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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          fib#_A(x1) = ((0,0),(1,0)) x1
          s_A(x1) = ((1,0),(1,0)) x1 + (1,3)
  
    precedence:
    
      fib# = s
    
    partial status:
  
      pi(fib#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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}


-- Reduction pair.

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

p1: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          fib#_A(x1) = ((0,0),(1,0)) x1 + (1,2)
          s_A(x1) = ((1,0),(0,0)) x1 + (2,1)
  
    precedence:
    
      fib# = s
    
    partial status:
  
      pi(fib#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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