Theorem disjssun 3279

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 3178	. . . . 5 ⊢ (A ∩ (B ∪ 𝐶)) = ((A ∩ B) ∪ (A ∩ 𝐶))
2	1	equncomi 3083	. . . 4 ⊢ (A ∩ (B ∪ 𝐶)) = ((A ∩ 𝐶) ∪ (A ∩ B))
3		uneq2 3085	. . . . 5 ⊢ ((A ∩ B) = ∅ → ((A ∩ 𝐶) ∪ (A ∩ B)) = ((A ∩ 𝐶) ∪ ∅))
4		un0 3245	. . . . 5 ⊢ ((A ∩ 𝐶) ∪ ∅) = (A ∩ 𝐶)
5	3, 4	syl6eq 2085	. . . 4 ⊢ ((A ∩ B) = ∅ → ((A ∩ 𝐶) ∪ (A ∩ B)) = (A ∩ 𝐶))
6	2, 5	syl5eq 2081	. . 3 ⊢ ((A ∩ B) = ∅ → (A ∩ (B ∪ 𝐶)) = (A ∩ 𝐶))
7	6	eqeq1d 2045	. 2 ⊢ ((A ∩ B) = ∅ → ((A ∩ (B ∪ 𝐶)) = A ↔ (A ∩ 𝐶) = A))
8		df-ss 2925	. 2 ⊢ (A ⊆ (B ∪ 𝐶) ↔ (A ∩ (B ∪ 𝐶)) = A)
9		df-ss 2925	. 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 1242   ∪ cun 2909   ∩ cin 2910   ⊆ wss 2911  ∅c0 3218

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 544  ax-in2 545  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-dif 2914  df-un 2916  df-in 2918  df-ss 2925  df-nul 3219

This theorem is referenced by: (None)