Theorem elintrab 3939

Description: Membership in the intersection of a class abstraction. (Contributed by NM, 17-Oct-1999.)

Hypothesis

Ref	Expression
inteqab.1	⊢ A ∈ V

Assertion

Ref	Expression
elintrab	⊢ (A ∈ ∩{x ∈ B ∣ φ} ↔ ∀x ∈ B (φ → A ∈ x))

Distinct variable group:   x,A

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

Proof of Theorem elintrab

Step	Hyp	Ref	Expression
1		inteqab.1	. . . 4 ⊢ A ∈ V
2	1	elintab 3938	. . 3 ⊢ (A ∈ ∩{x ∣ (x ∈ B ∧ φ)} ↔ ∀x((x ∈ B ∧ φ) → A ∈ x))
3		impexp 433	. . . 4 ⊢ (((x ∈ B ∧ φ) → A ∈ x) ↔ (x ∈ B → (φ → A ∈ x)))
4	3	albii 1566	. . 3 ⊢ (∀x((x ∈ B ∧ φ) → A ∈ x) ↔ ∀x(x ∈ B → (φ → A ∈ x)))
5	2, 4	bitri 240	. 2 ⊢ (A ∈ ∩{x ∣ (x ∈ B ∧ φ)} ↔ ∀x(x ∈ B → (φ → A ∈ x)))
6		df-rab 2624	. . . 4 ⊢ {x ∈ B ∣ φ} = {x ∣ (x ∈ B ∧ φ)}
7	6	inteqi 3931	. . 3 ⊢ ∩{x ∈ B ∣ φ} = ∩{x ∣ (x ∈ B ∧ φ)}
8	7	eleq2i 2417	. 2 ⊢ (A ∈ ∩{x ∈ B ∣ φ} ↔ A ∈ ∩{x ∣ (x ∈ B ∧ φ)})
9		df-ral 2620	. 2 ⊢ (∀x ∈ B (φ → A ∈ x) ↔ ∀x(x ∈ B → (φ → A ∈ x)))
10	5, 8, 9	3bitr4i 268	1 ⊢ (A ∈ ∩{x ∈ B ∣ φ} ↔ ∀x ∈ B (φ → A ∈ x))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540   ∈ wcel 1710  {cab 2339  ∀wral 2615  {crab 2619  Vcvv 2860  ∩cint 3927

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-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-ral 2620  df-rab 2624  df-v 2862  df-int 3928

This theorem is used by:  elintrabg  3940  intmin  3947