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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        n__f > g > cons > f > |0| > s > sel# > activate
      
      argument filter:
    
        pi(sel#) = 1
        pi(s) = 1
        pi(cons) = 1
        pi(activate) = [1]
        pi(g) = 1
        pi(|0|) = []
        pi(f) = 1
        pi(n__f) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        n__f > g > cons > f > sel# > s > |0| > activate
      
      argument filter:
    
        pi(sel#) = [1]
        pi(s) = [1]
        pi(cons) = 1
        pi(activate) = [1]
        pi(g) = [1]
        pi(|0|) = []
        pi(f) = [1]
        pi(n__f) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        n__f > g > cons > f > s > |0| > activate > sel#
      
      argument filter:
    
        pi(sel#) = 1
        pi(s) = [1]
        pi(cons) = [1]
        pi(activate) = [1]
        pi(g) = []
        pi(|0|) = []
        pi(f) = 1
        pi(n__f) = [1]
    

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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > g#
      
      argument filter:
    
        pi(g#) = [1]
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > g#
      
      argument filter:
    
        pi(g#) = [1]
        pi(s) = [1]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > g#
      
      argument filter:
    
        pi(g#) = [1]
        pi(s) = [1]
    

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.