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:

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

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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      add#_A(x1,x2) = ((1,0),(1,0)) x1
      s_A(x1) = ((1,1),(1,1)) x1 + (1,0)

The next rules are strictly ordered:

  p1

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