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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      f#_A(x1) = ((1,0,0),(1,1,0),(0,0,0)) x1
      +_A(x1,x2) = ((1,0,0),(0,0,0),(1,0,0)) x1 + ((1,0,0),(0,1,0),(0,1,0)) x2 + (1,1,1)
      s_A(x1) = (1,1,1)
      |0|_A() = (1,1,1)

The next rules are strictly ordered:

  p1, p2, p3

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