Metamath Proof Explorer


Theorem ssdifsym

Description: Symmetric class differences for subclasses. (Contributed by AV, 3-Jan-2022)

Ref Expression
Assertion ssdifsym ( ( 𝐴 ⊆ 𝑉 ∧ 𝐵 ⊆ 𝑉 ) → ( 𝐵 = ( 𝑉 ∖ 𝐴 ) ↔ 𝐴 = ( 𝑉 ∖ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 ssdifim ⊢ ( ( 𝐴 ⊆ 𝑉 ∧ 𝐵 = ( 𝑉 ∖ 𝐴 ) ) → 𝐴 = ( 𝑉 ∖ 𝐵 ) )
2 1 ex ⊢ ( 𝐴 ⊆ 𝑉 → ( 𝐵 = ( 𝑉 ∖ 𝐴 ) → 𝐴 = ( 𝑉 ∖ 𝐵 ) ) )
3 ssdifim ⊢ ( ( 𝐵 ⊆ 𝑉 ∧ 𝐴 = ( 𝑉 ∖ 𝐵 ) ) → 𝐵 = ( 𝑉 ∖ 𝐴 ) )
4 3 ex ⊢ ( 𝐵 ⊆ 𝑉 → ( 𝐴 = ( 𝑉 ∖ 𝐵 ) → 𝐵 = ( 𝑉 ∖ 𝐴 ) ) )
5 2 4 anbiim ⊢ ( ( 𝐴 ⊆ 𝑉 ∧ 𝐵 ⊆ 𝑉 ) → ( 𝐵 = ( 𝑉 ∖ 𝐴 ) ↔ 𝐴 = ( 𝑉 ∖ 𝐵 ) ) )