YES

We show the termination of the TRS R:

  fib(N) -> sel(N,fib1(s(|0|()),s(|0|())))
  fib1(X,Y) -> cons(X,n__fib1(Y,n__add(X,Y)))
  add(|0|(),X) -> X
  add(s(X),Y) -> s(add(X,Y))
  sel(|0|(),cons(X,XS)) -> X
  sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
  fib1(X1,X2) -> n__fib1(X1,X2)
  add(X1,X2) -> n__add(X1,X2)
  activate(n__fib1(X1,X2)) -> fib1(activate(X1),activate(X2))
  activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
  activate(X) -> X

-- SCC decomposition.

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

p1: fib#(N) -> sel#(N,fib1(s(|0|()),s(|0|())))
p2: fib#(N) -> fib1#(s(|0|()),s(|0|()))
p3: add#(s(X),Y) -> add#(X,Y)
p4: sel#(s(N),cons(X,XS)) -> sel#(N,activate(XS))
p5: sel#(s(N),cons(X,XS)) -> activate#(XS)
p6: activate#(n__fib1(X1,X2)) -> fib1#(activate(X1),activate(X2))
p7: activate#(n__fib1(X1,X2)) -> activate#(X1)
p8: activate#(n__fib1(X1,X2)) -> activate#(X2)
p9: activate#(n__add(X1,X2)) -> add#(activate(X1),activate(X2))
p10: activate#(n__add(X1,X2)) -> activate#(X1)
p11: activate#(n__add(X1,X2)) -> activate#(X2)

and R consists of:

r1: fib(N) -> sel(N,fib1(s(|0|()),s(|0|())))
r2: fib1(X,Y) -> cons(X,n__fib1(Y,n__add(X,Y)))
r3: add(|0|(),X) -> X
r4: add(s(X),Y) -> s(add(X,Y))
r5: sel(|0|(),cons(X,XS)) -> X
r6: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r7: fib1(X1,X2) -> n__fib1(X1,X2)
r8: add(X1,X2) -> n__add(X1,X2)
r9: activate(n__fib1(X1,X2)) -> fib1(activate(X1),activate(X2))
r10: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r11: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p4}
  {p7, p8, p10, p11}
  {p3}


-- 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: fib(N) -> sel(N,fib1(s(|0|()),s(|0|())))
r2: fib1(X,Y) -> cons(X,n__fib1(Y,n__add(X,Y)))
r3: add(|0|(),X) -> X
r4: add(s(X),Y) -> s(add(X,Y))
r5: sel(|0|(),cons(X,XS)) -> X
r6: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r7: fib1(X1,X2) -> n__fib1(X1,X2)
r8: add(X1,X2) -> n__add(X1,X2)
r9: activate(n__fib1(X1,X2)) -> fib1(activate(X1),activate(X2))
r10: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r11: activate(X) -> X

The set of usable rules consists of

  r2, r3, r4, r7, r8, r9, r10, r11

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      |0| > activate > add > n__add > fib1 > n__fib1 > cons > s > sel#
    
    argument filter:
  
      pi(sel#) = 1
      pi(s) = [1]
      pi(cons) = [2]
      pi(activate) = [1]
      pi(fib1) = []
      pi(n__fib1) = []
      pi(n__add) = [1, 2]
      pi(add) = [1, 2]
      pi(|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: activate#(n__add(X1,X2)) -> activate#(X2)
p2: activate#(n__add(X1,X2)) -> activate#(X1)
p3: activate#(n__fib1(X1,X2)) -> activate#(X2)
p4: activate#(n__fib1(X1,X2)) -> activate#(X1)

and R consists of:

r1: fib(N) -> sel(N,fib1(s(|0|()),s(|0|())))
r2: fib1(X,Y) -> cons(X,n__fib1(Y,n__add(X,Y)))
r3: add(|0|(),X) -> X
r4: add(s(X),Y) -> s(add(X,Y))
r5: sel(|0|(),cons(X,XS)) -> X
r6: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r7: fib1(X1,X2) -> n__fib1(X1,X2)
r8: add(X1,X2) -> n__add(X1,X2)
r9: activate(n__fib1(X1,X2)) -> fib1(activate(X1),activate(X2))
r10: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r11: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      activate# > n__fib1 > n__add
    
    argument filter:
  
      pi(activate#) = 1
      pi(n__add) = [1, 2]
      pi(n__fib1) = [1, 2]

The next rules are strictly ordered:

  p1, p2, p3, p4
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

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: add#(s(X),Y) -> add#(X,Y)

and R consists of:

r1: fib(N) -> sel(N,fib1(s(|0|()),s(|0|())))
r2: fib1(X,Y) -> cons(X,n__fib1(Y,n__add(X,Y)))
r3: add(|0|(),X) -> X
r4: add(s(X),Y) -> s(add(X,Y))
r5: sel(|0|(),cons(X,XS)) -> X
r6: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r7: fib1(X1,X2) -> n__fib1(X1,X2)
r8: add(X1,X2) -> n__add(X1,X2)
r9: activate(n__fib1(X1,X2)) -> fib1(activate(X1),activate(X2))
r10: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r11: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      s > add#
    
    argument filter:
  
      pi(add#) = [1, 2]
      pi(s) = [1]

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

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