YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          p#_A(x1,x2) = ((0,1),(0,0)) x1 + x2 + (1,1)
          a_A(x1) = (10,3)
          p_A(x1,x2) = x2 + (5,1)
          b_A(x1) = ((1,1),(1,1)) x1 + (11,4)
  
    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.