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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = x1 + (4,5)
            s_A(x1) = ((1,0),(1,1)) x1 + (3,6)
            f_A(x1,x2) = ((1,0),(1,1)) x1 + (1,2)
            |0|_A() = (1,1)
    
      precedence:
      
        f# = s = f = |0|
      
      partial status:
    
        pi(f#) = []
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = ((1,0),(0,0)) x1
            s_A(x1) = ((1,0),(1,1)) x1 + (3,3)
            f_A(x1,x2) = (2,2)
            |0|_A() = (1,1)
    
      precedence:
      
        f# > s = f = |0|
      
      partial status:
    
        pi(f#) = []
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2

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