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 0]     [0]
   [f](x0) = [0 0 0]x0 + [0]
             [0 0 1]     [1],
   
                 [1 0 0]     [1 0 1]  
   [+](x0, x1) = [0 0 0]x0 + [0 1 0]x1
                 [0 0 1]     [0 0 0]  
  orientation:
                 [1 0 0]    [1 0 1]    [1 0 1]     [1 0 0]    [1 0 1]    [1 0 1]               
   +(+(x,y),z) = [0 0 0]x + [0 0 0]y + [0 1 0]z >= [0 0 0]x + [0 0 0]y + [0 1 0]z = +(x,+(y,z))
                 [0 0 1]    [0 0 0]    [0 0 0]     [0 0 1]    [0 0 0]    [0 0 0]               
   
                  [1 1 0]    [1 1 1]    [1]    [1 0 0]    [1 1 1]    [0]            
   +(f(x),f(y)) = [0 0 0]x + [0 0 0]y + [0] >= [0 0 0]x + [0 0 0]y + [0] = f(+(x,y))
                  [0 0 1]    [0 0 0]    [1]    [0 0 1]    [0 0 0]    [1]            
   
                       [1 1 0]    [1 1 1]    [1 0 1]    [1]    [1 0 0]    [1 1 1]    [1 0 1]    [0]                 
   +(f(x),+(f(y),z)) = [0 0 0]x + [0 0 0]y + [0 1 0]z + [0] >= [0 0 0]x + [0 0 0]y + [0 1 0]z + [0] = +(f(+(x,y)),z)
                       [0 0 1]    [0 0 0]    [0 0 0]    [1]    [0 0 1]    [0 0 0]    [0 0 0]    [1]                 
  problem:
   +(+(x,y),z) -> +(x,+(y,z))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                  [1 0 1]     [1 0 0]     [1]
    [+](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0]
                  [1 0 0]     [0 0 1]     [1]
   orientation:
                  [2 0 1]    [1 0 1]    [1 0 0]    [3]    [1 0 1]    [1 0 1]    [1 0 0]    [2]              
    +(+(x,y),z) = [0 0 0]x + [0 0 0]y + [0 0 0]z + [0] >= [0 0 0]x + [0 0 0]y + [0 0 0]z + [0] = +(x,+(y,z))
                  [1 0 1]    [1 0 0]    [0 0 1]    [2]    [1 0 0]    [1 0 0]    [0 0 1]    [2]              
   problem:
    
   Qed