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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      sel#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,0,1,1)) x1
      s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,1,1,1)) x1 + (6,0,4,1)
      cons_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,0,0)) x2 + (1,0,0,4)
      activate_A(x1) = ((1,0,0,0),(1,0,0,0),(1,1,0,0),(0,1,1,0)) x1 + (3,1,0,1)
      from_A(x1) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,1,0)) x1 + (3,0,13,19)
      n__from_A(x1) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x1 + (1,14,6,0)
      n__s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(1,1,1,0)) x1 + (6,0,3,0)
      take_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,1,0)) x1 + (3,0,0,6)
      |0|_A() = (1,1,1,1)
      nil_A() = (2,2,2,9)
      n__take_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(1,0,0,0),(1,0,0,0)) x1 + ((0,0,0,0),(1,0,0,0),(1,1,0,0),(1,1,1,0)) x2 + (1,4,1,0)

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: 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      take#_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,0,0)) x2 + (0,1,0,0)
      s_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,1,0,0)) x1 + (0,1,1,0)
      cons_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,1,0,0)) x2 + (0,1,6,10)
      activate#_A(x1) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(1,1,1,0)) x1 + (0,0,2,2)
      n__take_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,0)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,0)) x2 + (2,1,1,1)
      activate_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,0,0,1)) x1 + (0,0,1,1)
      n__s_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,1,0,0)) x1 + (0,1,1,0)
      n__from_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,1,0,0)) x1 + (3,1,4,0)
      from_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,0,1,1)) x1 + (3,1,7,1)
      take_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,1,1,0),(0,1,1,0)) x1 + ((1,0,0,0),(1,1,0,0),(0,1,0,0),(0,1,1,0)) x2 + (2,2,2,3)
      |0|_A() = (1,1,1,0)
      nil_A() = (0,5,5,0)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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