Metamath Proof Explorer


Theorem bj-sscon

Description: Contraposition law for relative subclasses. Relative and generalized version of ssconb . Shortens ssconb , conss2 . (Contributed by BJ, 11-Nov-2021) This proof does not rely, even indirectly, on ssconb nor conss2 . (Proof modification is discouraged.)

Ref Expression
Assertion bj-sscon ⊢ A ∩ V ⊆ V ∖ B ↔ B ∩ V ⊆ V ∖ A

Proof

Step Hyp Ref Expression
1 incom ⊢ A ∩ B = B ∩ A
2 1 ineq1i ⊢ A ∩ B ∩ V = B ∩ A ∩ V
3 2 eqeq1i ⊢ A ∩ B ∩ V = ∅ ↔ B ∩ A ∩ V = ∅
4 bj-disj2r ⊢ A ∩ V ⊆ V ∖ B ↔ A ∩ B ∩ V = ∅
5 bj-disj2r ⊢ B ∩ V ⊆ V ∖ A ↔ B ∩ A ∩ V = ∅
6 3 4 5 3bitr4i ⊢ A ∩ V ⊆ V ∖ B ↔ B ∩ V ⊆ V ∖ A