YES

We show the termination of the TRS R:

  from(X) -> cons(X,n__from(s(X)))
  sel(|0|(),cons(X,Y)) -> X
  sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
  from(X) -> n__from(X)
  activate(n__from(X)) -> from(X)
  activate(X) -> X

-- SCC decomposition.

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

p1: sel#(s(X),cons(Y,Z)) -> sel#(X,activate(Z))
p2: sel#(s(X),cons(Y,Z)) -> activate#(Z)
p3: activate#(n__from(X)) -> from#(X)

and R consists of:

r1: from(X) -> cons(X,n__from(s(X)))
r2: sel(|0|(),cons(X,Y)) -> X
r3: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r4: from(X) -> n__from(X)
r5: activate(n__from(X)) -> from(X)
r6: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1}


-- 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: from(X) -> cons(X,n__from(s(X)))
r2: sel(|0|(),cons(X,Y)) -> X
r3: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r4: from(X) -> n__from(X)
r5: activate(n__from(X)) -> from(X)
r6: activate(X) -> X

The set of usable rules consists of

  r1, r4, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        n__from > s > cons > from > activate > sel#
      
      argument filter:
    
        pi(sel#) = [1]
        pi(s) = 1
        pi(cons) = 1
        pi(activate) = 1
        pi(from) = 1
        pi(n__from) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        n__from > s > cons > from > activate > sel#
      
      argument filter:
    
        pi(sel#) = [1]
        pi(s) = [1]
        pi(cons) = 1
        pi(activate) = [1]
        pi(from) = 1
        pi(n__from) = 1
    

The next rules are strictly ordered:

  p1

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