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 ( ( 𝐴𝐶𝐵𝐶 ) → ( 𝐴𝐵 ↔ ( 𝐶𝐵 ) ⊆ ( 𝐶𝐴 ) ) )

Proof

Step Hyp Ref Expression
1 sscon ( 𝐴𝐵 → ( 𝐶𝐵 ) ⊆ ( 𝐶𝐴 ) )
2 sscon ( ( 𝐶𝐵 ) ⊆ ( 𝐶𝐴 ) → ( 𝐶 ∖ ( 𝐶𝐴 ) ) ⊆ ( 𝐶 ∖ ( 𝐶𝐵 ) ) )
3 dfss4 ( 𝐴𝐶 ↔ ( 𝐶 ∖ ( 𝐶𝐴 ) ) = 𝐴 )
4 3 birani ( ( 𝐴𝐶𝐵𝐶 ) → ( 𝐶 ∖ ( 𝐶𝐴 ) ) = 𝐴 )
5 dfss4 ( 𝐵𝐶 ↔ ( 𝐶 ∖ ( 𝐶𝐵 ) ) = 𝐵 )
6 5 bilani ( ( 𝐴𝐶𝐵𝐶 ) → ( 𝐶 ∖ ( 𝐶𝐵 ) ) = 𝐵 )
7 4 6 sseq12d ( ( 𝐴𝐶𝐵𝐶 ) → ( ( 𝐶 ∖ ( 𝐶𝐴 ) ) ⊆ ( 𝐶 ∖ ( 𝐶𝐵 ) ) ↔ 𝐴𝐵 ) )
8 2 7 imbitrid ( ( 𝐴𝐶𝐵𝐶 ) → ( ( 𝐶𝐵 ) ⊆ ( 𝐶𝐴 ) → 𝐴𝐵 ) )
9 1 8 impbid2 ( ( 𝐴𝐶𝐵𝐶 ) → ( 𝐴𝐵 ↔ ( 𝐶𝐵 ) ⊆ ( 𝐶𝐴 ) ) )