YES

We show the termination of the TRS R:

  f(x,|0|()) -> s(|0|())
  f(s(x),s(y)) -> s(f(x,y))
  g(|0|(),x) -> g(f(x,x),x)

-- SCC decomposition.

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

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

and R consists of:

r1: f(x,|0|()) -> s(|0|())
r2: f(s(x),s(y)) -> s(f(x,y))
r3: g(|0|(),x) -> g(f(x,x),x)

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}


-- Reduction pair.

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

p1: g#(|0|(),x) -> g#(f(x,x),x)

and R consists of:

r1: f(x,|0|()) -> s(|0|())
r2: f(s(x),s(y)) -> s(f(x,y))
r3: g(|0|(),x) -> g(f(x,x),x)

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      g#_A(x1,x2) = ((1,0,0),(0,0,0),(0,1,0)) x1 + ((1,0,0),(0,0,0),(0,1,0)) x2
      |0|_A() = (4,3,0)
      f_A(x1,x2) = ((0,0,0),(0,0,0),(1,0,0)) x1 + (3,2,1)
      s_A(x1) = (2,1,2)

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: f#(s(x),s(y)) -> f#(x,y)

and R consists of:

r1: f(x,|0|()) -> s(|0|())
r2: f(s(x),s(y)) -> s(f(x,y))
r3: g(|0|(),x) -> g(f(x,x),x)

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,x2) = ((1,0,0),(0,0,0),(0,1,0)) x1
      s_A(x1) = ((1,0,0),(0,0,0),(1,1,0)) x1 + (1,1,1)

The next rules are strictly ordered:

  p1

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