Theorem abeqin 38961

Description: Intersection with class abstraction. (Contributed by Peter Mazsa, 21-Jul-2021.)

Hypotheses

Ref	Expression
abeqin.1	⊢ 𝐴 = (𝐵 ∩ 𝐶)
abeqin.2	⊢ 𝐵 = {𝑥 ∣ 𝜑}

Assertion

Ref	Expression
abeqin	⊢ 𝐴 = {𝑥 ∈ 𝐶 ∣ 𝜑}

Distinct variable group:   𝑥,𝐶

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

Proof of Theorem abeqin

Step	Hyp	Ref	Expression
1		abeqin.2	. . 3 ⊢ 𝐵 = {𝑥 ∣ 𝜑}
2	1	ineq1i 4169	. 2 ⊢ (𝐵 ∩ 𝐶) = ({𝑥 ∣ 𝜑} ∩ 𝐶)
3		abeqin.1	. 2 ⊢ 𝐴 = (𝐵 ∩ 𝐶)
4		dfrab2 4273	. 2 ⊢ {𝑥 ∈ 𝐶 ∣ 𝜑} = ({𝑥 ∣ 𝜑} ∩ 𝐶)
5	2, 3, 4	3eqtr4i 2798	1 ⊢ 𝐴 = {𝑥 ∈ 𝐶 ∣ 𝜑}

This proof depends on syntax axioms:   = wceq 1570  {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:  abeqinbi  38962  dfcnvrefrels3  39316  dffunsALTV  39475  dfdisjs  39500