YES

We show the termination of the TRS R:

  p(p(b(a(x0)),x1),p(x2,x3)) -> p(p(x3,a(x2)),p(b(a(x1)),b(x0)))

-- SCC decomposition.

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

p1: p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(p(x3,a(x2)),p(b(a(x1)),b(x0)))
p2: p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(x3,a(x2))
p3: p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(b(a(x1)),b(x0))

and R consists of:

r1: p(p(b(a(x0)),x1),p(x2,x3)) -> p(p(x3,a(x2)),p(b(a(x1)),b(x0)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: p#(p(b(a(x0)),x1),p(x2,x3)) -> p#(p(x3,a(x2)),p(b(a(x1)),b(x0)))

and R consists of:

r1: p(p(b(a(x0)),x1),p(x2,x3)) -> p(p(x3,a(x2)),p(b(a(x1)),b(x0)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            p#_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,1),(0,0)) x2 + (7,2)
            p_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (5,0)
            b_A(x1) = ((1,1),(0,0)) x1 + (6,3)
            a_A(x1) = ((0,0),(1,1)) x1 + (4,1)
    
      precedence:
      
        p# = p = b = a
      
      partial status:
    
        pi(p#) = []
        pi(p) = []
        pi(b) = []
        pi(a) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            p#_A(x1,x2) = x1
            p_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,0),(1,0)) x2 + (5,0)
            b_A(x1) = ((0,1),(0,0)) x1 + (0,1)
            a_A(x1) = (4,2)
    
      precedence:
      
        p# = p = b = a
      
      partial status:
    
        pi(p#) = [1]
        pi(p) = []
        pi(b) = []
        pi(a) = []
    

The next rules are strictly ordered:

  p1

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