YES

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

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