Theorem elintrab 3618

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 3617	. . 3 ⊢ (A ∈ ∩ {x ∣ (x ∈ B ∧ φ)} ↔ ∀x((x ∈ B ∧ φ) → A ∈ x))
3		impexp 250	. . . 4 ⊢ (((x ∈ B ∧ φ) → A ∈ x) ↔ (x ∈ B → (φ → A ∈ x)))
4	3	albii 1356	. . 3 ⊢ (∀x((x ∈ B ∧ φ) → A ∈ x) ↔ ∀x(x ∈ B → (φ → A ∈ x)))
5	2, 4	bitri 173	. 2 ⊢ (A ∈ ∩ {x ∣ (x ∈ B ∧ φ)} ↔ ∀x(x ∈ B → (φ → A ∈ x)))
6		df-rab 2309	. . . 4 ⊢ {x ∈ B ∣ φ} = {x ∣ (x ∈ B ∧ φ)}
7	6	inteqi 3610	. . 3 ⊢ ∩ {x ∈ B ∣ φ} = ∩ {x ∣ (x ∈ B ∧ φ)}
8	7	eleq2i 2101	. 2 ⊢ (A ∈ ∩ {x ∈ B ∣ φ} ↔ A ∈ ∩ {x ∣ (x ∈ B ∧ φ)})
9		df-ral 2305	. 2 ⊢ (∀x ∈ B (φ → A ∈ x) ↔ ∀x(x ∈ B → (φ → A ∈ x)))
10	5, 8, 9	3bitr4i 201	1 ⊢ (A ∈ ∩ {x ∈ B ∣ φ} ↔ ∀x ∈ B (φ → A ∈ x))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98  ∀wal 1240   ∈ wcel 1390  {cab 2023  ∀wral 2300  {crab 2304  Vcvv 2551  ∩ cint 3606

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-ral 2305  df-rab 2309  df-v 2553  df-int 3607

This theorem is referenced by:  elintrabg  3619  intmin  3626  bj-indint  9320