Theorem bj-inrab 32923

Description: Generalization of inrab 3899. (Contributed by BJ, 21-Apr-2019.)

Assertion

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

Proof of Theorem bj-inrab

Step	Hyp	Ref	Expression
1		an4 865	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ (𝜑 ∧ 𝜓)))
2		elin 3796	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
3	2	anbi1i 731	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ (𝜑 ∧ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ (𝜑 ∧ 𝜓)))
4	1, 3	bitr4i 267	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓)) ↔ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ (𝜑 ∧ 𝜓)))
5	4	abbii 2739	. 2 ⊢ {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓))} = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ (𝜑 ∧ 𝜓))}
6		df-rab 2921	. . . 4 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
7		df-rab 2921	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜓)}
8	6, 7	ineq12i 3812	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐵 ∣ 𝜓}) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜓)})
9		inab 3895	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜓)}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓))}
10	8, 9	eqtri 2644	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐵 ∣ 𝜓}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓))}
11		df-rab 2921	. 2 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ (𝜑 ∧ 𝜓)} = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ (𝜑 ∧ 𝜓))}
12	5, 10, 11	3eqtr4i 2654	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐵 ∣ 𝜓}) = {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ (𝜑 ∧ 𝜓)}

Syntax hints:   ∧ wa 384   = wceq 1483   ∈ wcel 1990  {cab 2608  {crab 2916   ∩ cin 3573

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-rab 2921  df-v 3202  df-in 3581

This theorem is referenced by:  bj-inrab2  32924