YES

We show the termination of the TRS R:

  |2nd|(cons(X,n__cons(Y,Z))) -> activate(Y)
  from(X) -> cons(X,n__from(s(X)))
  cons(X1,X2) -> n__cons(X1,X2)
  from(X) -> n__from(X)
  activate(n__cons(X1,X2)) -> cons(X1,X2)
  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,n__cons(Y,Z))) -> activate#(Y)
p2: from#(X) -> cons#(X,n__from(s(X)))
p3: activate#(n__cons(X1,X2)) -> cons#(X1,X2)
p4: activate#(n__from(X)) -> from#(X)

and R consists of:

r1: |2nd|(cons(X,n__cons(Y,Z))) -> activate(Y)
r2: from(X) -> cons(X,n__from(s(X)))
r3: cons(X1,X2) -> n__cons(X1,X2)
r4: from(X) -> n__from(X)
r5: activate(n__cons(X1,X2)) -> cons(X1,X2)
r6: activate(n__from(X)) -> from(X)
r7: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  (no SCCs)