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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        sel#_A(x1,x2) = x1 + 1
        s_A(x1) = max{12, x1 + 8}
        cons_A(x1,x2) = max{x1 + 3, x2 - 9}
        activate_A(x1) = x1 + 6
        from_A(x1) = max{11, x1 + 7}
        n__from_A(x1) = max{10, x1 + 7}
        take_A(x1,x2) = max{x1 - 1, x2 + 7}
        |0|_A = 2
        nil_A = 3
        n__take_A(x1,x2) = max{x1 - 7, x2 + 1}
  
    precedence:
    
      sel# = s = cons = activate = from = n__from = take = |0| = nil = n__take
    
    partial status:
  
      pi(sel#) = [1]
      pi(s) = [1]
      pi(cons) = []
      pi(activate) = [1]
      pi(from) = [1]
      pi(n__from) = [1]
      pi(take) = []
      pi(|0|) = []
      pi(nil) = []
      pi(n__take) = [2]

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        take#_A(x1,x2) = max{5, x1 - 1, x2 + 3}
        s_A(x1) = x1
        cons_A(x1,x2) = max{x1 - 1, x2 + 3}
        activate#_A(x1) = x1 + 5
        n__take_A(x1,x2) = max{x1 - 6, x2}
  
    precedence:
    
      take# = s = cons = activate# = n__take
    
    partial status:
  
      pi(take#) = []
      pi(s) = [1]
      pi(cons) = [2]
      pi(activate#) = [1]
      pi(n__take) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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:

  (no SCCs)