Theorem abeqinbi 39188

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 39187	. 2 ⊢ 𝐴 = {𝑥 ∈ 𝐶 ∣ 𝜑}
4		abeqinbi.3	. 2 ⊢ (𝑥 ∈ 𝐶 → (𝜑 ↔ 𝜓))
5	3, 4	rabimbieq 39186	1 ⊢ 𝐴 = {𝑥 ∈ 𝐶 ∣ 𝜓}

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569   ∈ wcel 2144  {cab 2738  {crab 3412   ∩ cin 4089

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-in 4091

This theorem is used by:  dfrefrels2  39526  dfcnvrefrels2  39541  dfsymrels2  39558  dftrrels2  39592