Theorem eqabrd 2880

Description: Equality of a class variable and a class abstraction (deduction form of eqabb 2878). (Contributed by NM, 16-Nov-1995.)

Hypothesis

Ref	Expression
eqabrd.1	⊢ (𝜑 → 𝐴 = {𝑥 ∣ 𝜓})

Assertion

Ref	Expression
eqabrd	⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝜓))

Proof of Theorem eqabrd

Step	Hyp	Ref	Expression
1		eqabrd.1	. . 3 ⊢ (𝜑 → 𝐴 = {𝑥 ∣ 𝜓})
2	1	eleq2d 2825	. 2 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ {𝑥 ∣ 𝜓}))
3		abid 2721	. 2 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜓} ↔ 𝜓)
4	2, 3	bitrdi 288	1 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 207   = wceq 1547   ∈ wcel 2119  {cab 2717

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-12 2189  ax-ext 2711

This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814

This theorem is referenced by:  eqabri  2881  fvelimab  6899  mapsnend  8973  nosupbnd2  27698  noinfbnd2  27713  fvineqsneu  37773  fvineqsneq  37774  ispridlc  38437  ac6s6  38539  dib1dim  41657  prprspr2  47993