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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