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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      p#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,0,0)) x2
      a_A(x1) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(0,1,1,0)) x1 + (2,1,1,1)
      p_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(1,0,1,0)) x2 + (1,1,1,1)
      b_A(x1) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,1,0,0)) x1 + (1,1,1,1)

The next rules are strictly ordered:

  p1

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