YES

Problem:
 a(a(f(x,y))) -> f(a(b(a(b(a(x))))),a(b(a(b(a(y))))))
 f(a(x),a(y)) -> a(f(x,y))
 f(b(x),b(y)) -> b(f(x,y))

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
             [1 0]  
   [b](x0) = [0 0]x0,
   
             [1 2]     [0]
   [a](x0) = [0 2]x0 + [2],
   
                        
   [f](x0, x1) = x0 + x1
  orientation:
                  [1 6]    [1 6]    [4]    [1 2]    [1 2]    [0]                                       
   a(a(f(x,y))) = [0 4]x + [0 4]y + [6] >= [0 0]x + [0 0]y + [4] = f(a(b(a(b(a(x))))),a(b(a(b(a(y))))))
   
                  [1 2]    [1 2]    [0]    [1 2]    [1 2]    [0]            
   f(a(x),a(y)) = [0 2]x + [0 2]y + [4] >= [0 2]x + [0 2]y + [2] = a(f(x,y))
   
                  [1 0]    [1 0]     [1 0]    [1 0]             
   f(b(x),b(y)) = [0 0]x + [0 0]y >= [0 0]x + [0 0]y = b(f(x,y))
  problem:
   f(a(x),a(y)) -> a(f(x,y))
   f(b(x),b(y)) -> b(f(x,y))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 1]     [0]
    [b](x0) = [0 0 0]x0 + [0]
              [0 0 0]     [1],
    
              [1 0 0]     [0]
    [a](x0) = [0 0 0]x0 + [0]
              [0 0 1]     [1],
    
                  [1 0 0]     [1 0 1]  
    [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                  [0 0 1]     [0 0 0]  
   orientation:
                   [1 0 0]    [1 0 1]    [1]    [1 0 0]    [1 0 1]    [0]            
    f(a(x),a(y)) = [0 0 0]x + [0 0 0]y + [0] >= [0 0 0]x + [0 0 0]y + [0] = a(f(x,y))
                   [0 0 1]    [0 0 0]    [1]    [0 0 1]    [0 0 0]    [1]            
    
                   [1 0 1]    [1 0 1]    [1]    [1 0 1]    [1 0 1]    [0]            
    f(b(x),b(y)) = [0 0 0]x + [0 0 0]y + [0] >= [0 0 0]x + [0 0 0]y + [0] = b(f(x,y))
                   [0 0 0]    [0 0 0]    [1]    [0 0 0]    [0 0 0]    [1]            
   problem:
    
   Qed