YES

We show the termination of the TRS R:

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1, p5}


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

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        sel#_A(x1,x2) = ((1,0,0),(1,1,0),(0,0,1)) x1 + ((1,0,0),(0,1,0),(1,0,1)) x2
        s_A(x1) = ((1,0,0),(1,1,0),(0,1,1)) x1 + (3,1,1)
        cons_A(x1,x2) = ((0,0,0),(1,0,0),(0,0,0)) x1 + ((1,0,0),(1,0,0),(1,1,0)) x2 + (1,1,1)
        activate_A(x1) = ((1,0,0),(0,1,0),(1,1,1)) x1 + (3,6,1)
        from_A(x1) = (3,0,0)
        n__from_A(x1) = ((0,0,0),(1,0,0),(1,1,0)) x1 + (1,1,1)
        first_A(x1,x2) = x2 + (4,5,2)
        |0|_A() = (1,1,1)
        nil_A() = (0,0,0)
        n__first_A(x1,x2) = ((0,0,0),(1,0,0),(1,1,0)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,6,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        first > s > n__first > nil > |0| > n__from > activate > cons > from > sel#
      
      argument filter:
    
        pi(sel#) = 1
        pi(s) = 1
        pi(cons) = []
        pi(activate) = []
        pi(from) = []
        pi(n__from) = []
        pi(first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(n__first) = []
    

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: first#(s(X),cons(Y,Z)) -> activate#(Z)
p2: activate#(n__first(X1,X2)) -> first#(X1,X2)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        first#_A(x1,x2) = ((1,0,0),(0,0,0),(1,0,0)) x2
        s_A(x1) = ((0,0,0),(1,0,0),(1,1,0)) x1 + (1,1,1)
        cons_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,1)) x1 + ((1,0,0),(0,0,0),(1,1,0)) x2 + (1,1,1)
        activate#_A(x1) = ((1,0,0),(0,1,0),(1,1,1)) x1 + (0,1,2)
        n__first_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        n__first > activate# > first# > cons > s
      
      argument filter:
    
        pi(first#) = []
        pi(s) = []
        pi(cons) = []
        pi(activate#) = [1]
        pi(n__first) = [1, 2]
    

The next rules are strictly ordered:

  p1, p2

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