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^3
      order: lexicographic order
      interpretations:
        p#_A(x1,x2) = ((0,0,0),(1,0,0),(1,0,0)) x2
        a_A(x1) = ((1,0,0),(0,0,0),(1,1,0)) x1 + (1,1,1)
        p_A(x1,x2) = ((1,0,0),(1,1,0),(0,0,1)) x1 + ((1,0,0),(1,0,0),(0,0,0)) x2
        b_A(x1) = ((1,0,0),(0,1,0),(0,1,1)) x1 + (1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        p# > b > p > a
      
      argument filter:
    
        pi(p#) = []
        pi(a) = []
        pi(p) = []
        pi(b) = [1]
    

The next rules are strictly ordered:

  p1

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