YES

Problem:
 f(g(X)) -> f(X)

Proof:
 DP Processor:
  DPs:
   f#(g(X)) -> f#(X)
  TRS:
   f(g(X)) -> f(X)
  SCC Processor:
   #sccs: 1
   #rules: 1
   #arcs: 1/1
   DPs:
    f#(g(X)) -> f#(X)
   TRS:
    f(g(X)) -> f(X)
   Matrix Interpretation Processor:
    dimension: 1
    interpretation:
     [f#](x0) = x0,
     
     [f](x0) = 0,
     
     [g](x0) = x0 + 1
    orientation:
     f#(g(X)) = X + 1 >= X = f#(X)
     
     f(g(X)) = 0 >= 0 = f(X)
    problem:
     DPs:
      
     TRS:
      f(g(X)) -> f(X)
    Qed