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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sel#_A(x1,x2) = max{42, x1 + 39}
          s_A(x1) = max{70, x1 + 29}
          cons_A(x1,x2) = 27
          activate_A(x1) = max{41, x1}
          from_A(x1) = 27
          n__from_A(x1) = 10
          n__s_A(x1) = max{70, x1 + 29}
          take_A(x1,x2) = max{82, x2 + 12}
          |0|_A = 1
          nil_A = 0
          n__take_A(x1,x2) = max{82, x2 + 12}
    
      precedence:
      
        sel# = activate > s > from = n__s > n__from = take > cons = |0| = nil = n__take
      
      partial status:
    
        pi(sel#) = [1]
        pi(s) = [1]
        pi(cons) = []
        pi(activate) = [1]
        pi(from) = []
        pi(n__from) = []
        pi(n__s) = [1]
        pi(take) = [2]
        pi(|0|) = []
        pi(nil) = []
        pi(n__take) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sel#_A(x1,x2) = max{4, x1}
          s_A(x1) = x1 + 17
          cons_A(x1,x2) = 4
          activate_A(x1) = max{5, x1 - 1}
          from_A(x1) = 4
          n__from_A(x1) = 6
          n__s_A(x1) = x1 + 23
          take_A(x1,x2) = max{4, x2}
          |0|_A = 4
          nil_A = 1
          n__take_A(x1,x2) = max{4, x2}
    
      precedence:
      
        activate > nil > from > sel# = cons = take > n__take > s = n__from = n__s = |0|
      
      partial status:
    
        pi(sel#) = [1]
        pi(s) = [1]
        pi(cons) = []
        pi(activate) = []
        pi(from) = []
        pi(n__from) = []
        pi(n__s) = []
        pi(take) = []
        pi(|0|) = []
        pi(nil) = []
        pi(n__take) = []
    

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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          take#_A(x1,x2) = x2 + 6
          s_A(x1) = max{15, x1 + 3}
          cons_A(x1,x2) = max{x1 + 7, x2 - 4}
          activate#_A(x1) = max{13, x1 + 2}
          n__take_A(x1,x2) = max{x1 + 11, x2 + 10}
          activate_A(x1) = x1
          n__s_A(x1) = max{15, x1 + 3}
          n__from_A(x1) = max{27, x1 + 16}
          from_A(x1) = max{27, x1 + 16}
          take_A(x1,x2) = max{x1 + 11, x2 + 10}
          |0|_A = 1
          nil_A = 0
    
      precedence:
      
        take# = s = activate = n__s = from > take = |0| > activate# = n__from > n__take > cons > nil
      
      partial status:
    
        pi(take#) = [2]
        pi(s) = [1]
        pi(cons) = []
        pi(activate#) = [1]
        pi(n__take) = [1, 2]
        pi(activate) = [1]
        pi(n__s) = []
        pi(n__from) = [1]
        pi(from) = [1]
        pi(take) = [1, 2]
        pi(|0|) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          take#_A(x1,x2) = max{59, x2 + 1}
          s_A(x1) = max{60, x1 + 22}
          cons_A(x1,x2) = 60
          activate#_A(x1) = 59
          n__take_A(x1,x2) = 0
          activate_A(x1) = 57
          n__s_A(x1) = 19
          n__from_A(x1) = x1 + 19
          from_A(x1) = max{61, x1 + 20}
          take_A(x1,x2) = max{57, x1}
          |0|_A = 29
          nil_A = 30
    
      precedence:
      
        from = |0| = nil > s > activate > activate# = take > take# = cons = n__take = n__s = n__from
      
      partial status:
    
        pi(take#) = [2]
        pi(s) = []
        pi(cons) = []
        pi(activate#) = []
        pi(n__take) = []
        pi(activate) = []
        pi(n__s) = []
        pi(n__from) = [1]
        pi(from) = []
        pi(take) = []
        pi(|0|) = []
        pi(nil) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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