Theorem eqabrd 2905

Description: Equality of a class variable and a class abstraction (deduction form of eqabb 2903). (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 2850	. 2 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ {𝑥 ∣ 𝜓}))
3		abid 2745	. 2 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜓} ↔ 𝜓)
4	2, 3	bitrdi 290	1 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570   ∈ wcel 2147  {cab 2741

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839

This theorem is used by:  eqabri  2906  fvelimab  6988  mapsnend  9102  nosupbnd2  28130  noinfbnd2  28145  fvineqsneu  38602  fvineqsneq  38603  ispridlc  39270  ac6s6  39370  dib1dim  42488  prprspr2  48873