YES

We show the termination of the TRS R:

  f(cons(nil(),y)) -> y
  f(cons(f(cons(nil(),y)),z)) -> copy(n(),y,z)
  copy(|0|(),y,z) -> f(z)
  copy(s(x),y,z) -> copy(x,y,cons(f(y),z))

-- SCC decomposition.

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

p1: f#(cons(f(cons(nil(),y)),z)) -> copy#(n(),y,z)
p2: copy#(|0|(),y,z) -> f#(z)
p3: copy#(s(x),y,z) -> copy#(x,y,cons(f(y),z))
p4: copy#(s(x),y,z) -> f#(y)

and R consists of:

r1: f(cons(nil(),y)) -> y
r2: f(cons(f(cons(nil(),y)),z)) -> copy(n(),y,z)
r3: copy(|0|(),y,z) -> f(z)
r4: copy(s(x),y,z) -> copy(x,y,cons(f(y),z))

The estimated dependency graph contains the following SCCs:

  {p3}


-- Reduction pair.

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

p1: copy#(s(x),y,z) -> copy#(x,y,cons(f(y),z))

and R consists of:

r1: f(cons(nil(),y)) -> y
r2: f(cons(f(cons(nil(),y)),z)) -> copy(n(),y,z)
r3: copy(|0|(),y,z) -> f(z)
r4: copy(s(x),y,z) -> copy(x,y,cons(f(y),z))

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:
      
        n > copy > nil > f > cons > s > copy#
      
      argument filter:
    
        pi(copy#) = 1
        pi(s) = [1]
        pi(cons) = [1, 2]
        pi(f) = 1
        pi(nil) = []
        pi(copy) = 2
        pi(n) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        n > copy > nil > f > cons > s > copy#
      
      argument filter:
    
        pi(copy#) = []
        pi(s) = [1]
        pi(cons) = [1, 2]
        pi(f) = [1]
        pi(nil) = []
        pi(copy) = 2
        pi(n) = []
    

The next rules are strictly ordered:

  p1

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