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,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)
  activate(n__fib1(X1,X2)) -> fib1(X1,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: fib1#(X,Y) -> add#(X,Y)
p4: add#(s(X),Y) -> add#(X,Y)
p5: sel#(s(N),cons(X,XS)) -> sel#(N,activate(XS))
p6: sel#(s(N),cons(X,XS)) -> activate#(XS)
p7: activate#(n__fib1(X1,X2)) -> fib1#(X1,X2)

and R consists of:

r1: fib(N) -> sel(N,fib1(s(|0|()),s(|0|())))
r2: fib1(X,Y) -> cons(X,n__fib1(Y,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: activate(n__fib1(X1,X2)) -> fib1(X1,X2)
r9: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p5}
  {p4}


-- 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,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: activate(n__fib1(X1,X2)) -> fib1(X1,X2)
r9: activate(X) -> X

The set of usable rules consists of

  r2, r3, r4, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        n__fib1 > add > cons > fib1 > |0| > activate > s > sel#
      
      argument filter:
    
        pi(sel#) = [1]
        pi(s) = [1]
        pi(cons) = 1
        pi(activate) = [1]
        pi(add) = [1, 2]
        pi(|0|) = []
        pi(fib1) = 1
        pi(n__fib1) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        n__fib1 > add > cons > fib1 > |0| > activate > s > sel#
      
      argument filter:
    
        pi(sel#) = []
        pi(s) = 1
        pi(cons) = 1
        pi(activate) = 1
        pi(add) = 1
        pi(|0|) = []
        pi(fib1) = 1
        pi(n__fib1) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        n__fib1 > add > cons > fib1 > |0| > activate > s > sel#
      
      argument filter:
    
        pi(sel#) = []
        pi(s) = [1]
        pi(cons) = 1
        pi(activate) = [1]
        pi(add) = 1
        pi(|0|) = []
        pi(fib1) = 1
        pi(n__fib1) = 1
    

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: 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,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: activate(n__fib1(X1,X2)) -> fib1(X1,X2)
r9: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        add# > s
      
      argument filter:
    
        pi(add#) = [1, 2]
        pi(s) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > add#
      
      argument filter:
    
        pi(add#) = 1
        pi(s) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > add#
      
      argument filter:
    
        pi(add#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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