YES

Problem:
 a() -> g(c())
 g(a()) -> b()
 f(g(X),b()) -> f(a(),X)

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
                 [1 0 0]     [1 0 0]     [0]
   [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0]
                 [0 0 0]     [0 0 0]     [1],
   
         [0]
   [c] = [0]
         [0],
   
         [1]
   [b] = [0]
         [0],
   
         [1]
   [a] = [0]
         [0],
   
             [1 0 0]  
   [g](x0) = [0 0 0]x0
             [0 0 0]  
  orientation:
         [1]    [0]         
   a() = [0] >= [0] = g(c())
         [0]    [0]         
   
            [1]    [1]      
   g(a()) = [0] >= [0] = b()
            [0]    [0]      
   
                 [1 0 0]    [1]    [1 0 0]    [1]           
   f(g(X),b()) = [0 0 0]X + [0] >= [0 0 0]X + [0] = f(a(),X)
                 [0 0 0]    [1]    [0 0 0]    [1]           
  problem:
   g(a()) -> b()
   f(g(X),b()) -> f(a(),X)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                  [1 0 0]     [1 1 1]  
    [f](x0, x1) = [0 1 1]x0 + [1 0 0]x1
                  [0 0 0]     [0 0 0]  ,
    
          [0]
    [b] = [1]
          [0],
    
          [0]
    [a] = [0]
          [1],
    
              [1 1 1]     [0]
    [g](x0) = [1 0 0]x0 + [1]
              [0 0 0]     [0]
   orientation:
             [1]    [0]      
    g(a()) = [1] >= [1] = b()
             [0]    [0]      
    
                  [1 1 1]    [1]    [1 1 1]    [0]           
    f(g(X),b()) = [1 0 0]X + [1] >= [1 0 0]X + [1] = f(a(),X)
                  [0 0 0]    [0]    [0 0 0]    [0]           
   problem:
    
   Qed