YES

Problem:
 or(x,x) -> x
 and(x,x) -> x
 not(not(x)) -> x
 not(and(x,y)) -> or(not(x),not(y))
 not(or(x,y)) -> and(not(x),not(y))

Proof:
 DP Processor:
  DPs:
   not#(and(x,y)) -> not#(y)
   not#(and(x,y)) -> not#(x)
   not#(and(x,y)) -> or#(not(x),not(y))
   not#(or(x,y)) -> not#(y)
   not#(or(x,y)) -> not#(x)
   not#(or(x,y)) -> and#(not(x),not(y))
  TRS:
   or(x,x) -> x
   and(x,x) -> x
   not(not(x)) -> x
   not(and(x,y)) -> or(not(x),not(y))
   not(or(x,y)) -> and(not(x),not(y))
  KBO Processor:
   argument filtering:
    pi(or) = [0,1]
    pi(and) = [0,1]
    pi(not) = [0]
    pi(or#) = 0
    pi(and#) = 0
    pi(not#) = 0
   weight function:
    w0 = 1
    w(not#) = w(or#) = 1
    w(and#) = w(not) = w(and) = w(or) = 0
   precedence:
    or# > not > not# ~ and# ~ and ~ or
   problem:
    DPs:
     
    TRS:
     or(x,x) -> x
     and(x,x) -> x
     not(not(x)) -> x
     not(and(x,y)) -> or(not(x),not(y))
     not(or(x,y)) -> and(not(x),not(y))
   Qed