Theorem abeqinbi 38966

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 38965	. 2 ⊢ 𝐴 = {𝑥 ∈ 𝐶 ∣ 𝜑}
4		abeqinbi.3	. 2 ⊢ (𝑥 ∈ 𝐶 → (𝜑 ↔ 𝜓))
5	3, 4	rabimbieq 38964	1 ⊢ 𝐴 = {𝑥 ∈ 𝐶 ∣ 𝜓}

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570   ∈ wcel 2146  {cab 2743  {crab 3418   ∩ cin 3905

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 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-v 3459  df-in 3913

This theorem is used by:  dfrefrels2  39304  dfcnvrefrels2  39319  dfsymrels2  39336  dftrrels2  39370