Theorem disjssun 3609

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 3502	. . . . 5 ⊢ (A ∩ (B ∪ C)) = ((A ∩ B) ∪ (A ∩ C))
2	1	equncomi 3411	. . . 4 ⊢ (A ∩ (B ∪ C)) = ((A ∩ C) ∪ (A ∩ B))
3		uneq2 3413	. . . . 5 ⊢ ((A ∩ B) = ∅ → ((A ∩ C) ∪ (A ∩ B)) = ((A ∩ C) ∪ ∅))
4		un0 3576	. . . . 5 ⊢ ((A ∩ C) ∪ ∅) = (A ∩ C)
5	3, 4	syl6eq 2401	. . . 4 ⊢ ((A ∩ B) = ∅ → ((A ∩ C) ∪ (A ∩ B)) = (A ∩ C))
6	2, 5	syl5eq 2397	. . 3 ⊢ ((A ∩ B) = ∅ → (A ∩ (B ∪ C)) = (A ∩ C))
7	6	eqeq1d 2361	. 2 ⊢ ((A ∩ B) = ∅ → ((A ∩ (B ∪ C)) = A ↔ (A ∩ C) = A))
8		df-ss 3260	. 2 ⊢ (A ⊆ (B ∪ C) ↔ (A ∩ (B ∪ C)) = A)
9		df-ss 3260	. 2 ⊢ (A ⊆ C ↔ (A ∩ C) = A)
10	7, 8, 9	3bitr4g 279	1 ⊢ ((A ∩ B) = ∅ → (A ⊆ (B ∪ C) ↔ A ⊆ C))

Syntax hints:   → wi 4   ↔ wb 176   = wceq 1642   ∪ cun 3208   ∩ cin 3209   ⊆ wss 3258  ∅c0 3551

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-un 3215  df-dif 3216  df-ss 3260  df-nul 3552

This theorem is referenced by: (None)