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 ) = ∅ )