Theorem iinuniss 3948

Description: A relationship involving union and indexed intersection. Exercise 23 of [Enderton] p. 33 but with equality changed to subset. (Contributed by Jim Kingdon, 19-Aug-2018.)

Assertion

Ref	Expression
iinuniss	⊢ (𝐴 ∪ ∩ 𝐵) ⊆ ∩ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥)

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem iinuniss

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		r19.32vr 2614	. . . 4 ⊢ ((𝑦 ∈ 𝐴 ∨ ∀𝑥 ∈ 𝐵 𝑦 ∈ 𝑥) → ∀𝑥 ∈ 𝐵 (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝑥))
2		vex 2729	. . . . . 6 ⊢ 𝑦 ∈ V
3	2	elint2 3831	. . . . 5 ⊢ (𝑦 ∈ ∩ 𝐵 ↔ ∀𝑥 ∈ 𝐵 𝑦 ∈ 𝑥)
4	3	orbi2i 752	. . . 4 ⊢ ((𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵) ↔ (𝑦 ∈ 𝐴 ∨ ∀𝑥 ∈ 𝐵 𝑦 ∈ 𝑥))
5		elun 3263	. . . . 5 ⊢ (𝑦 ∈ (𝐴 ∪ 𝑥) ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝑥))
6	5	ralbii 2472	. . . 4 ⊢ (∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥) ↔ ∀𝑥 ∈ 𝐵 (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝑥))
7	1, 4, 6	3imtr4i 200	. . 3 ⊢ ((𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵) → ∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥))
8	7	ss2abi 3214	. 2 ⊢ {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵)} ⊆ {𝑦 ∣ ∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥)}
9		df-un 3120	. 2 ⊢ (𝐴 ∪ ∩ 𝐵) = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵)}
10		df-iin 3869	. 2 ⊢ ∩ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = {𝑦 ∣ ∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥)}
11	8, 9, 10	3sstr4i 3183	1 ⊢ (𝐴 ∪ ∩ 𝐵) ⊆ ∩ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥)

Syntax hints:   ∨ wo 698   ∈ wcel 2136  {cab 2151  ∀wral 2444   ∪ cun 3114   ⊆ wss 3116  ∩ cint 3824  ∩ ciin 3867

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 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-ral 2449  df-v 2728  df-un 3120  df-in 3122  df-ss 3129  df-int 3825  df-iin 3869

This theorem is referenced by: (None)