Theorem elintrab 3783

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 3782	. . 3 ⊢ (𝐴 ∈ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↔ ∀𝑥((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ∈ 𝑥))
3		impexp 261	. . . 4 ⊢ (((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ∈ 𝑥) ↔ (𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ∈ 𝑥)))
4	3	albii 1446	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ∈ 𝑥) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ∈ 𝑥)))
5	2, 4	bitri 183	. 2 ⊢ (𝐴 ∈ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ∈ 𝑥)))
6		df-rab 2425	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
7	6	inteqi 3775	. . 3 ⊢ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} = ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
8	7	eleq2i 2206	. 2 ⊢ (𝐴 ∈ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ 𝐴 ∈ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
9		df-ral 2421	. 2 ⊢ (∀𝑥 ∈ 𝐵 (𝜑 → 𝐴 ∈ 𝑥) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ∈ 𝑥)))
10	5, 8, 9	3bitr4i 211	1 ⊢ (𝐴 ∈ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ ∀𝑥 ∈ 𝐵 (𝜑 → 𝐴 ∈ 𝑥))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  ∀wal 1329   ∈ wcel 1480  {cab 2125  ∀wral 2416  {crab 2420  Vcvv 2686  ∩ cint 3771

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-ral 2421  df-rab 2425  df-v 2688  df-int 3772

This theorem is referenced by:  elintrabg  3784  intmin  3791  bj-indint  13129