Theorem ssin0 46346

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 4144	. . . 4 ⊢ ((𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (𝐶 ∩ 𝐷) ⊆ (𝐴 ∩ 𝐵))
2	1	3adant1 1148	. . 3 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (𝐶 ∩ 𝐷) ⊆ (𝐴 ∩ 𝐵))
3		eqimss 3971	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∩ 𝐵) ⊆ ∅)
4	3	3ad2ant1 1151	. . 3 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (𝐴 ∩ 𝐵) ⊆ ∅)
5	2, 4	sstrd 3923	. 2 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (𝐶 ∩ 𝐷) ⊆ ∅)
6		ss0 4355	. 2 ⊢ ((𝐶 ∩ 𝐷) ⊆ ∅ → (𝐶 ∩ 𝐷) = ∅)
7	5, 6	syl 18	1 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (𝐶 ∩ 𝐷) = ∅)

This proof depends on syntax axioms:   → wi 4   ∧ w3a 1103   = wceq 1570   ⊆ wss 3900   ∩ cin 4092  ∅c0 4281

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-nul 4282

This theorem is used by:  sge0resplit  47689