Theorem iinuniss 4047

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 2679	. . . 4 ⊢ ((𝑦 ∈ 𝐴 ∨ ∀𝑥 ∈ 𝐵 𝑦 ∈ 𝑥) → ∀𝑥 ∈ 𝐵 (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝑥))
2		vex 2802	. . . . . 6 ⊢ 𝑦 ∈ V
3	2	elint2 3929	. . . . 5 ⊢ (𝑦 ∈ ∩ 𝐵 ↔ ∀𝑥 ∈ 𝐵 𝑦 ∈ 𝑥)
4	3	orbi2i 767	. . . 4 ⊢ ((𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵) ↔ (𝑦 ∈ 𝐴 ∨ ∀𝑥 ∈ 𝐵 𝑦 ∈ 𝑥))
5		elun 3345	. . . . 5 ⊢ (𝑦 ∈ (𝐴 ∪ 𝑥) ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝑥))
6	5	ralbii 2536	. . . 4 ⊢ (∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥) ↔ ∀𝑥 ∈ 𝐵 (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝑥))
7	1, 4, 6	3imtr4i 201	. . 3 ⊢ ((𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵) → ∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥))
8	7	ss2abi 3296	. 2 ⊢ {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵)} ⊆ {𝑦 ∣ ∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥)}
9		df-un 3201	. 2 ⊢ (𝐴 ∪ ∩ 𝐵) = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵)}
10		df-iin 3967	. 2 ⊢ ∩ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = {𝑦 ∣ ∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥)}
11	8, 9, 10	3sstr4i 3265	1 ⊢ (𝐴 ∪ ∩ 𝐵) ⊆ ∩ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥)

Syntax hints:   ∨ wo 713   ∈ wcel 2200  {cab 2215  ∀wral 2508   ∪ cun 3195   ⊆ wss 3197  ∩ cint 3922  ∩ ciin 3965

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 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-ral 2513  df-v 2801  df-un 3201  df-in 3203  df-ss 3210  df-int 3923  df-iin 3967

This theorem is referenced by: (None)