YES

Problem:
 g(b()) -> f(b())
 f(a()) -> g(a())
 b() -> a()

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
         [0]
   [a] = [0]
         [0],
   
             [1 0 0]     [0]
   [f](x0) = [0 0 0]x0 + [1]
             [0 0 0]     [0],
   
             [1 0 1]  
   [g](x0) = [0 0 1]x0
             [0 0 0]  ,
   
         [0]
   [b] = [0]
         [1]
  orientation:
            [1]    [0]         
   g(b()) = [1] >= [1] = f(b())
            [0]    [0]         
   
            [0]    [0]         
   f(a()) = [1] >= [0] = g(a())
            [0]    [0]         
   
         [0]    [0]      
   b() = [0] >= [0] = a()
         [1]    [0]      
  problem:
   f(a()) -> g(a())
   b() -> a()
  Matrix Interpretation Processor: dim=3
   
   interpretation:
          [0]
    [a] = [0]
          [0],
    
              [1 0 0]     [1]
    [f](x0) = [0 0 0]x0 + [0]
              [0 0 0]     [0],
    
              [1 0 0]  
    [g](x0) = [0 0 0]x0
              [0 0 0]  ,
    
          [1]
    [b] = [0]
          [0]
   orientation:
             [1]    [0]         
    f(a()) = [0] >= [0] = g(a())
             [0]    [0]         
    
          [1]    [0]      
    b() = [0] >= [0] = a()
          [0]    [0]      
   problem:
    
   Qed