Theorem uniin 3621

Description: The class union of the intersection of two classes. Exercise 4.12(n) of [Mendelson] p. 235. (Contributed by NM, 4-Dec-2003.) (Proof shortened by Andrew Salmon, 29-Jun-2011.)

Assertion

Ref	Expression
uniin	⊢ ∪ (𝐴 ∩ 𝐵) ⊆ (∪ 𝐴 ∩ ∪ 𝐵)

Proof of Theorem uniin

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		19.40 1562	. . . 4 ⊢ (∃𝑦((𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐴) ∧ (𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐵)) → (∃𝑦(𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐴) ∧ ∃𝑦(𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐵)))
2		elin 3155	. . . . . . 7 ⊢ (𝑦 ∈ (𝐴 ∩ 𝐵) ↔ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵))
3	2	anbi2i 444	. . . . . 6 ⊢ ((𝑥 ∈ 𝑦 ∧ 𝑦 ∈ (𝐴 ∩ 𝐵)) ↔ (𝑥 ∈ 𝑦 ∧ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)))
4		anandi 554	. . . . . 6 ⊢ ((𝑥 ∈ 𝑦 ∧ (𝑦 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) ↔ ((𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐴) ∧ (𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐵)))
5	3, 4	bitri 182	. . . . 5 ⊢ ((𝑥 ∈ 𝑦 ∧ 𝑦 ∈ (𝐴 ∩ 𝐵)) ↔ ((𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐴) ∧ (𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐵)))
6	5	exbii 1536	. . . 4 ⊢ (∃𝑦(𝑥 ∈ 𝑦 ∧ 𝑦 ∈ (𝐴 ∩ 𝐵)) ↔ ∃𝑦((𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐴) ∧ (𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐵)))
7		eluni 3604	. . . . 5 ⊢ (𝑥 ∈ ∪ 𝐴 ↔ ∃𝑦(𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐴))
8		eluni 3604	. . . . 5 ⊢ (𝑥 ∈ ∪ 𝐵 ↔ ∃𝑦(𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐵))
9	7, 8	anbi12i 447	. . . 4 ⊢ ((𝑥 ∈ ∪ 𝐴 ∧ 𝑥 ∈ ∪ 𝐵) ↔ (∃𝑦(𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐴) ∧ ∃𝑦(𝑥 ∈ 𝑦 ∧ 𝑦 ∈ 𝐵)))
10	1, 6, 9	3imtr4i 199	. . 3 ⊢ (∃𝑦(𝑥 ∈ 𝑦 ∧ 𝑦 ∈ (𝐴 ∩ 𝐵)) → (𝑥 ∈ ∪ 𝐴 ∧ 𝑥 ∈ ∪ 𝐵))
11		eluni 3604	. . 3 ⊢ (𝑥 ∈ ∪ (𝐴 ∩ 𝐵) ↔ ∃𝑦(𝑥 ∈ 𝑦 ∧ 𝑦 ∈ (𝐴 ∩ 𝐵)))
12		elin 3155	. . 3 ⊢ (𝑥 ∈ (∪ 𝐴 ∩ ∪ 𝐵) ↔ (𝑥 ∈ ∪ 𝐴 ∧ 𝑥 ∈ ∪ 𝐵))
13	10, 11, 12	3imtr4i 199	. 2 ⊢ (𝑥 ∈ ∪ (𝐴 ∩ 𝐵) → 𝑥 ∈ (∪ 𝐴 ∩ ∪ 𝐵))
14	13	ssriv 3003	1 ⊢ ∪ (𝐴 ∩ 𝐵) ⊆ (∪ 𝐴 ∩ ∪ 𝐵)

Syntax hints:   ∧ wa 102  ∃wex 1421   ∈ wcel 1433   ∩ cin 2972   ⊆ wss 2973  ∪ cuni 3601

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 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-in 2979  df-ss 2986  df-uni 3602

This theorem is referenced by: (None)