Theorem elunirab 3593

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 3592	. 2 ⊢ (𝐴 ∈ ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↔ ∃𝑥(𝐴 ∈ 𝑥 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)))
2		df-rab 2315	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
3	2	unieqi 3590	. . 3 ⊢ ∪ {𝑥 ∈ 𝐵 ∣ 𝜑} = ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
4	3	eleq2i 2104	. 2 ⊢ (𝐴 ∈ ∪ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ 𝐴 ∈ ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
5		df-rex 2312	. . 3 ⊢ (∃𝑥 ∈ 𝐵 (𝐴 ∈ 𝑥 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝐴 ∈ 𝑥 ∧ 𝜑)))
6		an12 495	. . . 4 ⊢ ((𝑥 ∈ 𝐵 ∧ (𝐴 ∈ 𝑥 ∧ 𝜑)) ↔ (𝐴 ∈ 𝑥 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)))
7	6	exbii 1496	. . 3 ⊢ (∃𝑥(𝑥 ∈ 𝐵 ∧ (𝐴 ∈ 𝑥 ∧ 𝜑)) ↔ ∃𝑥(𝐴 ∈ 𝑥 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)))
8	5, 7	bitri 173	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝐴 ∈ 𝑥 ∧ 𝜑) ↔ ∃𝑥(𝐴 ∈ 𝑥 ∧ (𝑥 ∈ 𝐵 ∧ 𝜑)))
9	1, 4, 8	3bitr4i 201	1 ⊢ (𝐴 ∈ ∪ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ ∃𝑥 ∈ 𝐵 (𝐴 ∈ 𝑥 ∧ 𝜑))

Syntax hints:   ∧ wa 97   ↔ wb 98  ∃wex 1381   ∈ wcel 1393  {cab 2026  ∃wrex 2307  {crab 2310  ∪ cuni 3580

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 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-rex 2312  df-rab 2315  df-v 2559  df-uni 3581

This theorem is referenced by: (None)