 ⊢ ({x∣φ} = A ↔ ∀x(φ ↔ x ∈ A))

Syntax hints:   ↔ wb 127  ∀wal 672  {cab 1090   = wceq 1091   ∈ wcel 1092

This theorem is referenced by:  biabldv 1185  disj 1733  eusn 1913  dm0rn0 2549

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

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