MAYBE

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , fibo(0()) -> fib(0())
  , fibo(s(0())) -> fib(s(0()))
  , fibo(s(s(x))) -> sum(fibo(s(x)), fibo(x))
  , fib(0()) -> s(0())
  , fib(s(0())) -> s(0())
  , fib(s(s(x))) -> if(true(), 0(), s(s(x)), 0(), 0())
  , sum(x, 0()) -> x
  , sum(x, s(y)) -> s(sum(x, y))
  , if(true(), c, s(s(x)), a, b) ->
    if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
  , if(false(), c, s(s(x)), a, b) -> sum(fibo(a), fibo(b)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { lt^#(x, 0()) -> c_1()
  , lt^#(0(), s(x)) -> c_2()
  , lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
  , fibo^#(0()) -> c_4(fib^#(0()))
  , fibo^#(s(0())) -> c_5(fib^#(s(0())))
  , fibo^#(s(s(x))) ->
    c_6(sum^#(fibo(s(x)), fibo(x)), fibo^#(s(x)), fibo^#(x))
  , fib^#(0()) -> c_7()
  , fib^#(s(0())) -> c_8()
  , fib^#(s(s(x))) -> c_9(if^#(true(), 0(), s(s(x)), 0(), 0()))
  , sum^#(x, 0()) -> c_10()
  , sum^#(x, s(y)) -> c_11(sum^#(x, y))
  , if^#(true(), c, s(s(x)), a, b) ->
    c_12(if^#(lt(s(c), s(s(x))), s(c), s(s(x)), b, c),
         lt^#(s(c), s(s(x))))
  , if^#(false(), c, s(s(x)), a, b) ->
    c_13(sum^#(fibo(a), fibo(b)), fibo^#(a), fibo^#(b)) }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { lt^#(x, 0()) -> c_1()
  , lt^#(0(), s(x)) -> c_2()
  , lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
  , fibo^#(0()) -> c_4(fib^#(0()))
  , fibo^#(s(0())) -> c_5(fib^#(s(0())))
  , fibo^#(s(s(x))) ->
    c_6(sum^#(fibo(s(x)), fibo(x)), fibo^#(s(x)), fibo^#(x))
  , fib^#(0()) -> c_7()
  , fib^#(s(0())) -> c_8()
  , fib^#(s(s(x))) -> c_9(if^#(true(), 0(), s(s(x)), 0(), 0()))
  , sum^#(x, 0()) -> c_10()
  , sum^#(x, s(y)) -> c_11(sum^#(x, y))
  , if^#(true(), c, s(s(x)), a, b) ->
    c_12(if^#(lt(s(c), s(s(x))), s(c), s(s(x)), b, c),
         lt^#(s(c), s(s(x))))
  , if^#(false(), c, s(s(x)), a, b) ->
    c_13(sum^#(fibo(a), fibo(b)), fibo^#(a), fibo^#(b)) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , fibo(0()) -> fib(0())
  , fibo(s(0())) -> fib(s(0()))
  , fibo(s(s(x))) -> sum(fibo(s(x)), fibo(x))
  , fib(0()) -> s(0())
  , fib(s(0())) -> s(0())
  , fib(s(s(x))) -> if(true(), 0(), s(s(x)), 0(), 0())
  , sum(x, 0()) -> x
  , sum(x, s(y)) -> s(sum(x, y))
  , if(true(), c, s(s(x)), a, b) ->
    if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
  , if(false(), c, s(s(x)), a, b) -> sum(fibo(a), fibo(b)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {1,2,7,8,10} by
applications of Pre({1,2,7,8,10}) = {3,4,5,6,11,13}. Here rules are
labeled as follows:

