Theorem bj-inrab 36862

Description: Generalization of inrab 4296. (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 656	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ (𝜑 ∧ 𝜓)))
2		elin 3947	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
3	2	anbi1i 624	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) ∧ (𝜑 ∧ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ (𝜑 ∧ 𝜓)))
4	1, 3	bitr4i 278	. . 3 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓)) ↔ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ (𝜑 ∧ 𝜓)))
5	4	abbii 2801	. 2 ⊢ {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓))} = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ (𝜑 ∧ 𝜓))}
6		df-rab 3420	. . . 4 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
7		df-rab 3420	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜓)}
8	6, 7	ineq12i 4198	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐵 ∣ 𝜓}) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜓)})
9		inab 4289	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜓)}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓))}
10	8, 9	eqtri 2757	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐵 ∣ 𝜓}) = {𝑥 ∣ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑥 ∈ 𝐵 ∧ 𝜓))}
11		df-rab 3420	. 2 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ (𝜑 ∧ 𝜓)} = {𝑥 ∣ (𝑥 ∈ (𝐴 ∩ 𝐵) ∧ (𝜑 ∧ 𝜓))}
12	5, 10, 11	3eqtr4i 2767	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐵 ∣ 𝜓}) = {𝑥 ∈ (𝐴 ∩ 𝐵) ∣ (𝜑 ∧ 𝜓)}

Syntax hints:   ∧ wa 395   = wceq 1539   ∈ wcel 2107  {cab 2712  {crab 3419   ∩ cin 3930

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1542  df-ex 1779  df-sb 2064  df-clab 2713  df-cleq 2726  df-clel 2808  df-rab 3420  df-v 3465  df-in 3938

This theorem is referenced by:  bj-inrab2  36863