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:
      
        nil > s > copy# > f > cons > n > copy
      
      argument filter:
    
        pi(copy#) = [1, 2, 3]
        pi(s) = [1]
        pi(cons) = [1, 2]
        pi(f) = [1]
        pi(nil) = []
        pi(copy) = 2
        pi(n) = []
    
    2. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        copy#_A(x1,x2,x3) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(0,1,1,0)) x1 + ((1,0,0,0),(1,1,0,0),(0,0,0,0),(1,0,1,0)) x3
        s_A(x1) = ((0,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (1,1,2,0)
        cons_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1 + ((1,0,0,0),(0,1,0,0),(1,0,1,0),(0,1,1,1)) x2
        f_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,0,1)) x1 + (2,1,0,1)
        nil_A() = (1,1,0,1)
        copy_A(x1,x2,x3) = (6,0,0,10)
        n_A() = (1,1,1,1)
    

The next rules are strictly ordered:

  p1

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