  DPs:
    { 1: lt^#(x, 0()) -> c_1()
    , 2: lt^#(0(), s(x)) -> c_2()
    , 3: lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
    , 4: fibo^#(0()) -> c_4(fib^#(0()))
    , 5: fibo^#(s(0())) -> c_5(fib^#(s(0())))
    , 6: fibo^#(s(s(x))) ->
         c_6(sum^#(fibo(s(x)), fibo(x)), fibo^#(s(x)), fibo^#(x))
    , 7: fib^#(0()) -> c_7()
    , 8: fib^#(s(0())) -> c_8()
    , 9: fib^#(s(s(x))) -> c_9(if^#(true(), 0(), s(s(x)), 0(), 0()))
    , 10: sum^#(x, 0()) -> c_10()
    , 11: sum^#(x, s(y)) -> c_11(sum^#(x, y))
    , 12: if^#(true(), c, s(s(x)), a, b) ->
          c_12(if^#(lt(s(c), s(s(x))), s(c), s(s(x)), b, c),
               lt^#(s(c), s(s(x))))
    , 13: if^#(false(), c, s(s(x)), a, b) ->
          c_13(sum^#(fibo(a), fibo(b)), fibo^#(a), fibo^#(b)) }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
  , fibo^#(0()) -> c_4(fib^#(0()))
  , fibo^#(s(0())) -> c_5(fib^#(s(0())))
  , fibo^#(s(s(x))) ->
    c_6(sum^#(fibo(s(x)), fibo(x)), fibo^#(s(x)), fibo^#(x))
  , fib^#(s(s(x))) -> c_9(if^#(true(), 0(), s(s(x)), 0(), 0()))
  , sum^#(x, s(y)) -> c_11(sum^#(x, y))
  , if^#(true(), c, s(s(x)), a, b) ->
    c_12(if^#(lt(s(c), s(s(x))), s(c), s(s(x)), b, c),
         lt^#(s(c), s(s(x))))
  , if^#(false(), c, s(s(x)), a, b) ->
    c_13(sum^#(fibo(a), fibo(b)), fibo^#(a), fibo^#(b)) }
Weak DPs:
  { lt^#(x, 0()) -> c_1()
  , lt^#(0(), s(x)) -> c_2()
  , fib^#(0()) -> c_7()
  , fib^#(s(0())) -> c_8()
  , sum^#(x, 0()) -> c_10() }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , fibo(0()) -> fib(0())
  , fibo(s(0())) -> fib(s(0()))
  , fibo(s(s(x))) -> sum(fibo(s(x)), fibo(x))
  , fib(0()) -> s(0())
  , fib(s(0())) -> s(0())
  , fib(s(s(x))) -> if(true(), 0(), s(s(x)), 0(), 0())
  , sum(x, 0()) -> x
  , sum(x, s(y)) -> s(sum(x, y))
  , if(true(), c, s(s(x)), a, b) ->
    if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
  , if(false(), c, s(s(x)), a, b) -> sum(fibo(a), fibo(b)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {2,3} by applications of
Pre({2,3}) = {4,8}. Here rules are labeled as follows:

