Theorem ssin0 45065

Description: If two classes are disjoint, two respective subclasses are disjoint. (Contributed by Glauco Siliprandi, 17-Aug-2020.)

Assertion

Ref	Expression
ssin0	⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (𝐶 ∩ 𝐷) = ∅)

Proof of Theorem ssin0

Step	Hyp	Ref	Expression
1		ss2in 4244	. . . 4 ⊢ ((𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (𝐶 ∩ 𝐷) ⊆ (𝐴 ∩ 𝐵))
2	1	3adant1 1130	. . 3 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (𝐶 ∩ 𝐷) ⊆ (𝐴 ∩ 𝐵))
3		eqimss 4041	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∩ 𝐵) ⊆ ∅)
4	3	3ad2ant1 1133	. . 3 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (𝐴 ∩ 𝐵) ⊆ ∅)
5	2, 4	sstrd 3993	. 2 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (𝐶 ∩ 𝐷) ⊆ ∅)
6		ss0 4401	. 2 ⊢ ((𝐶 ∩ 𝐷) ⊆ ∅ → (𝐶 ∩ 𝐷) = ∅)
7	5, 6	syl 17	1 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (𝐶 ∩ 𝐷) = ∅)

Syntax hints:   → wi 4   ∧ w3a 1086   = wceq 1539   ∩ cin 3949   ⊆ wss 3950  ∅c0 4332

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1779  df-sb 2064  df-clab 2714  df-cleq 2728  df-clel 2815  df-rab 3436  df-v 3481  df-dif 3953  df-in 3957  df-ss 3967  df-nul 4333

This theorem is referenced by:  sge0resplit  46426