MAYBE

Problem:
 p(a(x0),p(a(a(a(x1))),x2)) -> p(a(x2),p(a(a(b(x0))),x2))

Proof:
 DP Processor:
  DPs:
   p#(a(x0),p(a(a(a(x1))),x2)) -> p#(a(a(b(x0))),x2)
   p#(a(x0),p(a(a(a(x1))),x2)) -> p#(a(x2),p(a(a(b(x0))),x2))
  TRS:
   p(a(x0),p(a(a(a(x1))),x2)) -> p(a(x2),p(a(a(b(x0))),x2))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [p#](x0, x1) = x1,
    
    [b](x0) = 0,
    
    [p](x0, x1) = 1x1 + 0,
    
    [a](x0) = 2x0 + 0
   orientation:
    p#(a(x0),p(a(a(a(x1))),x2)) = 1x2 + 0 >= x2 = p#(a(a(b(x0))),x2)
    
    p#(a(x0),p(a(a(a(x1))),x2)) = 1x2 + 0 >= 1x2 + 0 = p#(a(x2),p(a(a(b(x0))),x2))
    
    p(a(x0),p(a(a(a(x1))),x2)) = 2x2 + 1 >= 2x2 + 1 = p(a(x2),p(a(a(b(x0))),x2))
   problem:
    DPs:
     p#(a(x0),p(a(a(a(x1))),x2)) -> p#(a(x2),p(a(a(b(x0))),x2))
    TRS:
     p(a(x0),p(a(a(a(x1))),x2)) -> p(a(x2),p(a(a(b(x0))),x2))
   Open