Theorem disjssun 3560

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 3456	. . . . 5 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶))
2	1	equncomi 3355	. . . 4 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵))
3		uneq2 3357	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) = ((𝐴 ∩ 𝐶) ∪ ∅))
4		un0 3530	. . . . 5 ⊢ ((𝐴 ∩ 𝐶) ∪ ∅) = (𝐴 ∩ 𝐶)
5	3, 4	eqtrdi 2280	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) = (𝐴 ∩ 𝐶))
6	2, 5	eqtrid 2276	. . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∩ (𝐵 ∪ 𝐶)) = (𝐴 ∩ 𝐶))
7	6	eqeq1d 2240	. 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴 ↔ (𝐴 ∩ 𝐶) = 𝐴))
8		df-ss 3214	. 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴)
9		df-ss 3214	. 2 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐴 ∩ 𝐶) = 𝐴)
10	7, 8, 9	3bitr4g 223	1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ 𝐶))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1398   ∪ cun 3199   ∩ cin 3200   ⊆ wss 3201  ∅c0 3496

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-v 2805  df-dif 3203  df-un 3205  df-in 3207  df-ss 3214  df-nul 3497

This theorem is referenced by: (None)