YES

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

Proof:
 DP Processor:
  DPs:
   *#(x,+(y,z)) -> *#(x,z)
   *#(x,+(y,z)) -> *#(x,y)
  TRS:
   *(x,+(y,z)) -> +(*(x,y),*(x,z))
  SCC Processor:
   #sccs: 1
   #rules: 2
   #arcs: 4/4
   DPs:
    *#(x,+(y,z)) -> *#(x,z)
    *#(x,+(y,z)) -> *#(x,y)
   TRS:
    *(x,+(y,z)) -> +(*(x,y),*(x,z))
   Matrix Interpretation Processor:
    dimension: 1
    interpretation:
     [*#](x0, x1) = x1 + 1,
     
     [*](x0, x1) = x1,
     
     [+](x0, x1) = x0 + x1 + 1
    orientation:
     *#(x,+(y,z)) = y + z + 2 >= z + 1 = *#(x,z)
     
     *#(x,+(y,z)) = y + z + 2 >= y + 1 = *#(x,y)
     
     *(x,+(y,z)) = y + z + 1 >= y + z + 1 = +(*(x,y),*(x,z))
    problem:
     DPs:
      
     TRS:
      *(x,+(y,z)) -> +(*(x,y),*(x,z))
    Qed