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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 biimpi ⊢ ( 𝐴 ⊆ 𝐶 → ( 𝐶 ∖ ( 𝐶 ∖ 𝐴 ) ) = 𝐴 )
5 4 adantr ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) → ( 𝐶 ∖ ( 𝐶 ∖ 𝐴 ) ) = 𝐴 )
6 dfss4 ⊢ ( 𝐵 ⊆ 𝐶 ↔ ( 𝐶 ∖ ( 𝐶 ∖ 𝐵 ) ) = 𝐵 )
7 6 biimpi ⊢ ( 𝐵 ⊆ 𝐶 → ( 𝐶 ∖ ( 𝐶 ∖ 𝐵 ) ) = 𝐵 )
8 7 adantl ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) → ( 𝐶 ∖ ( 𝐶 ∖ 𝐵 ) ) = 𝐵 )
9 5 8 sseq12d ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) → ( ( 𝐶 ∖ ( 𝐶 ∖ 𝐴 ) ) ⊆ ( 𝐶 ∖ ( 𝐶 ∖ 𝐵 ) ) ↔ 𝐴 ⊆ 𝐵 ) )
10 2 9 imbitrid ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) → ( ( 𝐶 ∖ 𝐵 ) ⊆ ( 𝐶 ∖ 𝐴 ) → 𝐴 ⊆ 𝐵 ) )
11 1 10 impbid2 ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) → ( 𝐴 ⊆ 𝐵 ↔ ( 𝐶 ∖ 𝐵 ) ⊆ ( 𝐶 ∖ 𝐴 ) ) )