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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          f#_A(x1,x2) = 1
          s_A(x1) = 2
          f_A(x1,x2) = 3
          |0|_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1,x2) = max{0, x1 - 5, x2 - 1}
          s_A(x1) = x1 + 3
          f_A(x1,x2) = max{1, x1 - 4}
          |0|_A = 1
          f#!_A(x1,x2) = max{2, x1}
          s!_A(x1) = max{0, x1 - 9}
          f!_A(x1,x2) = max{4, x1 - 4, x2 - 4}
          |0|!_A = 0
  
    partial status:
  
      pi(f#) = []
      pi(s) = []
      pi(f) = []
      pi(|0|) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.