Theorem eqrabi 38991

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 2897	. 2 ⊢ 𝐴 = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
3		df-rab 3415	. 2 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
4	2, 3	eqtr4i 2788	1 ⊢ 𝐴 = {𝑥 ∈ 𝐵 ∣ 𝜑}

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145  {cab 2740  {crab 3414

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-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415

This theorem is used by:  dfdisjs6  39677  dfdisjs7  39678