Theorem eqabrd 2868

Description: Equality of a class variable and a class abstraction (deduction form of eqabb 2865). (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 2811	. 2 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ {𝑥 ∣ 𝜓}))
3		abid 2705	. 2 ⊢ (𝑥 ∈ {𝑥 ∣ 𝜓} ↔ 𝜓)
4	2, 3	bitrdi 287	1 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 205   = wceq 1533   ∈ wcel 2098  {cab 2701

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-12 2163  ax-ext 2695

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1774  df-sb 2060  df-clab 2702  df-cleq 2716  df-clel 2802

This theorem is referenced by:  eqabri  2869  fvelimab  6955  mapsnend  9033  nosupbnd2  27590  noinfbnd2  27605  fvineqsneu  36793  fvineqsneq  36794  ispridlc  37442  ac6s6  37544  dib1dim  40540  prprspr2  46732