Theorem elintrab 4925

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

Hypothesis

Ref	Expression
elintab.ex	⊢ 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		elintab.ex	. . . 4 ⊢ 𝐴 ∈ V
2	1	elintab 4924	. . 3 ⊢ (𝐴 ∈ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↔ ∀𝑥((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ∈ 𝑥))
3		impexp 456	. . . 4 ⊢ (((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ∈ 𝑥) ↔ (𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ∈ 𝑥)))
4	3	albii 1852	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐵 ∧ 𝜑) → 𝐴 ∈ 𝑥) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ∈ 𝑥)))
5	2, 4	bitri 278	. 2 ⊢ (𝐴 ∈ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ∈ 𝑥)))
6		df-rab 3417	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
7	6	inteqi 4916	. . 3 ⊢ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} = ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
8	7	eleq2i 2856	. 2 ⊢ (𝐴 ∈ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ 𝐴 ∈ ∩ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
9		df-ral 3081	. 2 ⊢ (∀𝑥 ∈ 𝐵 (𝜑 → 𝐴 ∈ 𝑥) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝜑 → 𝐴 ∈ 𝑥)))
10	5, 8, 9	3bitr4i 306	1 ⊢ (𝐴 ∈ ∩ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ ∀𝑥 ∈ 𝐵 (𝜑 → 𝐴 ∈ 𝑥))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568   ∈ wcel 2147  {cab 2741  ∀wral 3080  {crab 3416  Vcvv 3454  ∩ cint 4912

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-int 4913

This theorem is used by:  elintrabg  4926  intmin  4933  naddelim  8734  rankunb  9901  isf34lem4  10494  ist1-3  23722  filufint  24294  elspani  32202  fldgensdrg  33980  evls1fldgencl  34413  ldsysgenld  34904  ldgenpisyslem1  34907  kur14lem9  36187  pclclN  41214  elpclN  41215  naddwordnexlem4  44664  lcosslsp  49881