YES

We show the termination of the TRS R:

  f(X) -> cons(X,n__f(g(X)))
  g(|0|()) -> s(|0|())
  g(s(X)) -> s(s(g(X)))
  sel(|0|(),cons(X,Y)) -> X
  sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
  f(X) -> n__f(X)
  activate(n__f(X)) -> f(X)
  activate(X) -> X

-- SCC decomposition.

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

p1: f#(X) -> g#(X)
p2: g#(s(X)) -> g#(X)
p3: sel#(s(X),cons(Y,Z)) -> sel#(X,activate(Z))
p4: sel#(s(X),cons(Y,Z)) -> activate#(Z)
p5: activate#(n__f(X)) -> f#(X)

and R consists of:

r1: f(X) -> cons(X,n__f(g(X)))
r2: g(|0|()) -> s(|0|())
r3: g(s(X)) -> s(s(g(X)))
r4: sel(|0|(),cons(X,Y)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: f(X) -> n__f(X)
r7: activate(n__f(X)) -> f(X)
r8: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p3}
  {p2}


-- Reduction pair.

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

p1: sel#(s(X),cons(Y,Z)) -> sel#(X,activate(Z))

and R consists of:

r1: f(X) -> cons(X,n__f(g(X)))
r2: g(|0|()) -> s(|0|())
r3: g(s(X)) -> s(s(g(X)))
r4: sel(|0|(),cons(X,Y)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: f(X) -> n__f(X)
r7: activate(n__f(X)) -> f(X)
r8: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r6, r7, r8

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      sel#_A(x1,x2) = ((1,1),(1,1)) x1
      s_A(x1) = ((0,1),(1,0)) x1 + (1,0)
      cons_A(x1,x2) = (0,0)
      activate_A(x1) = ((1,1),(0,1)) x1
      g_A(x1) = ((1,1),(1,1)) x1
      |0|_A() = (1,0)
      f_A(x1) = ((1,1),(0,0)) x1
      n__f_A(x1) = ((1,1),(0,0)) x1

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: g#(s(X)) -> g#(X)

and R consists of:

r1: f(X) -> cons(X,n__f(g(X)))
r2: g(|0|()) -> s(|0|())
r3: g(s(X)) -> s(s(g(X)))
r4: sel(|0|(),cons(X,Y)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: f(X) -> n__f(X)
r7: activate(n__f(X)) -> f(X)
r8: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      g#_A(x1) = ((1,0),(1,0)) x1
      s_A(x1) = ((1,1),(1,1)) x1 + (1,0)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8

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