YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        p#_A(x1,x2) = x1 + x2
        a_A(x1) = 1
        p_A(x1,x2) = x2 + 1
        b_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        p#_A(x1,x2) = x1
        a_A(x1) = 1
        p_A(x1,x2) = x2 + 1
        b_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

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