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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1,x2) = 0
              s_A(x1) = x1 + 2
              f_A(x1,x2) = x1 + 1
              |0|_A() = 1
              f#!_A(x1,x2) = x1
              s!_A(x1) = x1 + 2
              f!_A(x1,x2) = 1
              |0|!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1,x2) = 0
              s_A(x1) = x1 + 2
              f_A(x1,x2) = x1 + 1
              |0|_A() = 1
              f#!_A(x1,x2) = x1 + x2 + 3
              s!_A(x1) = 2
              f!_A(x1,x2) = 1
              |0|!_A() = 0
        
  
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      pi(f) = [1]
      pi(|0|) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.