Theorem elunirab 3905

Description: Membership in union of a class abstraction. (Contributed by NM, 4-Oct-2006.)

Assertion

Ref	Expression
elunirab	⊢ (A ∈ ∪{x ∈ B ∣ φ} ↔ ∃x ∈ B (A ∈ x ∧ φ))

Distinct variable group:   x,A

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

Proof of Theorem elunirab

Step	Hyp	Ref	Expression
1		eluniab 3904	. 2 ⊢ (A ∈ ∪{x ∣ (x ∈ B ∧ φ)} ↔ ∃x(A ∈ x ∧ (x ∈ B ∧ φ)))
2		df-rab 2624	. . . 4 ⊢ {x ∈ B ∣ φ} = {x ∣ (x ∈ B ∧ φ)}
3	2	unieqi 3902	. . 3 ⊢ ∪{x ∈ B ∣ φ} = ∪{x ∣ (x ∈ B ∧ φ)}
4	3	eleq2i 2417	. 2 ⊢ (A ∈ ∪{x ∈ B ∣ φ} ↔ A ∈ ∪{x ∣ (x ∈ B ∧ φ)})
5		df-rex 2621	. . 3 ⊢ (∃x ∈ B (A ∈ x ∧ φ) ↔ ∃x(x ∈ B ∧ (A ∈ x ∧ φ)))
6		an12 772	. . . 4 ⊢ ((x ∈ B ∧ (A ∈ x ∧ φ)) ↔ (A ∈ x ∧ (x ∈ B ∧ φ)))
7	6	exbii 1582	. . 3 ⊢ (∃x(x ∈ B ∧ (A ∈ x ∧ φ)) ↔ ∃x(A ∈ x ∧ (x ∈ B ∧ φ)))
8	5, 7	bitri 240	. 2 ⊢ (∃x ∈ B (A ∈ x ∧ φ) ↔ ∃x(A ∈ x ∧ (x ∈ B ∧ φ)))
9	1, 4, 8	3bitr4i 268	1 ⊢ (A ∈ ∪{x ∈ B ∣ φ} ↔ ∃x ∈ B (A ∈ x ∧ φ))

This proof depends on syntax axioms:   ↔ wb 176   ∧ wa 358  ∃wex 1541   ∈ wcel 1710  {cab 2339  ∃wrex 2616  {crab 2619  ∪cuni 3892

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-rex 2621  df-rab 2624  df-v 2862  df-uni 3893

This theorem is used by: (None)