NO

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

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