Metamath Proof Explorer


Theorem disjssi

Description: Subclass theorem for disjoints, inference version. (Contributed by Peter Mazsa, 28-Sep-2021)

Ref Expression
Hypothesis disjssi.1 ⊢ A ⊆ B
Assertion disjssi ⊢ Disj B → Disj A

Proof

Step Hyp Ref Expression
1 disjssi.1 ⊢ A ⊆ B
2 disjss ⊢ A ⊆ B → Disj B → Disj A
3 1 2 ax-mp ⊢ Disj B → Disj A