YES

We show the termination of the TRS R:

  from(X) -> cons(X,n__from(n__s(X)))
  head(cons(X,XS)) -> X
  |2nd|(cons(X,XS)) -> head(activate(XS))
  take(|0|(),XS) -> nil()
  take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
  sel(|0|(),cons(X,XS)) -> X
  sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
  from(X) -> n__from(X)
  s(X) -> n__s(X)
  take(X1,X2) -> n__take(X1,X2)
  activate(n__from(X)) -> from(activate(X))
  activate(n__s(X)) -> s(activate(X))
  activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
  activate(X) -> X

-- SCC decomposition.

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

p1: |2nd|#(cons(X,XS)) -> head#(activate(XS))
p2: |2nd|#(cons(X,XS)) -> activate#(XS)
p3: take#(s(N),cons(X,XS)) -> activate#(XS)
p4: sel#(s(N),cons(X,XS)) -> sel#(N,activate(XS))
p5: sel#(s(N),cons(X,XS)) -> activate#(XS)
p6: activate#(n__from(X)) -> from#(activate(X))
p7: activate#(n__from(X)) -> activate#(X)
p8: activate#(n__s(X)) -> s#(activate(X))
p9: activate#(n__s(X)) -> activate#(X)
p10: activate#(n__take(X1,X2)) -> take#(activate(X1),activate(X2))
p11: activate#(n__take(X1,X2)) -> activate#(X1)
p12: activate#(n__take(X1,X2)) -> activate#(X2)

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p4}
  {p3, p7, p9, p10, p11, p12}


-- Reduction pair.

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

p1: sel#(s(N),cons(X,XS)) -> sel#(N,activate(XS))

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> X

The set of usable rules consists of

  r1, r4, r5, r8, r9, r10, r11, r12, r13, r14

Take the reduction pair:

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

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: take#(s(N),cons(X,XS)) -> activate#(XS)
p2: activate#(n__take(X1,X2)) -> activate#(X2)
p3: activate#(n__take(X1,X2)) -> activate#(X1)
p4: activate#(n__take(X1,X2)) -> take#(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: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> X

The set of usable rules consists of

  r1, r4, r5, r8, r9, r10, r11, r12, r13, r14

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      take#_A(x1,x2) = max{x1 - 1, x2 + 3}
      s_A(x1) = x1 + 2
      cons_A(x1,x2) = max{x1 + 5, x2 - 2}
      activate#_A(x1) = x1
      n__take_A(x1,x2) = max{x1 + 5, x2 + 6}
      activate_A(x1) = x1
      n__s_A(x1) = x1 + 2
      n__from_A(x1) = x1 + 5
      from_A(x1) = x1 + 5
      take_A(x1,x2) = max{x1 + 5, x2 + 6}
      |0|_A = 0
      nil_A = 2
    
    2. max/plus interpretations on natural numbers:
    
      take#_A(x1,x2) = 1
      s_A(x1) = max{2, x1 + 1}
      cons_A(x1,x2) = 0
      activate#_A(x1) = max{0, x1 - 1}
      n__take_A(x1,x2) = 3
      activate_A(x1) = x1 + 2
      n__s_A(x1) = max{2, x1 + 1}
      n__from_A(x1) = max{3, x1 + 1}
      from_A(x1) = max{4, x1 + 1}
      take_A(x1,x2) = 4
      |0|_A = 0
      nil_A = 5
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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