MAYBE

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

Proof:
 DP Processor:
  DPs:
   f#(X,g(X)) -> f#(1(),g(X))
   g#(1()) -> g#(0())
  TRS:
   f(X,g(X)) -> f(1(),g(X))
   g(1()) -> g(0())
  TDG Processor:
   DPs:
    f#(X,g(X)) -> f#(1(),g(X))
    g#(1()) -> g#(0())
   TRS:
    f(X,g(X)) -> f(1(),g(X))
    g(1()) -> g(0())
   graph:
    g#(1()) -> g#(0()) -> g#(1()) -> g#(0())
    f#(X,g(X)) -> f#(1(),g(X)) -> f#(X,g(X)) -> f#(1(),g(X))
   SCC Processor:
    #sccs: 2
    #rules: 2
    #arcs: 2/4
    DPs:
     f#(X,g(X)) -> f#(1(),g(X))
    TRS:
     f(X,g(X)) -> f(1(),g(X))
     g(1()) -> g(0())
    Open
    
    DPs:
     g#(1()) -> g#(0())
    TRS:
     f(X,g(X)) -> f(1(),g(X))
     g(1()) -> g(0())
    Matrix Interpretation Processor:
     dimension: 1
     interpretation:
      [g#](x0) = x0,
      
      [0] = 0,
      
      [1] = 1,
      
      [f](x0, x1) = x1 + 1,
      
      [g](x0) = x0 + 1
     orientation:
      g#(1()) = 1 >= 0 = g#(0())
      
      f(X,g(X)) = X + 2 >= X + 2 = f(1(),g(X))
      
      g(1()) = 2 >= 1 = g(0())
     problem:
      DPs:
       
      TRS:
       f(X,g(X)) -> f(1(),g(X))
       g(1()) -> g(0())
     Qed