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=3
  
  interpretation:
             [1 1 1]     [0]
   [f](x0) = [0 0 0]x0 + [1]
             [0 0 0]     [1],
   
                      [1 1 0]  
   [+](x0, x1) = x0 + [0 0 0]x1
                      [0 0 0]  
  orientation:
                     [1 1 0]    [1 1 0]         [1 1 0]    [1 1 0]               
   +(+(x,y),z) = x + [0 0 0]y + [0 0 0]z >= x + [0 0 0]y + [0 0 0]z = +(x,+(y,z))
                     [0 0 0]    [0 0 0]         [0 0 0]    [0 0 0]               
   
                  [1 1 1]    [1 1 1]    [1]    [1 1 1]    [1 1 0]    [0]            
   +(f(x),f(y)) = [0 0 0]x + [0 0 0]y + [1] >= [0 0 0]x + [0 0 0]y + [1] = f(+(x,y))
                  [0 0 0]    [0 0 0]    [1]    [0 0 0]    [0 0 0]    [1]            
   
                       [1 1 1]    [1 1 1]    [1 1 0]    [1]    [1 1 1]    [1 1 0]    [1 1 0]    [0]                 
   +(f(x),+(f(y),z)) = [0 0 0]x + [0 0 0]y + [0 0 0]z + [1] >= [0 0 0]x + [0 0 0]y + [0 0 0]z + [1] = +(f(+(x,y)),z)
                       [0 0 0]    [0 0 0]    [0 0 0]    [1]    [0 0 0]    [0 0 0]    [0 0 0]    [1]                 
  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