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Theorem dblcompl 3227
 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.
StepHypRef Expression
1 vex 2862 . . . 4 x V
21elcompl 3225 . . 3 (x ∼ ∼ A ↔ ¬ x A)
31elcompl 3225 . . . 4 (x A ↔ ¬ x A)
43con2bii 322 . . 3 (x A ↔ ¬ x A)
52, 4bitr4i 243 . 2 (x ∼ ∼ Ax A)
65eqriv 2350 1 ∼ ∼ A = A
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   = wceq 1642   ∈ wcel 1710   ∼ ccompl 3205 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 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 2478  df-v 2861  df-nin 3211  df-compl 3212 This theorem is referenced by:  compleqb  3543  dfin5  3545  dfun4  3546  iunin  3547  iinun  3548  compl0  4071  sbthlem1  6203
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