Theorem elintrab 3896

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 3895	. . 3 ⊢ (𝐴 ∈ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↔ ∀𝑥((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ∈ 𝑥))
3		impexp 263	. . . 4 ⊢ (((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ∈ 𝑥) ↔ (𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ∈ 𝑥)))
4	3	albii 1492	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ∈ 𝑥) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ∈ 𝑥)))
5	2, 4	bitri 184	. 2 ⊢ (𝐴 ∈ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ∈ 𝑥)))
6		df-rab 2492	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
7	6	inteqi 3888	. . 3 ⊢ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} = ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
8	7	eleq2i 2271	. 2 ⊢ (𝐴 ∈ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ 𝐴 ∈ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
9		df-ral 2488	. 2 ⊢ (∀𝑥 ∈ 𝐵 (𝜑 → 𝐴 ∈ 𝑥) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ∈ 𝑥)))
10	5, 8, 9	3bitr4i 212	1 ⊢ (𝐴 ∈ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ ∀𝑥 ∈ 𝐵 (𝜑 → 𝐴 ∈ 𝑥))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105  ∀wal 1370   ∈ wcel 2175  {cab 2190  ∀wral 2483  {crab 2487  Vcvv 2771  ∩ cint 3884

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1469  ax-7 1470  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-10 1527  ax-11 1528  ax-i12 1529  ax-bndl 1531  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557  ax-ext 2186

This theorem depends on definitions:  df-bi 117  df-tru 1375  df-nf 1483  df-sb 1785  df-clab 2191  df-cleq 2197  df-clel 2200  df-nfc 2336  df-ral 2488  df-rab 2492  df-v 2773  df-int 3885

This theorem is referenced by:  elintrabg  3897  intmin  3904  bj-indint  15731