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=5
  
  interpretation:
                 [1 1 0 0 1]     [1 0 1 0 1]     [0]
                 [1 1 0 1 0]     [0 0 0 0 0]     [0]
   [f](x0, x1) = [1 0 0 0 0]x0 + [1 0 1 0 1]x1 + [0]
                 [0 0 0 0 0]     [0 0 0 0 0]     [0]
                 [1 1 0 0 0]     [0 0 0 0 0]     [1],
   
             [1 0 0 0 1]     [0]
             [0 1 0 1 1]     [1]
   [b](x0) = [0 0 0 0 0]x0 + [0]
             [0 0 0 1 1]     [0]
             [0 0 1 0 0]     [0],
   
             [1 0 0 0 0]     [0]
             [0 0 0 0 0]     [0]
   [a](x0) = [0 0 1 0 1]x0 + [1]
             [0 1 0 0 0]     [0]
             [0 0 0 0 0]     [0]
  orientation:
               [1 1 1 1 2]    [1 0 1 0 1]    [1]    [1 1 0 0 1]    [1 0 1 0 1]    [0]            
               [1 1 0 2 3]    [0 0 0 0 0]    [1]    [1 1 0 1 0]    [0 0 0 0 0]    [0]            
   f(b(x),y) = [1 0 0 0 1]x + [1 0 1 0 1]y + [0] >= [1 0 0 0 0]x + [1 0 1 0 1]y + [0] = f(x,b(y))
               [0 0 0 0 0]    [0 0 0 0 0]    [0]    [0 0 0 0 0]    [0 0 0 0 0]    [0]            
               [1 1 0 1 2]    [0 0 0 0 0]    [2]    [1 1 0 0 0]    [0 0 0 0 0]    [1]            
   
               [1 1 0 0 1]    [1 0 1 0 1]    [1]    [1 0 0 0 0]    [1 0 1 0 1]    [0]            
               [1 1 0 1 0]    [0 0 0 0 0]    [0]    [1 1 0 0 0]    [0 0 0 0 0]    [0]            
   f(x,a(y)) = [1 0 0 0 0]x + [1 0 1 0 1]y + [1] >= [1 0 0 0 0]x + [1 0 1 0 1]y + [0] = f(a(x),y)
               [0 0 0 0 0]    [0 0 0 0 0]    [0]    [0 0 0 0 0]    [0 0 0 0 0]    [0]            
               [1 1 0 0 0]    [0 0 0 0 0]    [1]    [1 0 0 0 0]    [0 0 0 0 0]    [1]            
   
            [1 1 0 0 1]    [1 0 1 0 1]    [0]    [1 0 0 0 0]    [1 0 1 0 1]    [0]            
            [1 1 0 1 0]    [0 0 0 0 0]    [0]    [1 1 0 0 0]    [0 0 0 0 0]    [0]            
   f(x,y) = [1 0 0 0 0]x + [1 0 1 0 1]y + [0] >= [1 0 0 0 0]x + [1 0 1 0 1]y + [0] = f(a(x),y)
            [0 0 0 0 0]    [0 0 0 0 0]    [0]    [0 0 0 0 0]    [0 0 0 0 0]    [0]            
            [1 1 0 0 0]    [0 0 0 0 0]    [1]    [1 0 0 0 0]    [0 0 0 0 0]    [1]            
   
            [1 1 0 0 1]    [1 0 1 0 1]    [0]    [1 1 0 0 1]    [1 0 1 0 1]    [0]            
            [1 1 0 1 0]    [0 0 0 0 0]    [0]    [1 1 0 1 0]    [0 0 0 0 0]    [0]            
   f(x,y) = [1 0 0 0 0]x + [1 0 1 0 1]y + [0] >= [1 0 0 0 0]x + [1 0 1 0 1]y + [0] = f(x,b(y))
            [0 0 0 0 0]    [0 0 0 0 0]    [0]    [0 0 0 0 0]    [0 0 0 0 0]    [0]            
            [1 1 0 0 0]    [0 0 0 0 0]    [1]    [1 1 0 0 0]    [0 0 0 0 0]    [1]            
  problem:
   strict:
    
   weak:
    f(x,y) -> f(a(x),y)
    f(x,y) -> f(x,b(y))
  Qed