Theorem eqabrd 2463

Description: Equality of a class variable and a class abstraction (deduction). (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 2420	. 2 ⊢ (φ → (x ∈ A ↔ x ∈ {x ∣ ψ}))
3		abid 2341	. 2 ⊢ (x ∈ {x ∣ ψ} ↔ ψ)
4	2, 3	syl6bb 252	1 ⊢ (φ → (x ∈ A ↔ ψ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   = wceq 1642   ∈ wcel 1710  {cab 2339

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-11 1746  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-an 360  df-ex 1542  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349

This theorem is used by:  fvelimab  5371