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Theorem ddif 3398
 Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.)
Assertion
Ref Expression
ddif (V (V A)) = A

Proof of Theorem ddif
Dummy variable x is distinct from all other variables.
StepHypRef Expression
1 vex 2862 . . . . 5 x V
2 eldif 3221 . . . . 5 (x (V A) ↔ (x V ¬ x A))
31, 2mpbiran 884 . . . 4 (x (V A) ↔ ¬ x A)
43con2bii 322 . . 3 (x A ↔ ¬ x (V A))
51biantrur 492 . . 3 x (V A) ↔ (x V ¬ x (V A)))
64, 5bitr2i 241 . 2 ((x V ¬ x (V A)) ↔ x A)
76difeqri 3387 1 (V (V A)) = A
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ∧ wa 358   = wceq 1642   ∈ wcel 1710  Vcvv 2859   ∖ cdif 3206 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  df-in 3213  df-dif 3215 This theorem is referenced by:  dfun3  3493  dfin3  3494  invdif  3496  ssindif0  3604  difdifdir  3637
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