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 monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      p#_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2
      a_A(x1) = ((1,1),(0,0)) x1
      p_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,0)
      b_A(x1) = ((1,1),(1,1)) x1

The next rules are strictly ordered:

  p1
  r1

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