YES

We show the termination of the TRS R:

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

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: p#(a(x0),p(a(b(x1)),x2)) -> p#(a(b(a(x2))),p(a(a(x1)),x2))
p2: p#(a(x0),p(a(b(x1)),x2)) -> p#(a(a(x1)),x2)

and R consists of:

r1: p(a(x0),p(a(b(x1)),x2)) -> p(a(b(a(x2))),p(a(a(x1)),x2))

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: p#(a(x0),p(a(b(x1)),x2)) -> p#(a(a(x1)),x2)

and R consists of:

r1: p(a(x0),p(a(b(x1)),x2)) -> p(a(b(a(x2))),p(a(a(x1)),x2))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        p#_A(x1,x2) = max{x1 - 7, x2 + 7}
        a_A(x1) = max{9, x1 + 4}
        p_A(x1,x2) = max{x1 - 8, x2 + 1}
        b_A(x1) = max{5, x1}
  
    precedence:
    
      p# = a = p = b
    
    partial status:
  
      pi(p#) = []
      pi(a) = []
      pi(p) = [2]
      pi(b) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.