Theorem dblcompl 3228

Description: Double complement law. (Contributed by SF, 10-Jan-2015.)

Assertion

Ref	Expression
dblcompl	⊢ ∼ ∼ A = A

Proof of Theorem dblcompl

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		vex 2863	. . . 4 ⊢ x ∈ V
2	1	elcompl 3226	. . 3 ⊢ (x ∈ ∼ ∼ A ↔ ¬ x ∈ ∼ A)
3	1	elcompl 3226	. . . 4 ⊢ (x ∈ ∼ A ↔ ¬ x ∈ A)
4	3	con2bii 322	. . 3 ⊢ (x ∈ A ↔ ¬ x ∈ ∼ A)
5	2, 4	bitr4i 243	. 2 ⊢ (x ∈ ∼ ∼ A ↔ x ∈ A)
6	5	eqriv 2350	1 ⊢ ∼ ∼ A = A

Syntax hints:  ¬ wn 3   = wceq 1642   ∈ wcel 1710   ∼ ccompl 3206

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213

This theorem is referenced by:  compleqb  3544  dfin5  3546  dfun4  3547  iunin  3548  iinun  3549  compl0  4072  sbthlem1  6204