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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          copy#_A(x1,x2,x3) = ((1,0),(1,1)) x1 + (1,2)
          s_A(x1) = ((1,0),(0,0)) x1 + (2,1)
          cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (1,16)
          f_A(x1) = ((1,0),(1,0)) x1 + (4,6)
          nil_A() = (2,11)
          copy_A(x1,x2,x3) = (1,15)
          n_A() = (1,10)
  
    precedence:
    
      cons = f > copy# > n > s > nil > copy
    
    partial status:
  
      pi(copy#) = [1]
      pi(s) = []
      pi(cons) = []
      pi(f) = []
      pi(nil) = []
      pi(copy) = []
      pi(n) = []

The next rules are strictly ordered:

  p1

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