NO

Problem:
 f(X,g(X)) -> f(1(),g(X))
 g(1()) -> g(0())

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
                 [1 0 1]     [1 0 0]  
   [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                 [1 0 0]     [1 0 1]  ,
   
         [0]
   [0] = [0]
         [0],
   
             [1 1 1]  
   [g](x0) = [0 0 0]x0
             [0 0 1]  ,
   
         [0]
   [1] = [1]
         [0]
  orientation:
               [2 1 2]     [1 1 1]               
   f(X,g(X)) = [0 0 0]X >= [0 0 0]X = f(1(),g(X))
               [2 1 2]     [1 1 2]               
   
            [1]    [0]         
   g(1()) = [0] >= [0] = g(0())
            [0]    [0]         
  problem:
   f(X,g(X)) -> f(1(),g(X))
  Unfolding Processor:
   loop length: 1
   terms:
    f(1(),g(1()))
   context: []
   substitution:
    
   Qed