Metamath Proof Explorer


Theorem ssdisjd

Description: Subset preserves disjointness. Deduction form of ssdisj . (Contributed by Zhi Wang, 7-Sep-2024)

Ref Expression
Hypotheses ssdisjd.1 ⊢ φ → A ⊆ B
ssdisjd.2 ⊢ φ → B ∩ C = ∅
Assertion ssdisjd ⊢ φ → A ∩ C = ∅

Proof

Step Hyp Ref Expression
1 ssdisjd.1 ⊢ φ → A ⊆ B
2 ssdisjd.2 ⊢ φ → B ∩ C = ∅
3 1 ssrind ⊢ φ → A ∩ C ⊆ B ∩ C
4 sseq0 ⊢ A ∩ C ⊆ B ∩ C ∧ B ∩ C = ∅ → A ∩ C = ∅
5 3 2 4 syl2anc ⊢ φ → A ∩ C = ∅