Theorem bj-inrab3 37842

Description: Generalization of dfrab3ss 4269. Shortens dfrab3ss 4269. (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 4265	. . 3 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = (𝐵 ∩ {𝑥 ∣ 𝜑})
2	1	ineq2i 4163	. 2 ⊢ (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}) = (𝐴 ∩ (𝐵 ∩ {𝑥 ∣ 𝜑}))
3		dfrab3 4265	. . . 4 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = (𝐴 ∩ {𝑥 ∣ 𝜑})
4	3	ineq2i 4163	. . 3 ⊢ (𝐵 ∩ {𝑥 ∈ 𝐴 ∣ 𝜑}) = (𝐵 ∩ (𝐴 ∩ {𝑥 ∣ 𝜑}))
5		incom 4155	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵) = (𝐵 ∩ {𝑥 ∈ 𝐴 ∣ 𝜑})
6		in12 4174	. . 3 ⊢ (𝐴 ∩ (𝐵 ∩ {𝑥 ∣ 𝜑})) = (𝐵 ∩ (𝐴 ∩ {𝑥 ∣ 𝜑}))
7	4, 5, 6	3eqtr4i 2794	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵) = (𝐴 ∩ (𝐵 ∩ {𝑥 ∣ 𝜑}))
8	2, 7	eqtr4i 2787	1 ⊢ (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}) = ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ 𝐵)

This proof depends on syntax axioms:   = wceq 1570  {cab 2739  {crab 3413   ∩ cin 3898

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 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-rab 3414  df-v 3453  df-in 3906

This theorem is used by: (None)