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))

-- SCC decomposition.

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

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

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