MAYBE

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

Proof:
 DP Processor:
  DPs:
   f#(x,f(a(),f(f(a(),a()),a()))) -> f#(a(),x)
   f#(x,f(a(),f(f(a(),a()),a()))) -> f#(f(a(),x),x)
  TRS:
   f(x,f(a(),f(f(a(),a()),a()))) -> f(f(a(),x),x)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0, x1) = 2x0 + 1x1 + 0,
    
    [f](x0, x1) = 1x0 + x1 + 0,
    
    [a] = 0
   orientation:
    f#(x,f(a(),f(f(a(),a()),a()))) = 2x + 3 >= 1x + 2 = f#(a(),x)
    
    f#(x,f(a(),f(f(a(),a()),a()))) = 2x + 3 >= 2x + 3 = f#(f(a(),x),x)
    
    f(x,f(a(),f(f(a(),a()),a()))) = 1x + 2 >= 1x + 2 = f(f(a(),x),x)
   problem:
    DPs:
     f#(x,f(a(),f(f(a(),a()),a()))) -> f#(f(a(),x),x)
    TRS:
     f(x,f(a(),f(f(a(),a()),a()))) -> f(f(a(),x),x)
   Open