YES

We show the termination of the TRS R:

  fib(|0|()) -> |0|()
  fib(s(|0|())) -> s(|0|())
  fib(s(s(x))) -> +(fib(s(x)),fib(x))
  +(x,|0|()) -> x
  +(x,s(y)) -> s(+(x,y))

-- SCC decomposition.

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

p1: fib#(s(s(x))) -> +#(fib(s(x)),fib(x))
p2: fib#(s(s(x))) -> fib#(s(x))
p3: fib#(s(s(x))) -> fib#(x)
p4: +#(x,s(y)) -> +#(x,y)

and R consists of:

r1: fib(|0|()) -> |0|()
r2: fib(s(|0|())) -> s(|0|())
r3: fib(s(s(x))) -> +(fib(s(x)),fib(x))
r4: +(x,|0|()) -> x
r5: +(x,s(y)) -> s(+(x,y))

The estimated dependency graph contains the following SCCs:

  {p2, p3}
  {p4}


-- Reduction pair.

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

p1: fib#(s(s(x))) -> fib#(x)
p2: fib#(s(s(x))) -> fib#(s(x))

and R consists of:

r1: fib(|0|()) -> |0|()
r2: fib(s(|0|())) -> s(|0|())
r3: fib(s(s(x))) -> +(fib(s(x)),fib(x))
r4: +(x,|0|()) -> x
r5: +(x,s(y)) -> s(+(x,y))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5

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: +#(x,s(y)) -> +#(x,y)

and R consists of:

r1: fib(|0|()) -> |0|()
r2: fib(s(|0|())) -> s(|0|())
r3: fib(s(s(x))) -> +(fib(s(x)),fib(x))
r4: +(x,|0|()) -> x
r5: +(x,s(y)) -> s(+(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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