Theorem abeqinbi 38932

Description: Intersection with class abstraction and equivalent wff's. (Contributed by Peter Mazsa, 21-Jul-2021.)

Hypotheses

Ref	Expression
abeqinbi.1	⊢ 𝐴 = (𝐵 ∩ 𝐶)
abeqinbi.2	⊢ 𝐵 = {𝑥 ∣ 𝜑}
abeqinbi.3	⊢ (𝑥 ∈ 𝐶 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
abeqinbi	⊢ 𝐴 = {𝑥 ∈ 𝐶 ∣ 𝜓}

Distinct variable group:   𝑥,𝐶

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

Proof of Theorem abeqinbi

Step	Hyp	Ref	Expression
1		abeqinbi.1	. . 3 ⊢ 𝐴 = (𝐵 ∩ 𝐶)
2		abeqinbi.2	. . 3 ⊢ 𝐵 = {𝑥 ∣ 𝜑}
3	1, 2	abeqin 38931	. 2 ⊢ 𝐴 = {𝑥 ∈ 𝐶 ∣ 𝜑}
4		abeqinbi.3	. 2 ⊢ (𝑥 ∈ 𝐶 → (𝜑 ↔ 𝜓))
5	3, 4	rabimbieq 38930	1 ⊢ 𝐴 = {𝑥 ∈ 𝐶 ∣ 𝜓}

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570   ∈ wcel 2143  {cab 2741  {crab 3416   ∩ cin 3904

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1573  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3417  df-v 3457  df-in 3912

This theorem is used by:  dfrefrels2  39270  dfcnvrefrels2  39285  dfsymrels2  39302  dftrrels2  39336