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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          sel#_A(x1,x2) = x1 + 2
          s_A(x1) = x1 + 3
          cons_A(x1,x2) = 0
          activate_A(x1) = x1 + 4
          add_A(x1,x2) = x1 + x2 + 4
          |0|_A() = 1
          fib1_A(x1,x2) = x2 + 2
          n__fib1_A(x1,x2) = x2 + 1
  
    precedence:
    
      sel# = s = activate = add = |0| = fib1 = n__fib1 > cons
    
    partial status:
  
      pi(sel#) = [1]
      pi(s) = []
      pi(cons) = []
      pi(activate) = [1]
      pi(add) = []
      pi(|0|) = []
      pi(fib1) = [2]
      pi(n__fib1) = [2]

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          add#_A(x1,x2) = x1
          s_A(x1) = x1 + 1
  
    precedence:
    
      add# = s
    
    partial status:
  
      pi(add#) = [1]
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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