YES

We show the termination of the TRS R:

  a(f(),|0|()) -> a(s(),|0|())
  a(d(),|0|()) -> |0|()
  a(d(),a(s(),x)) -> a(s(),a(s(),a(d(),a(p(),a(s(),x)))))
  a(f(),a(s(),x)) -> a(d(),a(f(),a(p(),a(s(),x))))
  a(p(),a(s(),x)) -> x

-- SCC decomposition.

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

p1: a#(f(),|0|()) -> a#(s(),|0|())
p2: a#(d(),a(s(),x)) -> a#(s(),a(s(),a(d(),a(p(),a(s(),x)))))
p3: a#(d(),a(s(),x)) -> a#(s(),a(d(),a(p(),a(s(),x))))
p4: a#(d(),a(s(),x)) -> a#(d(),a(p(),a(s(),x)))
p5: a#(d(),a(s(),x)) -> a#(p(),a(s(),x))
p6: a#(f(),a(s(),x)) -> a#(d(),a(f(),a(p(),a(s(),x))))
p7: a#(f(),a(s(),x)) -> a#(f(),a(p(),a(s(),x)))
p8: a#(f(),a(s(),x)) -> a#(p(),a(s(),x))

and R consists of:

r1: a(f(),|0|()) -> a(s(),|0|())
r2: a(d(),|0|()) -> |0|()
r3: a(d(),a(s(),x)) -> a(s(),a(s(),a(d(),a(p(),a(s(),x)))))
r4: a(f(),a(s(),x)) -> a(d(),a(f(),a(p(),a(s(),x))))
r5: a(p(),a(s(),x)) -> x

The estimated dependency graph contains the following SCCs:

  {p7}
  {p4}


-- Reduction pair.

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

p1: a#(f(),a(s(),x)) -> a#(f(),a(p(),a(s(),x)))

and R consists of:

r1: a(f(),|0|()) -> a(s(),|0|())
r2: a(d(),|0|()) -> |0|()
r3: a(d(),a(s(),x)) -> a(s(),a(s(),a(d(),a(p(),a(s(),x)))))
r4: a(f(),a(s(),x)) -> a(d(),a(f(),a(p(),a(s(),x))))
r5: a(p(),a(s(),x)) -> x

The set of usable rules consists of

  r5

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      a#_A(x1,x2) = x2
      f_A() = (0,0)
      a_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (0,1)
      s_A() = (1,1)
      p_A() = (0,1)

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: a#(d(),a(s(),x)) -> a#(d(),a(p(),a(s(),x)))

and R consists of:

r1: a(f(),|0|()) -> a(s(),|0|())
r2: a(d(),|0|()) -> |0|()
r3: a(d(),a(s(),x)) -> a(s(),a(s(),a(d(),a(p(),a(s(),x)))))
r4: a(f(),a(s(),x)) -> a(d(),a(f(),a(p(),a(s(),x))))
r5: a(p(),a(s(),x)) -> x

The set of usable rules consists of

  r5

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      a#_A(x1,x2) = x2
      d_A() = (0,0)
      a_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (0,1)
      s_A() = (1,1)
      p_A() = (0,1)

The next rules are strictly ordered:

  p1

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