Theorem eqrabi 39379

Description: Class element of a restricted class abstraction. (Contributed by Peter Mazsa, 24-Jul-2021.)

Hypothesis

Ref	Expression
eqrabi.1	⊢ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐵 ∧ 𝜑))

Assertion

Ref	Expression
eqrabi	⊢ 𝐴 = {𝑥 ∈ 𝐵 ∣ 𝜑}

Distinct variable group:   𝑥,𝐴

Allowed substitution hints:   𝜑(𝑥)   𝐵(𝑥)

Proof of Theorem eqrabi

Step	Hyp	Ref	Expression
1		eqrabi.1	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐵 ∧ 𝜑))
2	1	eqabi 2899	. 2 ⊢ 𝐴 = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
3		df-rab 3417	. 2 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
4	2, 3	eqtr4i 2789	1 ⊢ 𝐴 = {𝑥 ∈ 𝐵 ∣ 𝜑}

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2147  {cab 2741  {crab 3416

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-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417

This theorem is used by:  dfdisjs6  40065  dfdisjs7  40066