Metamath Proof Explorer


Theorem sscon34b

Description: Relative complementation reverses inclusion of subclasses. Relativized version of complss . (Contributed by RP, 3-Jun-2021)

Ref Expression
Assertion sscon34b ⊢ A ⊆ C ∧ B ⊆ C → A ⊆ B ↔ C ∖ B ⊆ C ∖ A

Proof

Step Hyp Ref Expression
1 sscon ⊢ A ⊆ B → C ∖ B ⊆ C ∖ A
2 sscon ⊢ C ∖ B ⊆ C ∖ A → C ∖ C ∖ A ⊆ C ∖ C ∖ B
3 dfss4 ⊢ A ⊆ C ↔ C ∖ C ∖ A = A
4 3 birani ⊢ A ⊆ C ∧ B ⊆ C → C ∖ C ∖ A = A
5 dfss4 ⊢ B ⊆ C ↔ C ∖ C ∖ B = B
6 5 bilani ⊢ A ⊆ C ∧ B ⊆ C → C ∖ C ∖ B = B
7 4 6 sseq12d ⊢ A ⊆ C ∧ B ⊆ C → C ∖ C ∖ A ⊆ C ∖ C ∖ B ↔ A ⊆ B
8 2 7 imbitrid ⊢ A ⊆ C ∧ B ⊆ C → C ∖ B ⊆ C ∖ A → A ⊆ B
9 1 8 impbid2 ⊢ A ⊆ C ∧ B ⊆ C → A ⊆ B ↔ C ∖ B ⊆ C ∖ A