Theorem inrab2 3529

Description: Intersection with a restricted class abstraction. (Contributed by NM, 19-Nov-2007.)

Assertion

Ref	Expression
inrab2	⊢ ({x ∈ A ∣ φ} ∩ B) = {x ∈ (A ∩ B) ∣ φ}

Distinct variable group:   x,B

Allowed substitution hints:   φ(x)   A(x)

Proof of Theorem inrab2

Step	Hyp	Ref	Expression
1		df-rab 2624	. . 3 ⊢ {x ∈ A ∣ φ} = {x ∣ (x ∈ A ∧ φ)}
2		abid2 2471	. . . 4 ⊢ {x ∣ x ∈ B} = B
3	2	eqcomi 2357	. . 3 ⊢ B = {x ∣ x ∈ B}
4	1, 3	ineq12i 3456	. 2 ⊢ ({x ∈ A ∣ φ} ∩ B) = ({x ∣ (x ∈ A ∧ φ)} ∩ {x ∣ x ∈ B})
5		df-rab 2624	. . 3 ⊢ {x ∈ (A ∩ B) ∣ φ} = {x ∣ (x ∈ (A ∩ B) ∧ φ)}
6		inab 3523	. . . 4 ⊢ ({x ∣ (x ∈ A ∧ φ)} ∩ {x ∣ x ∈ B}) = {x ∣ ((x ∈ A ∧ φ) ∧ x ∈ B)}
7		elin 3220	. . . . . . 7 ⊢ (x ∈ (A ∩ B) ↔ (x ∈ A ∧ x ∈ B))
8	7	anbi1i 676	. . . . . 6 ⊢ ((x ∈ (A ∩ B) ∧ φ) ↔ ((x ∈ A ∧ x ∈ B) ∧ φ))
9		an32 773	. . . . . 6 ⊢ (((x ∈ A ∧ x ∈ B) ∧ φ) ↔ ((x ∈ A ∧ φ) ∧ x ∈ B))
10	8, 9	bitri 240	. . . . 5 ⊢ ((x ∈ (A ∩ B) ∧ φ) ↔ ((x ∈ A ∧ φ) ∧ x ∈ B))
11	10	abbii 2466	. . . 4 ⊢ {x ∣ (x ∈ (A ∩ B) ∧ φ)} = {x ∣ ((x ∈ A ∧ φ) ∧ x ∈ B)}
12	6, 11	eqtr4i 2376	. . 3 ⊢ ({x ∣ (x ∈ A ∧ φ)} ∩ {x ∣ x ∈ B}) = {x ∣ (x ∈ (A ∩ B) ∧ φ)}
13	5, 12	eqtr4i 2376	. 2 ⊢ {x ∈ (A ∩ B) ∣ φ} = ({x ∣ (x ∈ A ∧ φ)} ∩ {x ∣ x ∈ B})
14	4, 13	eqtr4i 2376	1 ⊢ ({x ∈ A ∣ φ} ∩ B) = {x ∈ (A ∩ B) ∣ φ}

This proof depends on syntax axioms:   ∧ wa 358   = wceq 1642   ∈ wcel 1710  {cab 2339  {crab 2619   ∩ cin 3209

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-rab 2624  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214

This theorem is used by: (None)