Theorem eqabcb 2883

Description: Equality of a class variable and a class abstraction. Commuted form of eqabb 2881. (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 2881	. 2 ⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑))
2		eqcom 2744	. 2 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ 𝐴 = {𝑥 ∣ 𝜑})
3		bicom 222	. . 3 ⊢ ((𝜑 ↔ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐴 ↔ 𝜑))
4	3	albii 1819	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝑥 ∈ 𝐴) ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑))
5	1, 2, 4	3bitr4i 303	1 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ ∀𝑥(𝜑 ↔ 𝑥 ∈ 𝐴))

Syntax hints:   ↔ wb 206  ∀wal 1538   = wceq 1540   ∈ wcel 2108  {cab 2714

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1543  df-ex 1780  df-nf 1784  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816

This theorem is referenced by:  rabeqcOLD  3690  dm0rn0  5935  dffo3  7122  dffo3f  7126  dfsup2  9484  rankf  9834  fmla0xp  35388  dfon3  35893  dfiota3  35924  onsupmaxb  43251  scottabf  44259