MAYBE

Problem:
 f(x,f(f(a(),a()),a())) -> f(f(a(),f(a(),a())),x)

Proof:
 DP Processor:
  DPs:
   f#(x,f(f(a(),a()),a())) -> f#(a(),f(a(),a()))
   f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x)
  TRS:
   f(x,f(f(a(),a()),a())) -> f(f(a(),f(a(),a())),x)
  Usable Rule Processor:
   DPs:
    f#(x,f(f(a(),a()),a())) -> f#(a(),f(a(),a()))
    f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x)
   TRS:
    
   Restore Modifier:
    DPs:
     f#(x,f(f(a(),a()),a())) -> f#(a(),f(a(),a()))
     f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x)
    TRS:
     f(x,f(f(a(),a()),a())) -> f(f(a(),f(a(),a())),x)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 4/4
     DPs:
      f#(x,f(f(a(),a()),a())) -> f#(a(),f(a(),a()))
      f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x)
     TRS:
      f(x,f(f(a(),a()),a())) -> f(f(a(),f(a(),a())),x)
     Matrix Interpretation Processor:
      dimension: 1
      interpretation:
       [f#](x0, x1) = x0 + x1,
       
       [f](x0, x1) = x0 + x1 + 1,
       
       [a] = 0
      orientation:
       f#(x,f(f(a(),a()),a())) = x + 2 >= 1 = f#(a(),f(a(),a()))
       
       f#(x,f(f(a(),a()),a())) = x + 2 >= x + 2 = f#(f(a(),f(a(),a())),x)
       
       f(x,f(f(a(),a()),a())) = x + 3 >= x + 3 = f(f(a(),f(a(),a())),x)
      problem:
       DPs:
        f#(x,f(f(a(),a()),a())) -> f#(f(a(),f(a(),a())),x)
       TRS:
        f(x,f(f(a(),a()),a())) -> f(f(a(),f(a(),a())),x)
      Open