Theorem eqabrd 2902

Description: Equality of a class variable and a class abstraction (deduction form of eqabb 2900). (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 2847	. 2 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ {𝑥 ∣ 𝜓}))
3		abid 2743	. 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 2145  {cab 2739

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-12 2213  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836

This theorem is used by:  eqabri  2903  fvelimab  6957  mapsnend  9064  nosupbnd2  28073  noinfbnd2  28088  fvineqsneu  38334  fvineqsneq  38335  ispridlc  39004  ac6s6  39104  dib1dim  42222  prprspr2  48599