Theorem disjssun 3262

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 3161	. . . . 5 ⊢ (A ∩ (B ∪ 𝐶)) = ((A ∩ B) ∪ (A ∩ 𝐶))
2	1	equncomi 3066	. . . 4 ⊢ (A ∩ (B ∪ 𝐶)) = ((A ∩ 𝐶) ∪ (A ∩ B))
3		uneq2 3068	. . . . 5 ⊢ ((A ∩ B) = ∅ → ((A ∩ 𝐶) ∪ (A ∩ B)) = ((A ∩ 𝐶) ∪ ∅))
4		un0 3228	. . . . 5 ⊢ ((A ∩ 𝐶) ∪ ∅) = (A ∩ 𝐶)
5	3, 4	syl6eq 2070	. . . 4 ⊢ ((A ∩ B) = ∅ → ((A ∩ 𝐶) ∪ (A ∩ B)) = (A ∩ 𝐶))
6	2, 5	syl5eq 2066	. . 3 ⊢ ((A ∩ B) = ∅ → (A ∩ (B ∪ 𝐶)) = (A ∩ 𝐶))
7	6	eqeq1d 2030	. 2 ⊢ ((A ∩ B) = ∅ → ((A ∩ (B ∪ 𝐶)) = A ↔ (A ∩ 𝐶) = A))
8		df-ss 2908	. 2 ⊢ (A ⊆ (B ∪ 𝐶) ↔ (A ∩ (B ∪ 𝐶)) = A)
9		df-ss 2908	. 2 ⊢ (A ⊆ 𝐶 ↔ (A ∩ 𝐶) = A)
10	7, 8, 9	3bitr4g 212	1 ⊢ ((A ∩ B) = ∅ → (A ⊆ (B ∪ 𝐶) ↔ A ⊆ 𝐶))

Syntax hints:   → wi 4   ↔ wb 98   = wceq 1228   ∪ cun 2892   ∩ cin 2893   ⊆ wss 2894  ∅c0 3201

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 532  ax-in2 533  ax-io 617  ax-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-10 1377  ax-11 1378  ax-i12 1379  ax-bnd 1380  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004

This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-nfc 2149  df-v 2537  df-dif 2897  df-un 2899  df-in 2901  df-ss 2908  df-nul 3202

This theorem is referenced by: (None)