Theorem abeq1 2460

Description: Equality of a class variable and a class abstraction. (Contributed by NM, 20-Aug-1993.)

Assertion

Ref	Expression
abeq1	⊢ ({x ∣ φ} = A ↔ ∀x(φ ↔ x ∈ A))

Distinct variable group:   x,A

Allowed substitution hint:   φ(x)

Proof of Theorem abeq1

Step	Hyp	Ref	Expression
1		abeq2 2459	. 2 ⊢ (A = {x ∣ φ} ↔ ∀x(x ∈ A ↔ φ))
2		eqcom 2355	. 2 ⊢ ({x ∣ φ} = A ↔ A = {x ∣ φ})
3		bicom 191	. . 3 ⊢ ((φ ↔ x ∈ A) ↔ (x ∈ A ↔ φ))
4	3	albii 1566	. 2 ⊢ (∀x(φ ↔ x ∈ A) ↔ ∀x(x ∈ A ↔ φ))
5	1, 2, 4	3bitr4i 268	1 ⊢ ({x ∣ φ} = A ↔ ∀x(φ ↔ x ∈ A))

Syntax hints:   ↔ wb 176  ∀wal 1540   = wceq 1642   ∈ wcel 1710  {cab 2339

This theorem was proved from 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 theorem 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 referenced by:  abbi1dv  2470  euabsn2  3792  phidisjnn  4616  dm0rn0  4922  dffo3  5423