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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        copy#_A(x1,x2,x3) = x1 + x2 + x3
        s_A(x1) = x1 + 2
        cons_A(x1,x2) = x2 + 1
        f_A(x1) = x1 + 1
        nil_A() = 1
        copy_A(x1,x2,x3) = 0
        n_A() = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        copy#_A(x1,x2,x3) = x2 + x3
        s_A(x1) = x1 + 1
        cons_A(x1,x2) = 1
        f_A(x1) = 1
        nil_A() = 1
        copy_A(x1,x2,x3) = 2
        n_A() = 1
    

The next rules are strictly ordered:

  p1

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