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))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [minus#](x0) = x0,
    
    [f](x0, x1) = x0 + x1 + 1,
    
    [h](x0) = x0 + 3,
    
    [minus](x0) = x0
   orientation:
    minus#(h(x)) = x + 3 >= x = minus#(x)
    
    minus#(f(x,y)) = x + y + 1 >= x = minus#(x)
    
    minus#(f(x,y)) = x + y + 1 >= y = minus#(y)
    
    minus(minus(x)) = x >= x = x
    
    minus(h(x)) = x + 3 >= x + 3 = h(minus(x))
    
    minus(f(x,y)) = x + y + 1 >= x + y + 1 = f(minus(y),minus(x))
   problem:
    DPs:
     
    TRS:
     minus(minus(x)) -> x
     minus(h(x)) -> h(minus(x))
     minus(f(x,y)) -> f(minus(y),minus(x))
   Qed