Theorem bj-inrab3 37969

Description: Generalization of dfrab3ss 4270. Shortens dfrab3ss 4270. (Contributed by BJ, 21-Apr-2019.) (Revised by OpenAI, 7-Jul-2020.)

Assertion

Ref	Expression
bj-inrab3	⊢ (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}) = ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵)

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem bj-inrab3

Step	Hyp	Ref	Expression
1		dfrab3 4266	. . 3 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = (𝐵 ∩ {𝑥 ∣ 𝜑})
2	1	ineq2i 4116	. 2 ⊢ (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}) = (𝐴 ∩ (𝐵 ∩ {𝑥 ∣ 𝜑}))
3		dfrab3 4266	. . . 4 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = (𝐴 ∩ {𝑥 ∣ 𝜑})
4	3	ineq2i 4116	. . 3 ⊢ (𝐵 ∩ {𝑥 ∈ 𝐴 ∣ 𝜑}) = (𝐵 ∩ (𝐴 ∩ {𝑥 ∣ 𝜑}))
5		incom 4108	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵) = (𝐵 ∩ {𝑥 ∈ 𝐴 ∣ 𝜑})
6		in12 4127	. . 3 ⊢ (𝐴 ∩ (𝐵 ∩ {𝑥 ∣ 𝜑})) = (𝐵 ∩ (𝐴 ∩ {𝑥 ∣ 𝜑}))
7	4, 5, 6	3eqtr4i 2796	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵) = (𝐴 ∩ (𝐵 ∩ {𝑥 ∣ 𝜑}))
8	2, 7	eqtr4i 2789	1 ⊢ (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}) = ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵)

This proof depends on syntax axioms:   = wceq 1569  {cab 2741  {crab 3415   ∩ cin 4091

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 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3416  df-v 3455  df-in 4093

This theorem is used by: (None)