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:

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

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:

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

The next rules are strictly ordered:

  p1

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