Metamath Proof Explorer


Theorem rcompleq

Description: Two subclasses are equal if and only if their relative complements are equal. Relativized version of compleq . (Contributed by RP, 10-Jun-2021)

Ref Expression
Assertion rcompleq ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) → ( 𝐴 = 𝐵 ↔ ( 𝐶 ∖ 𝐴 ) = ( 𝐶 ∖ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 ancom ⊢ ( ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴 ) ↔ ( 𝐵 ⊆ 𝐴 ∧ 𝐴 ⊆ 𝐵 ) )
2 sscon34b ⊢ ( ( 𝐵 ⊆ 𝐶 ∧ 𝐴 ⊆ 𝐶 ) → ( 𝐵 ⊆ 𝐴 ↔ ( 𝐶 ∖ 𝐴 ) ⊆ ( 𝐶 ∖ 𝐵 ) ) )
3 2 ancoms ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) → ( 𝐵 ⊆ 𝐴 ↔ ( 𝐶 ∖ 𝐴 ) ⊆ ( 𝐶 ∖ 𝐵 ) ) )
4 sscon34b ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) → ( 𝐴 ⊆ 𝐵 ↔ ( 𝐶 ∖ 𝐵 ) ⊆ ( 𝐶 ∖ 𝐴 ) ) )
5 3 4 anbi12d ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) → ( ( 𝐵 ⊆ 𝐴 ∧ 𝐴 ⊆ 𝐵 ) ↔ ( ( 𝐶 ∖ 𝐴 ) ⊆ ( 𝐶 ∖ 𝐵 ) ∧ ( 𝐶 ∖ 𝐵 ) ⊆ ( 𝐶 ∖ 𝐴 ) ) ) )
6 1 5 bitrid ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) → ( ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴 ) ↔ ( ( 𝐶 ∖ 𝐴 ) ⊆ ( 𝐶 ∖ 𝐵 ) ∧ ( 𝐶 ∖ 𝐵 ) ⊆ ( 𝐶 ∖ 𝐴 ) ) ) )
7 eqss ⊢ ( 𝐴 = 𝐵 ↔ ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴 ) )
8 eqss ⊢ ( ( 𝐶 ∖ 𝐴 ) = ( 𝐶 ∖ 𝐵 ) ↔ ( ( 𝐶 ∖ 𝐴 ) ⊆ ( 𝐶 ∖ 𝐵 ) ∧ ( 𝐶 ∖ 𝐵 ) ⊆ ( 𝐶 ∖ 𝐴 ) ) )
9 6 7 8 3bitr4g ⊢ ( ( 𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐶 ) → ( 𝐴 = 𝐵 ↔ ( 𝐶 ∖ 𝐴 ) = ( 𝐶 ∖ 𝐵 ) ) )