Theorem eqrabi 38760

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 2790	1 ⊢ 𝐴 = {𝑥 ∈ 𝐵 ∣ 𝜑}

Syntax hints:   ↔ wb 208   ∧ wa 399   = wceq 1562   ∈ wcel 2144  {cab 2742  {crab 3416

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1817  ax-4 1831  ax-5 1932  ax-6 1989  ax-7 2030  ax-8 2146  ax-9 2154  ax-ext 2736

This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1565  df-ex 1802  df-sb 2093  df-clab 2743  df-cleq 2756  df-clel 2839  df-rab 3417

This theorem is referenced by:  dfdisjs6  39446  dfdisjs7  39447