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Metamath Proof Explorer

Theorem ddif

Description: Double complement under universal class. Exercise 4.10(s) of Mendelson p. 231. (Contributed by NM, 8-Jan-2002)

Ref Expression
Assertion ddif ⊢ V ∖ V ∖ A = A

Proof

Step Hyp Ref Expression
1 velcomp ⊢ x ∈ V ∖ A ↔ ¬ x ∈ A
2 1 con2bii ⊢ x ∈ A ↔ ¬ x ∈ V ∖ A
3 vex ⊢ x ∈ V
4 3 biantrur ⊢ ¬ x ∈ V ∖ A ↔ x ∈ V ∧ ¬ x ∈ V ∖ A
5 2 4 bitr2i ⊢ x ∈ V ∧ ¬ x ∈ V ∖ A ↔ x ∈ A
6 5 difeqri ⊢ V ∖ V ∖ A = A