YES

We show the termination of the TRS R:

  natsFrom(N) -> cons(N,n__natsFrom(s(N)))
  fst(pair(XS,YS)) -> XS
  snd(pair(XS,YS)) -> YS
  splitAt(|0|(),XS) -> pair(nil(),XS)
  splitAt(s(N),cons(X,XS)) -> u(splitAt(N,activate(XS)),N,X,activate(XS))
  u(pair(YS,ZS),N,X,XS) -> pair(cons(activate(X),YS),ZS)
  head(cons(N,XS)) -> N
  tail(cons(N,XS)) -> activate(XS)
  sel(N,XS) -> head(afterNth(N,XS))
  take(N,XS) -> fst(splitAt(N,XS))
  afterNth(N,XS) -> snd(splitAt(N,XS))
  natsFrom(X) -> n__natsFrom(X)
  activate(n__natsFrom(X)) -> natsFrom(X)
  activate(X) -> X

-- SCC decomposition.

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

p1: splitAt#(s(N),cons(X,XS)) -> u#(splitAt(N,activate(XS)),N,X,activate(XS))
p2: splitAt#(s(N),cons(X,XS)) -> splitAt#(N,activate(XS))
p3: splitAt#(s(N),cons(X,XS)) -> activate#(XS)
p4: u#(pair(YS,ZS),N,X,XS) -> activate#(X)
p5: tail#(cons(N,XS)) -> activate#(XS)
p6: sel#(N,XS) -> head#(afterNth(N,XS))
p7: sel#(N,XS) -> afterNth#(N,XS)
p8: take#(N,XS) -> fst#(splitAt(N,XS))
p9: take#(N,XS) -> splitAt#(N,XS)
p10: afterNth#(N,XS) -> snd#(splitAt(N,XS))
p11: afterNth#(N,XS) -> splitAt#(N,XS)
p12: activate#(n__natsFrom(X)) -> natsFrom#(X)

and R consists of:

r1: natsFrom(N) -> cons(N,n__natsFrom(s(N)))
r2: fst(pair(XS,YS)) -> XS
r3: snd(pair(XS,YS)) -> YS
r4: splitAt(|0|(),XS) -> pair(nil(),XS)
r5: splitAt(s(N),cons(X,XS)) -> u(splitAt(N,activate(XS)),N,X,activate(XS))
r6: u(pair(YS,ZS),N,X,XS) -> pair(cons(activate(X),YS),ZS)
r7: head(cons(N,XS)) -> N
r8: tail(cons(N,XS)) -> activate(XS)
r9: sel(N,XS) -> head(afterNth(N,XS))
r10: take(N,XS) -> fst(splitAt(N,XS))
r11: afterNth(N,XS) -> snd(splitAt(N,XS))
r12: natsFrom(X) -> n__natsFrom(X)
r13: activate(n__natsFrom(X)) -> natsFrom(X)
r14: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

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

and R consists of:

r1: natsFrom(N) -> cons(N,n__natsFrom(s(N)))
r2: fst(pair(XS,YS)) -> XS
r3: snd(pair(XS,YS)) -> YS
r4: splitAt(|0|(),XS) -> pair(nil(),XS)
r5: splitAt(s(N),cons(X,XS)) -> u(splitAt(N,activate(XS)),N,X,activate(XS))
r6: u(pair(YS,ZS),N,X,XS) -> pair(cons(activate(X),YS),ZS)
r7: head(cons(N,XS)) -> N
r8: tail(cons(N,XS)) -> activate(XS)
r9: sel(N,XS) -> head(afterNth(N,XS))
r10: take(N,XS) -> fst(splitAt(N,XS))
r11: afterNth(N,XS) -> snd(splitAt(N,XS))
r12: natsFrom(X) -> n__natsFrom(X)
r13: activate(n__natsFrom(X)) -> natsFrom(X)
r14: activate(X) -> X

The set of usable rules consists of

  r1, r12, r13, r14

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        splitAt#_A(x1,x2) = ((1,1),(1,1)) x1
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        cons_A(x1,x2) = (1,1)
        activate_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        natsFrom_A(x1) = ((1,1),(1,1)) x1 + (2,1)
        n__natsFrom_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        splitAt#_A(x1,x2) = ((0,0),(1,0)) x1
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        cons_A(x1,x2) = (3,3)
        activate_A(x1) = (1,1)
        natsFrom_A(x1) = x1 + (2,2)
        n__natsFrom_A(x1) = ((1,0),(1,1)) x1 + (3,3)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        splitAt#_A(x1,x2) = (0,0)
        s_A(x1) = (1,1)
        cons_A(x1,x2) = (1,3)
        activate_A(x1) = (1,1)
        natsFrom_A(x1) = ((1,1),(0,1)) x1 + (2,2)
        n__natsFrom_A(x1) = (3,3)
    

The next rules are strictly ordered:

  p1

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