YES

We show the termination of the TRS R:

  from(X) -> cons(X,n__from(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)
  take(X1,X2) -> n__take(X1,X2)
  activate(n__from(X)) -> from(X)
  activate(n__take(X1,X2)) -> take(X1,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#(X)
p7: activate#(n__take(X1,X2)) -> take#(X1,X2)

and R consists of:

r1: from(X) -> cons(X,n__from(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: take(X1,X2) -> n__take(X1,X2)
r10: activate(n__from(X)) -> from(X)
r11: activate(n__take(X1,X2)) -> take(X1,X2)
r12: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p4}
  {p3, p7}


-- 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(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: take(X1,X2) -> n__take(X1,X2)
r10: activate(n__from(X)) -> from(X)
r11: activate(n__take(X1,X2)) -> take(X1,X2)
r12: activate(X) -> X

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sel#_A(x1,x2) = x1 + (1,1)
            s_A(x1) = ((1,0),(1,1)) x1 + (3,5)
            cons_A(x1,x2) = ((1,0),(0,0)) x2 + (2,3)
            activate_A(x1) = x1 + (7,2)
            from_A(x1) = x1 + (7,4)
            n__from_A(x1) = x1 + (1,3)
            take_A(x1,x2) = x1 + (5,1)
            |0|_A() = (2,1)
            nil_A() = (1,3)
            n__take_A(x1,x2) = x1
    
      precedence:
      
        sel# > s = activate > from > n__from = nil > |0| > cons = take = n__take
      
      partial status:
    
        pi(sel#) = [1]
        pi(s) = [1]
        pi(cons) = []
        pi(activate) = [1]
        pi(from) = [1]
        pi(n__from) = [1]
        pi(take) = [1]
        pi(|0|) = []
        pi(nil) = []
        pi(n__take) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sel#_A(x1,x2) = ((1,0),(1,1)) x1 + (1,1)
            s_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            cons_A(x1,x2) = (0,0)
            activate_A(x1) = (2,6)
            from_A(x1) = (0,0)
            n__from_A(x1) = ((1,0),(1,0)) x1
            take_A(x1,x2) = (0,1)
            |0|_A() = (2,3)
            nil_A() = (1,2)
            n__take_A(x1,x2) = (0,0)
    
      precedence:
      
        |0| = nil > sel# > s = cons = activate = from = n__from = take = n__take
      
      partial status:
    
        pi(sel#) = [1]
        pi(s) = []
        pi(cons) = []
        pi(activate) = []
        pi(from) = []
        pi(n__from) = []
        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)) -> take#(X1,X2)

and R consists of:

r1: from(X) -> cons(X,n__from(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: take(X1,X2) -> n__take(X1,X2)
r10: activate(n__from(X)) -> from(X)
r11: activate(n__take(X1,X2)) -> take(X1,X2)
r12: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            take#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (1,1)
            s_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            cons_A(x1,x2) = x1 + x2 + (2,2)
            activate#_A(x1) = ((1,0),(1,1)) x1 + (1,1)
            n__take_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (2,2)
    
      precedence:
      
        n__take > take# = s = cons = activate#
      
      partial status:
    
        pi(take#) = [2]
        pi(s) = [1]
        pi(cons) = [1, 2]
        pi(activate#) = [1]
        pi(n__take) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            take#_A(x1,x2) = (0,0)
            s_A(x1) = x1 + (1,1)
            cons_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,1)
            activate#_A(x1) = (0,0)
            n__take_A(x1,x2) = (1,1)
    
      precedence:
      
        n__take > s > cons = activate# > take#
      
      partial status:
    
        pi(take#) = []
        pi(s) = []
        pi(cons) = []
        pi(activate#) = []
        pi(n__take) = []
    

The next rules are strictly ordered:

  p1, p2

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