Metamath Proof Explorer


Theorem ssdifsn

Description: Subset of a set with an element removed. (Contributed by Emmett Weisz, 7-Jul-2021) (Proof shortened by JJ, 31-May-2022)

Ref Expression
Assertion ssdifsn ⊢ A ⊆ B ∖ C ↔ A ⊆ B ∧ ¬ C ∈ A

Proof

Step Hyp Ref Expression
1 difss2 ⊢ A ⊆ B ∖ C → A ⊆ B
2 reldisj ⊢ A ⊆ B → A ∩ C = ∅ ↔ A ⊆ B ∖ C
3 2 bicomd ⊢ A ⊆ B → A ⊆ B ∖ C ↔ A ∩ C = ∅
4 1 3 biadanii ⊢ A ⊆ B ∖ C ↔ A ⊆ B ∧ A ∩ C = ∅
5 disjsn ⊢ A ∩ C = ∅ ↔ ¬ C ∈ A
6 5 anbi2i ⊢ A ⊆ B ∧ A ∩ C = ∅ ↔ A ⊆ B ∧ ¬ C ∈ A
7 4 6 bitri ⊢ A ⊆ B ∖ C ↔ A ⊆ B ∧ ¬ C ∈ A