YES

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

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