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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            copy#_A(x1,x2,x3) = ((1,0),(0,0)) x1
            s_A(x1) = x1 + (4,1)
            cons_A(x1,x2) = ((1,0),(1,1)) x2 + (3,2)
            f_A(x1) = x1 + (5,2)
            nil_A() = (1,2)
            copy_A(x1,x2,x3) = x1 + (2,1)
            n_A() = (0,0)
    
      precedence:
      
        copy# > s = nil > copy > cons = f = n
      
      partial status:
    
        pi(copy#) = []
        pi(s) = []
        pi(cons) = [2]
        pi(f) = [1]
        pi(nil) = []
        pi(copy) = [1]
        pi(n) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            copy#_A(x1,x2,x3) = (0,0)
            s_A(x1) = (5,5)
            cons_A(x1,x2) = ((1,0),(1,1)) x2 + (6,2)
            f_A(x1) = (4,4)
            nil_A() = (2,2)
            copy_A(x1,x2,x3) = x1 + (2,2)
            n_A() = (1,1)
    
      precedence:
      
        nil > cons > copy# = s = f = copy = n
      
      partial status:
    
        pi(copy#) = []
        pi(s) = []
        pi(cons) = [2]
        pi(f) = []
        pi(nil) = []
        pi(copy) = [1]
        pi(n) = []
    

The next rules are strictly ordered:

  p1

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