Theorem eqabcb 2877

Description: Equality of a class variable and a class abstraction. Commuted form of eqabb 2876. (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 2876	. 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 1821	. 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 1540   = wceq 1542   ∈ wcel 2114  {cab 2715

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-10 2147  ax-11 2163  ax-12 2185  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-nf 1786  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812

This theorem is referenced by:  dm0rn0OLD  5882  dffo3  7056  dffo3f  7060  dfsup2  9359  rankf  9718  fmla0xp  35599  dfon3  36106  dfiota3  36137  onsupmaxb  43596  scottabf  44596