Theorem eqabcb 2868

Description: Equality of a class variable and a class abstraction. Commuted form of eqabb 2866. (Contributed by NM, 20-Aug-1993.)

Assertion

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

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem eqabcb

Step	Hyp	Ref	Expression
1		eqabb 2866	. 2 ⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑))
2		eqcom 2733	. 2 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ 𝐴 = {𝑥 ∣ 𝜑})
3		bicom 221	. . 3 ⊢ ((𝜑 ↔ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐴 ↔ 𝜑))
4	3	albii 1814	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝑥 ∈ 𝐴) ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑))
5	1, 2, 4	3bitr4i 302	1 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ ∀𝑥(𝜑 ↔ 𝑥 ∈ 𝐴))

Syntax hints:   ↔ wb 205  ∀wal 1532   = wceq 1534   ∈ wcel 2099  {cab 2703

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-10 2130  ax-11 2147  ax-12 2167  ax-ext 2697

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-tru 1537  df-ex 1775  df-nf 1779  df-sb 2061  df-clab 2704  df-cleq 2718  df-clel 2803

This theorem is referenced by:  rabeqcOLD  3678  dm0rn0  5923  dffo3  7108  dffo3f  7112  dfsup2  9480  rankf  9830  fmla0xp  35224  dfon3  35729  dfiota3  35760  onsupmaxb  42941  scottabf  43951