YES

Problem:
 minus(minus(x)) -> x
 minus(h(x)) -> h(minus(x))
 minus(f(x,y)) -> f(minus(y),minus(x))

Proof:
 DP Processor:
  DPs:
   minus#(h(x)) -> minus#(x)
   minus#(f(x,y)) -> minus#(x)
   minus#(f(x,y)) -> minus#(y)
  TRS:
   minus(minus(x)) -> x
   minus(h(x)) -> h(minus(x))
   minus(f(x,y)) -> f(minus(y),minus(x))
  CDG Processor:
   DPs:
    minus#(h(x)) -> minus#(x)
    minus#(f(x,y)) -> minus#(x)
    minus#(f(x,y)) -> minus#(y)
   TRS:
    minus(minus(x)) -> x
    minus(h(x)) -> h(minus(x))
    minus(f(x,y)) -> f(minus(y),minus(x))
   graph:
    
   Qed