MAYBE

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

Proof:
 DP Processor:
  DPs:
   f#(x,f(y,x)) -> f#(a(),y)
   f#(x,f(y,x)) -> f#(x,x)
   f#(x,f(y,x)) -> f#(f(x,x),f(a(),y))
  TRS:
   f(x,f(y,x)) -> f(f(x,x),f(a(),y))
  Arctic Interpretation Processor:
   dimension: 1
   usable rules:
    f(x,f(y,x)) -> f(f(x,x),f(a(),y))
   interpretation:
    [f#](x0, x1) = x1 + -12,
    
    [a] = 1,
    
    [f](x0, x1) = 3x0 + 3x1 + 4
   orientation:
    f#(x,f(y,x)) = 3x + 3y + 4 >= y + -12 = f#(a(),y)
    
    f#(x,f(y,x)) = 3x + 3y + 4 >= x + -12 = f#(x,x)
    
    f#(x,f(y,x)) = 3x + 3y + 4 >= 3y + 4 = f#(f(x,x),f(a(),y))
    
    f(x,f(y,x)) = 6x + 6y + 7 >= 6x + 6y + 7 = f(f(x,x),f(a(),y))
   problem:
    DPs:
     f#(x,f(y,x)) -> f#(f(x,x),f(a(),y))
    TRS:
     f(x,f(y,x)) -> f(f(x,x),f(a(),y))
   Restore Modifier:
    DPs:
     f#(x,f(y,x)) -> f#(f(x,x),f(a(),y))
    TRS:
     f(x,f(y,x)) -> f(f(x,x),f(a(),y))
    Open