Theorem eqrabi 38427

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 2870	. 2 ⊢ 𝐴 = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
3		df-rab 3399	. 2 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
4	2, 3	eqtr4i 2761	1 ⊢ 𝐴 = {𝑥 ∈ 𝐵 ∣ 𝜑}

Syntax hints:   ↔ wb 206   ∧ wa 395   = wceq 1542   ∈ wcel 2114  {cab 2713  {crab 3398

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2714  df-cleq 2727  df-clel 2810  df-rab 3399

This theorem is referenced by:  dfdisjs6  39112  dfdisjs7  39113