YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [f](x0) = 4x0 + 7,
   
   [+](x0, x1) = x0 + x1 + 2
  orientation:
   +(+(x,y),z) = x + y + z + 4 >= x + y + z + 4 = +(x,+(y,z))
   
   +(f(x),f(y)) = 4x + 4y + 16 >= 4x + 4y + 15 = f(+(x,y))
   
   +(f(x),+(f(y),z)) = 4x + 4y + z + 18 >= 4x + 4y + z + 17 = +(f(+(x,y)),z)
  problem:
   +(+(x,y),z) -> +(x,+(y,z))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+](x0, x1) = 2x0 + x1 + 1
   orientation:
    +(+(x,y),z) = 4x + 2y + z + 3 >= 2x + 2y + z + 2 = +(x,+(y,z))
   problem:
    
   Qed