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 combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        sel#_A(x1,x2) = x1
        s_A(x1) = x1
        cons_A(x1,x2) = 0
        activate_A(x1) = x1
        fib1_A(x1,x2) = 1
        n__fib1_A(x1,x2) = 1
        n__add_A(x1,x2) = x1 + x2 + 1
        add_A(x1,x2) = x1 + x2 + 1
        |0|_A() = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        sel#_A(x1,x2) = x1
        s_A(x1) = x1 + 1
        cons_A(x1,x2) = 0
        activate_A(x1) = x1 + 2
        fib1_A(x1,x2) = 1
        n__fib1_A(x1,x2) = 0
        n__add_A(x1,x2) = x1
        add_A(x1,x2) = x1
        |0|_A() = 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: 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 combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        activate#_A(x1) = x1
        n__add_A(x1,x2) = x1 + x2 + 1
        n__fib1_A(x1,x2) = x1 + x2 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        activate#_A(x1) = x1
        n__add_A(x1,x2) = x1 + x2 + 1
        n__fib1_A(x1,x2) = x1 + x2 + 1
    

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 combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        add#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        add#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 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.