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:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          p#_A(x1,x2) = x2
          a_A(x1) = 1
          p_A(x1,x2) = x2 + 2
          b_A(x1) = 2
  
    precedence:
    
      p# = a = p = b
    
    partial status:
  
      pi(p#) = [2]
      pi(a) = []
      pi(p) = [2]
      pi(b) = []

The next rules are strictly ordered:

  p1

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