  DPs:
    { 1: lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
    , 2: fibo^#(0()) -> c_4(fib^#(0()))
    , 3: fibo^#(s(0())) -> c_5(fib^#(s(0())))
    , 4: fibo^#(s(s(x))) ->
         c_6(sum^#(fibo(s(x)), fibo(x)), fibo^#(s(x)), fibo^#(x))
    , 5: fib^#(s(s(x))) -> c_9(if^#(true(), 0(), s(s(x)), 0(), 0()))
    , 6: sum^#(x, s(y)) -> c_11(sum^#(x, y))
    , 7: if^#(true(), c, s(s(x)), a, b) ->
         c_12(if^#(lt(s(c), s(s(x))), s(c), s(s(x)), b, c),
              lt^#(s(c), s(s(x))))
    , 8: if^#(false(), c, s(s(x)), a, b) ->
         c_13(sum^#(fibo(a), fibo(b)), fibo^#(a), fibo^#(b))
    , 9: lt^#(x, 0()) -> c_1()
    , 10: lt^#(0(), s(x)) -> c_2()
    , 11: fib^#(0()) -> c_7()
    , 12: fib^#(s(0())) -> c_8()
    , 13: sum^#(x, 0()) -> c_10() }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
  , fibo^#(s(s(x))) ->
    c_6(sum^#(fibo(s(x)), fibo(x)), fibo^#(s(x)), fibo^#(x))
  , fib^#(s(s(x))) -> c_9(if^#(true(), 0(), s(s(x)), 0(), 0()))
  , sum^#(x, s(y)) -> c_11(sum^#(x, y))
  , if^#(true(), c, s(s(x)), a, b) ->
    c_12(if^#(lt(s(c), s(s(x))), s(c), s(s(x)), b, c),
         lt^#(s(c), s(s(x))))
  , if^#(false(), c, s(s(x)), a, b) ->
    c_13(sum^#(fibo(a), fibo(b)), fibo^#(a), fibo^#(b)) }
Weak DPs:
  { lt^#(x, 0()) -> c_1()
  , lt^#(0(), s(x)) -> c_2()
  , fibo^#(0()) -> c_4(fib^#(0()))
  , fibo^#(s(0())) -> c_5(fib^#(s(0())))
  , fib^#(0()) -> c_7()
  , fib^#(s(0())) -> c_8()
  , sum^#(x, 0()) -> c_10() }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , fibo(0()) -> fib(0())
  , fibo(s(0())) -> fib(s(0()))
  , fibo(s(s(x))) -> sum(fibo(s(x)), fibo(x))
  , fib(0()) -> s(0())
  , fib(s(0())) -> s(0())
  , fib(s(s(x))) -> if(true(), 0(), s(s(x)), 0(), 0())
  , sum(x, 0()) -> x
  , sum(x, s(y)) -> s(sum(x, y))
  , if(true(), c, s(s(x)), a, b) ->
    if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
  , if(false(), c, s(s(x)), a, b) -> sum(fibo(a), fibo(b)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ lt^#(x, 0()) -> c_1()
, lt^#(0(), s(x)) -> c_2()
, fibo^#(0()) -> c_4(fib^#(0()))
, fibo^#(s(0())) -> c_5(fib^#(s(0())))
, fib^#(0()) -> c_7()
, fib^#(s(0())) -> c_8()
, sum^#(x, 0()) -> c_10() }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
  , fibo^#(s(s(x))) ->
    c_6(sum^#(fibo(s(x)), fibo(x)), fibo^#(s(x)), fibo^#(x))
  , fib^#(s(s(x))) -> c_9(if^#(true(), 0(), s(s(x)), 0(), 0()))
  , sum^#(x, s(y)) -> c_11(sum^#(x, y))
  , if^#(true(), c, s(s(x)), a, b) ->
    c_12(if^#(lt(s(c), s(s(x))), s(c), s(s(x)), b, c),
         lt^#(s(c), s(s(x))))
  , if^#(false(), c, s(s(x)), a, b) ->
    c_13(sum^#(fibo(a), fibo(b)), fibo^#(a), fibo^#(b)) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , fibo(0()) -> fib(0())
  , fibo(s(0())) -> fib(s(0()))
  , fibo(s(s(x))) -> sum(fibo(s(x)), fibo(x))
  , fib(0()) -> s(0())
  , fib(s(0())) -> s(0())
  , fib(s(s(x))) -> if(true(), 0(), s(s(x)), 0(), 0())
  , sum(x, 0()) -> x
  , sum(x, s(y)) -> s(sum(x, y))
  , if(true(), c, s(s(x)), a, b) ->
    if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
  , if(false(), c, s(s(x)), a, b) -> sum(fibo(a), fibo(b)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
     -->_1 lt^#(s(x), s(y)) -> c_3(lt^#(x, y)) :1
  
