MAYBE

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

Proof:
 DP Processor:
  DPs:
   f#(x,f(a(),a())) -> f#(x,a())
   f#(x,f(a(),a())) -> f#(f(x,a()),x)
  TRS:
   f(x,f(a(),a())) -> f(f(x,a()),x)
  Usable Rule Processor:
   DPs:
    f#(x,f(a(),a())) -> f#(x,a())
    f#(x,f(a(),a())) -> f#(f(x,a()),x)
   TRS:
    f3(x,y) -> x
    f3(x,y) -> y
   Restore Modifier:
    DPs:
     f#(x,f(a(),a())) -> f#(x,a())
     f#(x,f(a(),a())) -> f#(f(x,a()),x)
    TRS:
     f(x,f(a(),a())) -> f(f(x,a()),x)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 4/4
     DPs:
      f#(x,f(a(),a())) -> f#(x,a())
      f#(x,f(a(),a())) -> f#(f(x,a()),x)
     TRS:
      f(x,f(a(),a())) -> f(f(x,a()),x)
     Matrix Interpretation Processor:
      dimension: 1
      interpretation:
       [f#](x0, x1) = x0 + x1,
       
       [f](x0, x1) = 1,
       
       [a] = 0
      orientation:
       f#(x,f(a(),a())) = x + 1 >= x = f#(x,a())
       
       f#(x,f(a(),a())) = x + 1 >= x + 1 = f#(f(x,a()),x)
       
       f(x,f(a(),a())) = 1 >= 1 = f(f(x,a()),x)
      problem:
       DPs:
        f#(x,f(a(),a())) -> f#(f(x,a()),x)
       TRS:
        f(x,f(a(),a())) -> f(f(x,a()),x)
      Open