Theorem abeq1 1566

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

Assertion

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

Distinct variable group:   x,A

Proof of Theorem abeq1

Step	Hyp	Ref	Expression
1		abeq2 1565	. 2 ⊢ (A = {x∣φ} ↔ ∀x(x ∈ A ↔ φ))
2		eqcom 1474	. 2 ⊢ ({x∣φ} = A ↔ A = {x∣φ})
3		bicom 519	. . 3 ⊢ ((φ ↔ x ∈ A) ↔ (x ∈ A ↔ φ))
4	3	albii 997	. 2 ⊢ (∀x(φ ↔ x ∈ A) ↔ ∀x(x ∈ A ↔ φ))
5	1, 2, 4	3bitr4 183	1 ⊢ ({x∣φ} = A ↔ ∀x(φ ↔ x ∈ A))

Syntax hints:   ↔ wb 146  ∀wal 952   = wceq 954   ∈ wcel 956  {cab 1461

This theorem is referenced by:  abbi1dv 1576  disj 2307  eusn 2442  dm0rn0 3325  dffo3 3810  homcard 10462

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-10 964  ax-12 966  ax-17 969  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-16 1208  ax-11o 1216  ax-ext 1457

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 979  df-sb 1170  df-clab 1462  df-cleq 1467  df-clel 1470

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