YES

We show the termination of the TRS R:

  from(X) -> cons(X,n__from(n__s(X)))
  sel(|0|(),cons(X,Y)) -> X
  sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
  from(X) -> n__from(X)
  s(X) -> n__s(X)
  activate(n__from(X)) -> from(activate(X))
  activate(n__s(X)) -> s(activate(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#(activate(X))
p4: activate#(n__from(X)) -> activate#(X)
p5: activate#(n__s(X)) -> s#(activate(X))
p6: activate#(n__s(X)) -> activate#(X)

and R consists of:

r1: from(X) -> cons(X,n__from(n__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: s(X) -> n__s(X)
r6: activate(n__from(X)) -> from(activate(X))
r7: activate(n__s(X)) -> s(activate(X))
r8: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1}
  {p4, p6}


-- 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(n__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: s(X) -> n__s(X)
r6: activate(n__from(X)) -> from(activate(X))
r7: activate(n__s(X)) -> s(activate(X))
r8: activate(X) -> X

The set of usable rules consists of

  r1, r4, r5, r6, r7, r8

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      activate > cons > s > n__s > from > n__from > sel#
    
    argument filter:
  
      pi(sel#) = 1
      pi(s) = [1]
      pi(cons) = 1
      pi(activate) = [1]
      pi(from) = [1]
      pi(n__from) = [1]
      pi(n__s) = [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: activate#(n__s(X)) -> activate#(X)
p2: activate#(n__from(X)) -> activate#(X)

and R consists of:

r1: from(X) -> cons(X,n__from(n__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: s(X) -> n__s(X)
r6: activate(n__from(X)) -> from(activate(X))
r7: activate(n__s(X)) -> s(activate(X))
r8: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      activate# > n__from > n__s
    
    argument filter:
  
      pi(activate#) = 1
      pi(n__s) = [1]
      pi(n__from) = [1]

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8

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