YES

We show the termination of the TRS R:

  |2nd|(cons1(X,cons(Y,Z))) -> Y
  |2nd|(cons(X,X1)) -> |2nd|(cons1(X,activate(X1)))
  from(X) -> cons(X,n__from(s(X)))
  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: |2nd|#(cons(X,X1)) -> |2nd|#(cons1(X,activate(X1)))
p2: |2nd|#(cons(X,X1)) -> activate#(X1)
p3: activate#(n__from(X)) -> from#(X)

and R consists of:

r1: |2nd|(cons1(X,cons(Y,Z))) -> Y
r2: |2nd|(cons(X,X1)) -> |2nd|(cons1(X,activate(X1)))
r3: from(X) -> cons(X,n__from(s(X)))
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:

  (no SCCs)