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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a#_A(x1,x2) = x2
        f_A() = 0
        a_A(x1,x2) = x1 + x2
        s_A() = 1
        p_A() = 0
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > f > a > p > a#
      
      argument filter:
    
        pi(a#) = [2]
        pi(f) = []
        pi(a) = [1]
        pi(s) = []
        pi(p) = []
    

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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a#_A(x1,x2) = x2
        d_A() = 0
        a_A(x1,x2) = x1 + x2
        s_A() = 1
        p_A() = 0
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > d > a > p > a#
      
      argument filter:
    
        pi(a#) = [2]
        pi(d) = []
        pi(a) = [1]
        pi(s) = []
        pi(p) = []
    

The next rules are strictly ordered:

  p1

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