Theorem disjssun 3307

Description: Subset relation for disjoint classes. (Contributed by NM, 25-Oct-2005.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

Ref	Expression
disjssun	⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ 𝐶))

Proof of Theorem disjssun

Step	Hyp	Ref	Expression
1		indi 3211	. . . . 5 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶))
2	1	equncomi 3118	. . . 4 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵))
3		uneq2 3120	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) = ((𝐴 ∩ 𝐶) ∪ ∅))
4		un0 3278	. . . . 5 ⊢ ((𝐴 ∩ 𝐶) ∪ ∅) = (𝐴 ∩ 𝐶)
5	3, 4	syl6eq 2129	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) = (𝐴 ∩ 𝐶))
6	2, 5	syl5eq 2125	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∩ (𝐵 ∪ 𝐶)) = (𝐴 ∩ 𝐶))
7	6	eqeq1d 2089	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴 ↔ (𝐴 ∩ 𝐶) = 𝐴))
8		df-ss 2986	. 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴)
9		df-ss 2986	. 2 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐴 ∩ 𝐶) = 𝐴)
10	7, 8, 9	3bitr4g 221	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ 𝐶))

Syntax hints:   → wi 4   ↔ wb 103   = wceq 1284   ∪ cun 2971   ∩ cin 2972   ⊆ wss 2973  ∅c0 3251

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-in1 576  ax-in2 577  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-dif 2975  df-un 2977  df-in 2979  df-ss 2986  df-nul 3252

This theorem is referenced by: (None)