YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: f#(s(x),y) -> f#(f(x,y),y)
p2: f#(s(x),y) -> f#(x,y)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: f#(s(x),y) -> f#(f(x,y),y)
p2: f#(s(x),y) -> f#(x,y)

and R consists of:

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

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

The next rules are strictly ordered:

  p1, p2

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