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))
  Restore Modifier:
   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))
   SCC Processor:
    #sccs: 1
    #rules: 2
    #arcs: 4/4
    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))
    Matrix Interpretation Processor:
     dimension: 1
     interpretation:
      [p#](x0, x1) = x1 + 1,
      
      [b](x0) = 0,
      
      [p](x0, x1) = x1 + 1,
      
      [a](x0) = 0
     orientation:
      p#(a(x0),p(a(a(a(x1))),x2)) = x2 + 2 >= x2 + 1 = p#(a(a(b(x0))),x2)
      
      p#(a(x0),p(a(a(a(x1))),x2)) = x2 + 2 >= x2 + 2 = p#(a(x2),p(a(a(b(x0))),x2))
      
      p(a(x0),p(a(a(a(x1))),x2)) = x2 + 2 >= x2 + 2 = 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