  2: fibo^#(s(s(x))) ->
     c_6(sum^#(fibo(s(x)), fibo(x)), fibo^#(s(x)), fibo^#(x))
     -->_1 sum^#(x, s(y)) -> c_11(sum^#(x, y)) :4
     -->_3 fibo^#(s(s(x))) ->
           c_6(sum^#(fibo(s(x)), fibo(x)), fibo^#(s(x)), fibo^#(x)) :2
     -->_2 fibo^#(s(s(x))) ->
           c_6(sum^#(fibo(s(x)), fibo(x)), fibo^#(s(x)), fibo^#(x)) :2
  
  3: fib^#(s(s(x))) -> c_9(if^#(true(), 0(), s(s(x)), 0(), 0()))
     -->_1 if^#(true(), c, s(s(x)), a, b) ->
           c_12(if^#(lt(s(c), s(s(x))), s(c), s(s(x)), b, c),
                lt^#(s(c), s(s(x)))) :5
  
  4: sum^#(x, s(y)) -> c_11(sum^#(x, y))
     -->_1 sum^#(x, s(y)) -> c_11(sum^#(x, y)) :4
  
  5: if^#(true(), c, s(s(x)), a, b) ->
     c_12(if^#(lt(s(c), s(s(x))), s(c), s(s(x)), b, c),
          lt^#(s(c), s(s(x))))
     -->_1 if^#(false(), c, s(s(x)), a, b) ->
           c_13(sum^#(fibo(a), fibo(b)), fibo^#(a), fibo^#(b)) :6
     -->_1 if^#(true(), c, s(s(x)), a, b) ->
           c_12(if^#(lt(s(c), s(s(x))), s(c), s(s(x)), b, c),
                lt^#(s(c), s(s(x)))) :5
     -->_2 lt^#(s(x), s(y)) -> c_3(lt^#(x, y)) :1
  
  6: if^#(false(), c, s(s(x)), a, b) ->
     c_13(sum^#(fibo(a), fibo(b)), fibo^#(a), fibo^#(b))
     -->_1 sum^#(x, s(y)) -> c_11(sum^#(x, y)) :4
     -->_3 fibo^#(s(s(x))) ->
           c_6(sum^#(fibo(s(x)), fibo(x)), fibo^#(s(x)), fibo^#(x)) :2
     -->_2 fibo^#(s(s(x))) ->
           c_6(sum^#(fibo(s(x)), fibo(x)), fibo^#(s(x)), fibo^#(x)) :2
  

Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).

  { fib^#(s(s(x))) -> c_9(if^#(true(), 0(), s(s(x)), 0(), 0())) }


We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { lt^#(s(x), s(y)) -> c_3(lt^#(x, y))
  , fibo^#(s(s(x))) ->
    c_6(sum^#(fibo(s(x)), fibo(x)), fibo^#(s(x)), fibo^#(x))
  , sum^#(x, s(y)) -> c_11(sum^#(x, y))
  , if^#(true(), c, s(s(x)), a, b) ->
    c_12(if^#(lt(s(c), s(s(x))), s(c), s(s(x)), b, c),
         lt^#(s(c), s(s(x))))
  , if^#(false(), c, s(s(x)), a, b) ->
    c_13(sum^#(fibo(a), fibo(b)), fibo^#(a), fibo^#(b)) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , fibo(0()) -> fib(0())
  , fibo(s(0())) -> fib(s(0()))
  , fibo(s(s(x))) -> sum(fibo(s(x)), fibo(x))
  , fib(0()) -> s(0())
  , fib(s(0())) -> s(0())
  , fib(s(s(x))) -> if(true(), 0(), s(s(x)), 0(), 0())
  , sum(x, 0()) -> x
  , sum(x, s(y)) -> s(sum(x, y))
  , if(true(), c, s(s(x)), a, b) ->
    if(lt(s(c), s(s(x))), s(c), s(s(x)), b, c)
  , if(false(), c, s(s(x)), a, b) -> sum(fibo(a), fibo(b)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..