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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        fib#_A(x1) = ((0,1),(1,0)) x1
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        fib#_A(x1) = (0,0)
        s_A(x1) = ((1,1),(1,0)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        fib#_A(x1) = (0,0)
        s_A(x1) = x1 + (1,1)
    

The next rules are strictly ordered:

  p1, p2

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