YES

We show the termination of the TRS R:

  first(|0|(),X) -> nil()
  first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
  from(X) -> cons(X,n__from(n__s(X)))
  first(X1,X2) -> n__first(X1,X2)
  from(X) -> n__from(X)
  s(X) -> n__s(X)
  activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
  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: first#(s(X),cons(Y,Z)) -> activate#(Z)
p2: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p3: activate#(n__first(X1,X2)) -> activate#(X1)
p4: activate#(n__first(X1,X2)) -> activate#(X2)
p5: activate#(n__from(X)) -> from#(activate(X))
p6: activate#(n__from(X)) -> activate#(X)
p7: activate#(n__s(X)) -> s#(activate(X))
p8: activate#(n__s(X)) -> activate#(X)

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: s(X) -> n__s(X)
r7: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r8: activate(n__from(X)) -> from(activate(X))
r9: activate(n__s(X)) -> s(activate(X))
r10: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p6, p8}


-- Reduction pair.

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

p1: first#(s(X),cons(Y,Z)) -> activate#(Z)
p2: activate#(n__s(X)) -> activate#(X)
p3: activate#(n__from(X)) -> activate#(X)
p4: activate#(n__first(X1,X2)) -> activate#(X2)
p5: activate#(n__first(X1,X2)) -> activate#(X1)
p6: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: s(X) -> n__s(X)
r7: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r8: activate(n__from(X)) -> from(activate(X))
r9: activate(n__s(X)) -> s(activate(X))
r10: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: lexicographic order
      interpretations:
        first#_A(x1,x2) = ((1,0),(1,1)) x2 + (1,2)
        s_A(x1) = ((1,0),(1,1)) x1 + (0,1)
        cons_A(x1,x2) = ((1,0),(1,0)) x2 + (0,1)
        activate#_A(x1) = ((1,0),(1,1)) x1
        n__s_A(x1) = ((1,0),(1,1)) x1 + (0,1)
        n__from_A(x1) = ((1,0),(1,1)) x1 + (2,3)
        n__first_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (2,2)
        activate_A(x1) = ((1,0),(1,1)) x1 + (0,1)
        first_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (2,3)
        |0|_A() = (0,0)
        nil_A() = (0,0)
        from_A(x1) = ((1,0),(1,1)) x1 + (2,4)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        activate# > activate > from > |0| > first > nil > s > cons > n__first > n__from > n__s > first#
      
      argument filter:
    
        pi(first#) = [2]
        pi(s) = 1
        pi(cons) = []
        pi(activate#) = []
        pi(n__s) = 1
        pi(n__from) = []
        pi(n__first) = []
        pi(activate) = []
        pi(first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(from) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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