Theorem ssin0 45889

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 4190	. . . 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 3989	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∩ 𝐵) ⊆ ∅)
4	3	3ad2ant1 1151	. . 3 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (𝐴 ∩ 𝐵) ⊆ ∅)
5	2, 4	sstrd 3941	. 2 ⊢ (((𝐴 ∩ 𝐵) = ∅ ∧ 𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (𝐶 ∩ 𝐷) ⊆ ∅)
6		ss0 4352	. 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   ∩ cin 3898   ⊆ wss 3899  ∅c0 4279

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

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 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-dif 3902  df-in 3906  df-ss 3916  df-nul 4280

This theorem is used by:  sge0resplit  47234