YES

Problem:
 +(-(x,y),z) -> -(+(x,z),y)
 -(+(x,y),y) -> x

Proof:
 DP Processor:
  DPs:
   +#(-(x,y),z) -> +#(x,z)
   +#(-(x,y),z) -> -#(+(x,z),y)
  TRS:
   +(-(x,y),z) -> -(+(x,z),y)
   -(+(x,y),y) -> x
  Matrix Interpretation Processor: dim=3
   
   usable rules:
    
   interpretation:
    [-#](x0, x1) = [0],
    
    [+#](x0, x1) = [0 0 1]x0 + [1 1 1]x1 + [1],
    
                  [1 0 0]     [0 0 0]     [0]
    [+](x0, x1) = [1 1 0]x0 + [1 0 0]x1 + [1]
                  [0 0 0]     [0 0 0]     [0],
    
                  [1 0 1]     [0 0 0]     [0]
    [-](x0, x1) = [1 0 0]x0 + [1 0 0]x1 + [0]
                  [1 1 1]     [0 0 0]     [1]
   orientation:
    +#(-(x,y),z) = [1 1 1]x + [1 1 1]z + [2] >= [0 0 1]x + [1 1 1]z + [1] = +#(x,z)
    
    +#(-(x,y),z) = [1 1 1]x + [1 1 1]z + [2] >= [0] = -#(+(x,z),y)
    
                  [1 0 1]    [0 0 0]    [0 0 0]    [0]    [1 0 0]    [0 0 0]    [0 0 0]    [0]              
    +(-(x,y),z) = [2 0 1]x + [1 0 0]y + [1 0 0]z + [1] >= [1 0 0]x + [1 0 0]y + [0 0 0]z + [0] = -(+(x,z),y)
                  [0 0 0]    [0 0 0]    [0 0 0]    [0]    [2 1 0]    [0 0 0]    [1 0 0]    [2]              
    
                  [1 0 0]    [0 0 0]    [0]         
    -(+(x,y),y) = [1 0 0]x + [1 0 0]y + [0] >= x = x
                  [2 1 0]    [1 0 0]    [2]         
   problem:
    DPs:
     
    TRS:
     +(-(x,y),z) -> -(+(x,z),y)
     -(+(x,y),y) -> x
   Qed