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		vex	⊢ 𝑥 ∈ V
2		eldif	⊢ ( 𝑥 ∈ ( V ∖ 𝐴 ) ↔ ( 𝑥 ∈ V ∧ ¬ 𝑥 ∈ 𝐴 ) )
3	1 2	mpbiran	⊢ ( 𝑥 ∈ ( V ∖ 𝐴 ) ↔ ¬ 𝑥 ∈ 𝐴 )
4	3	con2bii	⊢ ( 𝑥 ∈ 𝐴 ↔ ¬ 𝑥 ∈ ( V ∖ 𝐴 ) )
5	1	biantrur	⊢ ( ¬ 𝑥 ∈ ( V ∖ 𝐴 ) ↔ ( 𝑥 ∈ V ∧ ¬ 𝑥 ∈ ( V ∖ 𝐴 ) ) )
6	4 5	bitr2i	⊢ ( ( 𝑥 ∈ V ∧ ¬ 𝑥 ∈ ( V ∖ 𝐴 ) ) ↔ 𝑥 ∈ 𝐴 )
7	6	difeqri	⊢ ( V ∖ ( V ∖ 𝐴 ) ) = 𝐴