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 ) )