Theorem bj-inrab 34240

Description: Generalization of inrab 4274. (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 654	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ (𝜑 ∧ 𝜓)))
2		elin 4168	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
3	2	anbi1i 625	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ (𝜑 ∧ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ (𝜑 ∧ 𝜓)))
4	1, 3	bitr4i 280	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓)) ↔ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ (𝜑 ∧ 𝜓)))
5	4	abbii 2886	. 2 ⊢ {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓))} = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ (𝜑 ∧ 𝜓))}
6		df-rab 3147	. . . 4 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
7		df-rab 3147	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜓)}
8	6, 7	ineq12i 4186	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐵 ∣ 𝜓}) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜓)})
9		inab 4270	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜓)}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓))}
10	8, 9	eqtri 2844	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐵 ∣ 𝜓}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓))}
11		df-rab 3147	. 2 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ (𝜑 ∧ 𝜓)} = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ (𝜑 ∧ 𝜓))}
12	5, 10, 11	3eqtr4i 2854	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐵 ∣ 𝜓}) = {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ (𝜑 ∧ 𝜓)}

Syntax hints:   ∧ wa 398   = wceq 1533   ∈ wcel 2110  {cab 2799  {crab 3142   ∩ cin 3934

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-rab 3147  df-v 3496  df-in 3942

This theorem is referenced by:  bj-inrab2  34241