YES

Problem:
 *(x,+(y,z)) -> +(*(x,y),*(x,z))
 *(+(x,y),z) -> +(*(x,z),*(y,z))
 *(x,1()) -> x
 *(1(),y) -> y

Proof:
 DP Processor:
  DPs:
   *#(x,+(y,z)) -> *#(x,z)
   *#(x,+(y,z)) -> *#(x,y)
   *#(+(x,y),z) -> *#(y,z)
   *#(+(x,y),z) -> *#(x,z)
  TRS:
   *(x,+(y,z)) -> +(*(x,y),*(x,z))
   *(+(x,y),z) -> +(*(x,z),*(y,z))
   *(x,1()) -> x
   *(1(),y) -> y
  Usable Rule Processor:
   DPs:
    *#(x,+(y,z)) -> *#(x,z)
    *#(x,+(y,z)) -> *#(x,y)
    *#(+(x,y),z) -> *#(y,z)
    *#(+(x,y),z) -> *#(x,z)
   TRS:
    
   Matrix Interpretation Processor: dim=4
    
    usable rules:
     
    interpretation:
     [*#](x0, x1) = [1 0 1 0]x0 + [1 1 0 0]x1 + [1],
     
                   [1 0 0 0]     [1 0 0 0]     [1]
                   [1 1 0 0]     [0 1 0 0]     [1]
     [+](x0, x1) = [0 0 1 0]x0 + [0 0 1 0]x1 + [1]
                   [0 0 0 0]     [0 0 0 0]     [0]
    orientation:
     *#(x,+(y,z)) = [1 0 1 0]x + [2 1 0 0]y + [1 1 0 0]z + [3] >= [1 0 1 0]x + [1 1 0 0]z + [1] = *#(x,z)
     
     *#(x,+(y,z)) = [1 0 1 0]x + [2 1 0 0]y + [1 1 0 0]z + [3] >= [1 0 1 0]x + [1 1 0 0]y + [1] = *#(x,y)
     
     *#(+(x,y),z) = [1 0 1 0]x + [1 0 1 0]y + [1 1 0 0]z + [3] >= [1 0 1 0]y + [1 1 0 0]z + [1] = *#(y,z)
     
     *#(+(x,y),z) = [1 0 1 0]x + [1 0 1 0]y + [1 1 0 0]z + [3] >= [1 0 1 0]x + [1 1 0 0]z + [1] = *#(x,z)
    problem:
     DPs:
      
     TRS:
      
    Qed