MAYBE

Problem:
 g(c(x,s(y))) -> g(c(s(x),y))
 f(c(s(x),y)) -> f(c(x,s(y)))
 f(f(x)) -> f(d(f(x)))
 f(x) -> x

Proof:
 DP Processor:
  DPs:
   g#(c(x,s(y))) -> g#(c(s(x),y))
   f#(c(s(x),y)) -> f#(c(x,s(y)))
   f#(f(x)) -> f#(d(f(x)))
  TRS:
   g(c(x,s(y))) -> g(c(s(x),y))
   f(c(s(x),y)) -> f(c(x,s(y)))
   f(f(x)) -> f(d(f(x)))
   f(x) -> x
  Usable Rule Processor:
   DPs:
    g#(c(x,s(y))) -> g#(c(s(x),y))
    f#(c(s(x),y)) -> f#(c(x,s(y)))
    f#(f(x)) -> f#(d(f(x)))
   TRS:
    f(c(s(x),y)) -> f(c(x,s(y)))
    f(f(x)) -> f(d(f(x)))
    f(x) -> x
   EDG Processor:
    DPs:
     g#(c(x,s(y))) -> g#(c(s(x),y))
     f#(c(s(x),y)) -> f#(c(x,s(y)))
     f#(f(x)) -> f#(d(f(x)))
    TRS:
     f(c(s(x),y)) -> f(c(x,s(y)))
     f(f(x)) -> f(d(f(x)))
     f(x) -> x
    graph:
     f#(c(s(x),y)) -> f#(c(x,s(y))) -> f#(c(s(x),y)) -> f#(c(x,s(y)))
     g#(c(x,s(y))) -> g#(c(s(x),y)) -> g#(c(x,s(y))) -> g#(c(s(x),y))
    Restore Modifier:
     DPs:
      g#(c(x,s(y))) -> g#(c(s(x),y))
      f#(c(s(x),y)) -> f#(c(x,s(y)))
      f#(f(x)) -> f#(d(f(x)))
     TRS:
      g(c(x,s(y))) -> g(c(s(x),y))
      f(c(s(x),y)) -> f(c(x,s(y)))
      f(f(x)) -> f(d(f(x)))
      f(x) -> x
     SCC Processor:
      #sccs: 2
      #rules: 2
      #arcs: 2/9
      DPs:
       g#(c(x,s(y))) -> g#(c(s(x),y))
      TRS:
       g(c(x,s(y))) -> g(c(s(x),y))
       f(c(s(x),y)) -> f(c(x,s(y)))
       f(f(x)) -> f(d(f(x)))
       f(x) -> x
      Open
      
      DPs:
       f#(c(s(x),y)) -> f#(c(x,s(y)))
      TRS:
       g(c(x,s(y))) -> g(c(s(x),y))
       f(c(s(x),y)) -> f(c(x,s(y)))
       f(f(x)) -> f(d(f(x)))
       f(x) -> x
      Matrix Interpretation Processor:
       dimension: 1
       interpretation:
        [f#](x0) = x0,
        
        [d](x0) = x0,
        
        [f](x0) = x0,
        
        [g](x0) = 0,
        
        [c](x0, x1) = x0,
        
        [s](x0) = x0 + 1
       orientation:
        f#(c(s(x),y)) = x + 1 >= x = f#(c(x,s(y)))
        
        g(c(x,s(y))) = 0 >= 0 = g(c(s(x),y))
        
        f(c(s(x),y)) = x + 1 >= x = f(c(x,s(y)))
        
        f(f(x)) = x >= x = f(d(f(x)))
        
        f(x) = x >= x = x
       problem:
        DPs:
         
        TRS:
         g(c(x,s(y))) -> g(c(s(x),y))
         f(c(s(x),y)) -> f(c(x,s(y)))
         f(f(x)) -> f(d(f(x)))
         f(x) -> x
       Qed