Metamath Proof Explorer


Theorem disjss

Description: Subclass theorem for disjoints. (Contributed by Peter Mazsa, 28-Oct-2020) (Revised by Peter Mazsa, 22-Sep-2021)

Ref Expression
Assertion disjss ⊢ A ⊆ B → Disj B → Disj A

Proof

Step Hyp Ref Expression
1 cnvss ⊢ A ⊆ B → A -1 ⊆ B -1
2 funALTVss ⊢ A -1 ⊆ B -1 → FunALTV B -1 → FunALTV A -1
3 1 2 syl ⊢ A ⊆ B → FunALTV B -1 → FunALTV A -1
4 relss ⊢ A ⊆ B → Rel ⁡ B → Rel ⁡ A
5 3 4 anim12d ⊢ A ⊆ B → FunALTV B -1 ∧ Rel ⁡ B → FunALTV A -1 ∧ Rel ⁡ A
6 dfdisjALTV ⊢ Disj B ↔ FunALTV B -1 ∧ Rel ⁡ B
7 dfdisjALTV ⊢ Disj A ↔ FunALTV A -1 ∧ Rel ⁡ A
8 5 6 7 3imtr4g ⊢ A ⊆ B → Disj B → Disj A