Theorem iinrab 4029

Description: Indexed intersection of a restricted class builder. (Contributed by NM, 6-Dec-2011.)

Assertion

Ref	Expression
iinrab	⊢ (A ≠ ∅ → ∩x ∈ A {y ∈ B ∣ φ} = {y ∈ B ∣ ∀x ∈ A φ})

Distinct variable groups:   y,A,x   x,B

Allowed substitution hints:   φ(x, y)   B(y)

Proof of Theorem iinrab

Step	Hyp	Ref	Expression
1		r19.28zv 3646	. . 3 ⊢ (A ≠ ∅ → (∀x ∈ A (y ∈ B ∧ φ) ↔ (y ∈ B ∧ ∀x ∈ A φ)))
2	1	abbidv 2468	. 2 ⊢ (A ≠ ∅ → {y ∣ ∀x ∈ A (y ∈ B ∧ φ)} = {y ∣ (y ∈ B ∧ ∀x ∈ A φ)})
3		df-rab 2624	. . . . 5 ⊢ {y ∈ B ∣ φ} = {y ∣ (y ∈ B ∧ φ)}
4	3	a1i 10	. . . 4 ⊢ (x ∈ A → {y ∈ B ∣ φ} = {y ∣ (y ∈ B ∧ φ)})
5	4	iineq2i 3989	. . 3 ⊢ ∩x ∈ A {y ∈ B ∣ φ} = ∩x ∈ A {y ∣ (y ∈ B ∧ φ)}
6		iinab 4028	. . 3 ⊢ ∩x ∈ A {y ∣ (y ∈ B ∧ φ)} = {y ∣ ∀x ∈ A (y ∈ B ∧ φ)}
7	5, 6	eqtri 2373	. 2 ⊢ ∩x ∈ A {y ∈ B ∣ φ} = {y ∣ ∀x ∈ A (y ∈ B ∧ φ)}
8		df-rab 2624	. 2 ⊢ {y ∈ B ∣ ∀x ∈ A φ} = {y ∣ (y ∈ B ∧ ∀x ∈ A φ)}
9	2, 7, 8	3eqtr4g 2410	1 ⊢ (A ≠ ∅ → ∩x ∈ A {y ∈ B ∣ φ} = {y ∈ B ∣ ∀x ∈ A φ})

This proof depends on syntax axioms:   → wi 4   ∧ wa 358   = wceq 1642   ∈ wcel 1710  {cab 2339   ≠ wne 2517  ∀wral 2615  {crab 2619  ∅c0 3551  ∩ciin 3971

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-ne 2519  df-ral 2620  df-rab 2624  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216  df-nul 3552  df-iin 3973

This theorem is used by:  iinrab2  4030  riinrab  4042