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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1,x2) = max{1, x1}
        s_A(x1) = x1 + 4
        f_A(x1,x2) = x1 + 3
        |0|_A = 1
  
    base order 2:
  
      precedence:
      
        f > s = |0| > f#
      
  
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      pi(f) = [1]
      pi(|0|) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

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


-- Reduction pair.

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

p1: f#(s(x),y) -> f#(f(x,y),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, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1,x2) = x1 + 2
        s_A(x1) = 6
        f_A(x1,x2) = 3
        |0|_A = 1
  
    base order 2:
  
      precedence:
      
        s > f# > f > |0|
      
  
    partial status:
  
      pi(f#) = [1]
      pi(s) = []
      pi(f) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.