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:

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

The next rules are strictly ordered:

  p1

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