Theorem riinrab 4042

Description: Relative intersection of a relative abstraction. (Contributed by Stefan O'Rear, 3-Apr-2015.)

Assertion

Ref	Expression
riinrab	⊢ (A ∩ ∩x ∈ X {y ∈ A ∣ φ}) = {y ∈ A ∣ ∀x ∈ X φ}

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

Allowed substitution hints:   φ(x, y)

Proof of Theorem riinrab

Step	Hyp	Ref	Expression
1		riin0 4040	. . 3 ⊢ (X = ∅ → (A ∩ ∩x ∈ X {y ∈ A ∣ φ}) = A)
2		rzal 3652	. . . . 5 ⊢ (X = ∅ → ∀x ∈ X φ)
3	2	ralrimivw 2699	. . . 4 ⊢ (X = ∅ → ∀y ∈ A ∀x ∈ X φ)
4		rabid2 2789	. . . 4 ⊢ (A = {y ∈ A ∣ ∀x ∈ X φ} ↔ ∀y ∈ A ∀x ∈ X φ)
5	3, 4	sylibr 203	. . 3 ⊢ (X = ∅ → A = {y ∈ A ∣ ∀x ∈ X φ})
6	1, 5	eqtrd 2385	. 2 ⊢ (X = ∅ → (A ∩ ∩x ∈ X {y ∈ A ∣ φ}) = {y ∈ A ∣ ∀x ∈ X φ})
7		ssrab2 3352	. . . . 5 ⊢ {y ∈ A ∣ φ} ⊆ A
8	7	rgenw 2682	. . . 4 ⊢ ∀x ∈ X {y ∈ A ∣ φ} ⊆ A
9		riinn0 4041	. . . 4 ⊢ ((∀x ∈ X {y ∈ A ∣ φ} ⊆ A ∧ X ≠ ∅) → (A ∩ ∩x ∈ X {y ∈ A ∣ φ}) = ∩x ∈ X {y ∈ A ∣ φ})
10	8, 9	mpan 651	. . 3 ⊢ (X ≠ ∅ → (A ∩ ∩x ∈ X {y ∈ A ∣ φ}) = ∩x ∈ X {y ∈ A ∣ φ})
11		iinrab 4029	. . 3 ⊢ (X ≠ ∅ → ∩x ∈ X {y ∈ A ∣ φ} = {y ∈ A ∣ ∀x ∈ X φ})
12	10, 11	eqtrd 2385	. 2 ⊢ (X ≠ ∅ → (A ∩ ∩x ∈ X {y ∈ A ∣ φ}) = {y ∈ A ∣ ∀x ∈ X φ})
13	6, 12	pm2.61ine 2593	1 ⊢ (A ∩ ∩x ∈ X {y ∈ A ∣ φ}) = {y ∈ A ∣ ∀x ∈ X φ}

This proof depends on syntax axioms:   = wceq 1642   ≠ wne 2517  ∀wral 2615  {crab 2619   ∩ cin 3209   ⊆ wss 3258  ∅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-rex 2621  df-rab 2624  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216  df-ss 3260  df-nul 3552  df-iin 3973

This theorem is used by: (None)