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(n__s(X)))
  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: |2nd|#(cons(X,X1)) -> |2nd|#(cons1(X,activate(X1)))
p2: |2nd|#(cons(X,X1)) -> activate#(X1)
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: |2nd|(cons1(X,cons(Y,Z))) -> Y
r2: |2nd|(cons(X,X1)) -> |2nd|(cons1(X,activate(X1)))
r3: from(X) -> cons(X,n__from(n__s(X)))
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:

  {p4, p6}


-- 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: |2nd|(cons1(X,cons(Y,Z))) -> Y
r2: |2nd|(cons(X,X1)) -> |2nd|(cons1(X,activate(X1)))
r3: from(X) -> cons(X,n__from(n__s(X)))
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 combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        activate#_A(x1) = ((1,1),(1,1)) x1
        n__s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        n__from_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        activate#_A(x1) = ((1,1),(1,1)) x1
        n__s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        n__from_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        activate#_A(x1) = ((1,1),(1,1)) x1
        n__s_A(x1) = ((1,1),(0,1)) x1 + (1,1)
        n__from_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    

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.