YES

We show the termination of the TRS R:

  f(|0|()) -> s(|0|())
  f(s(|0|())) -> s(s(|0|()))
  f(s(|0|())) -> *(s(s(|0|())),f(|0|()))
  f(+(x,s(|0|()))) -> +(s(s(|0|())),f(x))
  f(+(x,y)) -> *(f(x),f(y))

-- SCC decomposition.

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

p1: f#(s(|0|())) -> f#(|0|())
p2: f#(+(x,s(|0|()))) -> f#(x)
p3: f#(+(x,y)) -> f#(x)
p4: f#(+(x,y)) -> f#(y)

and R consists of:

r1: f(|0|()) -> s(|0|())
r2: f(s(|0|())) -> s(s(|0|()))
r3: f(s(|0|())) -> *(s(s(|0|())),f(|0|()))
r4: f(+(x,s(|0|()))) -> +(s(s(|0|())),f(x))
r5: f(+(x,y)) -> *(f(x),f(y))

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4}


-- Reduction pair.

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

p1: f#(+(x,s(|0|()))) -> f#(x)
p2: f#(+(x,y)) -> f#(y)
p3: f#(+(x,y)) -> f#(x)

and R consists of:

r1: f(|0|()) -> s(|0|())
r2: f(s(|0|())) -> s(s(|0|()))
r3: f(s(|0|())) -> *(s(s(|0|())),f(|0|()))
r4: f(+(x,s(|0|()))) -> +(s(s(|0|())),f(x))
r5: f(+(x,y)) -> *(f(x),f(y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        f# > |0| > s > +
      
      argument filter:
    
        pi(f#) = [1]
        pi(+) = [1, 2]
        pi(s) = []
        pi(|0|) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        f# > |0| > s > +
      
      argument filter:
    
        pi(f#) = []
        pi(+) = 1
        pi(s) = []
        pi(|0|) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        f# > |0| > s > +
      
      argument filter:
    
        pi(f#) = []
        pi(+) = [1]
        pi(s) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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