Theorem iinuniss 3964

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 2623	. . . 4 ⊢ ((𝑦 ∈ 𝐴 ∨ ∀𝑥 ∈ 𝐵 𝑦 ∈ 𝑥) → ∀𝑥 ∈ 𝐵 (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝑥))
2		vex 2738	. . . . . 6 ⊢ 𝑦 ∈ V
3	2	elint2 3847	. . . . 5 ⊢ (𝑦 ∈ ∩ 𝐵 ↔ ∀𝑥 ∈ 𝐵 𝑦 ∈ 𝑥)
4	3	orbi2i 762	. . . 4 ⊢ ((𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵) ↔ (𝑦 ∈ 𝐴 ∨ ∀𝑥 ∈ 𝐵 𝑦 ∈ 𝑥))
5		elun 3274	. . . . 5 ⊢ (𝑦 ∈ (𝐴 ∪ 𝑥) ↔ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ 𝑥))
6	5	ralbii 2481	. . . 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 3225	. 2 ⊢ {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵)} ⊆ {𝑦 ∣ ∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥)}
9		df-un 3131	. 2 ⊢ (𝐴 ∪ ∩ 𝐵) = {𝑦 ∣ (𝑦 ∈ 𝐴 ∨ 𝑦 ∈ ∩ 𝐵)}
10		df-iin 3885	. 2 ⊢ ∩ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥) = {𝑦 ∣ ∀𝑥 ∈ 𝐵 𝑦 ∈ (𝐴 ∪ 𝑥)}
11	8, 9, 10	3sstr4i 3194	1 ⊢ (𝐴 ∪ ∩ 𝐵) ⊆ ∩ 𝑥 ∈ 𝐵 (𝐴 ∪ 𝑥)

Syntax hints:   ∨ wo 708   ∈ wcel 2146  {cab 2161  ∀wral 2453   ∪ cun 3125   ⊆ wss 3127  ∩ cint 3840  ∩ ciin 3883

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 709  ax-5 1445  ax-7 1446  ax-gen 1447  ax-ie1 1491  ax-ie2 1492  ax-8 1502  ax-10 1503  ax-11 1504  ax-i12 1505  ax-bndl 1507  ax-4 1508  ax-17 1524  ax-i9 1528  ax-ial 1532  ax-i5r 1533  ax-ext 2157

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1459  df-sb 1761  df-clab 2162  df-cleq 2168  df-clel 2171  df-nfc 2306  df-ral 2458  df-v 2737  df-un 3131  df-in 3133  df-ss 3140  df-int 3841  df-iin 3885

This theorem is referenced by: (None)