Theorem iinuniss 3787

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 2510	. . . 4 ⊢ ((𝑦 ∈ 𝐴 ∨ ∀𝑥 ∈ 𝐵 𝑦 ∈ 𝑥) → ∀𝑥 ∈ 𝐵 (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝑥))
2		vex 2617	. . . . . 6 ⊢ 𝑦 ∈ V
3	2	elint2 3672	. . . . 5 ⊢ (𝑦 ∈ ∩ 𝐵 ↔ ∀𝑥 ∈ 𝐵 𝑦 ∈ 𝑥)
4	3	orbi2i 712	. . . 4 ⊢ ((𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵) ↔ (𝑦 ∈ 𝐴 ∨ ∀𝑥 ∈ 𝐵 𝑦 ∈ 𝑥))
5		elun 3127	. . . . 5 ⊢ (𝑦 ∈ (𝐴 ∪ 𝑥) ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝑥))
6	5	ralbii 2380	. . . 4 ⊢ (∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥) ↔ ∀𝑥 ∈ 𝐵 (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝑥))
7	1, 4, 6	3imtr4i 199	. . 3 ⊢ ((𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵) → ∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥))
8	7	ss2abi 3079	. 2 ⊢ {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵)} ⊆ {𝑦 ∣ ∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥)}
9		df-un 2990	. 2 ⊢ (𝐴 ∪ ∩ 𝐵) = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵)}
10		df-iin 3710	. 2 ⊢ ∩ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = {𝑦 ∣ ∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥)}
11	8, 9, 10	3sstr4i 3051	1 ⊢ (𝐴 ∪ ∩ 𝐵) ⊆ ∩ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥)

Syntax hints:   ∨ wo 662   ∈ wcel 1436  {cab 2071  ∀wral 2355   ∪ cun 2984   ⊆ wss 2986  ∩ cint 3665  ∩ ciin 3708

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 663  ax-5 1379  ax-7 1380  ax-gen 1381  ax-ie1 1425  ax-ie2 1426  ax-8 1438  ax-10 1439  ax-11 1440  ax-i12 1441  ax-bndl 1442  ax-4 1443  ax-17 1462  ax-i9 1466  ax-ial 1470  ax-i5r 1471  ax-ext 2067

This theorem depends on definitions:  df-bi 115  df-tru 1290  df-nf 1393  df-sb 1690  df-clab 2072  df-cleq 2078  df-clel 2081  df-nfc 2214  df-ral 2360  df-v 2616  df-un 2990  df-in 2992  df-ss 2999  df-int 3666  df-iin 3710

This theorem is referenced by: (None)