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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          p#_A(x1,x2) = max{x1 + 10, x2 + 15}
          a_A(x1) = x1 + 2
          p_A(x1,x2) = max{x1 + 3, x2 + 4}
          b_A(x1) = x1 + 5
    
      precedence:
      
        p# = a = p = b
      
      partial status:
    
        pi(p#) = [1, 2]
        pi(a) = [1]
        pi(p) = [1, 2]
        pi(b) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          p#_A(x1,x2) = max{x1 - 4, x2 + 2}
          a_A(x1) = max{6, x1 + 2}
          p_A(x1,x2) = x2
          b_A(x1) = max{4, 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.