YES

We show the termination of the TRS R:

  from(X) -> cons(X,n__from(n__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)
  s(X) -> n__s(X)
  first(X1,X2) -> n__first(X1,X2)
  activate(n__from(X)) -> from(activate(X))
  activate(n__s(X)) -> s(activate(X))
  activate(n__first(X1,X2)) -> first(activate(X1),activate(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#(activate(X))
p5: activate#(n__from(X)) -> activate#(X)
p6: activate#(n__s(X)) -> s#(activate(X))
p7: activate#(n__s(X)) -> activate#(X)
p8: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p9: activate#(n__first(X1,X2)) -> activate#(X1)
p10: activate#(n__first(X1,X2)) -> activate#(X2)

and R consists of:

r1: from(X) -> cons(X,n__from(n__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: s(X) -> n__s(X)
r8: first(X1,X2) -> n__first(X1,X2)
r9: activate(n__from(X)) -> from(activate(X))
r10: activate(n__s(X)) -> s(activate(X))
r11: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r12: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1, p5, p7, p8, p9, p10}


-- 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(n__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: s(X) -> n__s(X)
r8: first(X1,X2) -> n__first(X1,X2)
r9: activate(n__from(X)) -> from(activate(X))
r10: activate(n__s(X)) -> s(activate(X))
r11: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r12: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r6, r7, r8, r9, r10, r11, r12

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      sel#_A(x1,x2) = max{11, x1 + 1}
      s_A(x1) = x1 + 11
      cons_A(x1,x2) = max{12, x2 + 8}
      activate_A(x1) = x1 + 10
      from_A(x1) = 13
      n__from_A(x1) = 4
      n__s_A(x1) = x1 + 11
      first_A(x1,x2) = x1 + 3
      |0|_A = 1
      nil_A = 3
      n__first_A(x1,x2) = x1 + 3
    
    2. max/plus interpretations on natural numbers:
    
      sel#_A(x1,x2) = 0
      s_A(x1) = max{2, x1 - 3}
      cons_A(x1,x2) = 6
      activate_A(x1) = max{2, x1 - 2}
      from_A(x1) = 5
      n__from_A(x1) = 6
      n__s_A(x1) = max{2, x1 - 3}
      first_A(x1,x2) = 1
      |0|_A = 0
      nil_A = 0
      n__first_A(x1,x2) = 0
    

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)) -> activate#(X2)
p3: activate#(n__first(X1,X2)) -> activate#(X1)
p4: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p5: activate#(n__s(X)) -> activate#(X)
p6: activate#(n__from(X)) -> activate#(X)

and R consists of:

r1: from(X) -> cons(X,n__from(n__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: s(X) -> n__s(X)
r8: first(X1,X2) -> n__first(X1,X2)
r9: activate(n__from(X)) -> from(activate(X))
r10: activate(n__s(X)) -> s(activate(X))
r11: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r12: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r6, r7, r8, r9, r10, r11, r12

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      first#_A(x1,x2) = max{x1 + 2, x2 + 5}
      s_A(x1) = x1 + 1
      cons_A(x1,x2) = max{2, x1 - 2, x2 - 3}
      activate#_A(x1) = x1 + 1
      n__first_A(x1,x2) = max{x1 + 5, x2 + 6}
      activate_A(x1) = x1
      n__s_A(x1) = x1 + 1
      n__from_A(x1) = x1 + 2
      from_A(x1) = x1 + 2
      first_A(x1,x2) = max{x1 + 5, x2 + 6}
      |0|_A = 3
      nil_A = 7
    
    2. max/plus interpretations on natural numbers:
    
      first#_A(x1,x2) = 1
      s_A(x1) = 11
      cons_A(x1,x2) = 2
      activate#_A(x1) = max{0, x1 - 2}
      n__first_A(x1,x2) = 4
      activate_A(x1) = max{14, x1 + 8}
      n__s_A(x1) = 0
      n__from_A(x1) = max{10, x1 + 3}
      from_A(x1) = max{10, x1 + 3}
      first_A(x1,x2) = 13
      |0|_A = 0
      nil_A = 0
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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