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)
  , times(0(), y) -> 0()
  , times(s(x), y) -> plus(y, times(x, y))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y))
  , fac(x) -> loop(x, s(0()), s(0()))
  , loop(x, c, y) -> if(lt(x, c), x, c, y)
  , if(true(), x, c, y) -> y
  , if(false(), x, c, y) -> loop(x, s(c), times(y, s(c))) }
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))
  , times^#(0(), y) -> c_4()
  , times^#(s(x), y) -> c_5(plus^#(y, times(x, y)), times^#(x, y))
  , plus^#(0(), y) -> c_6()
  , plus^#(s(x), y) -> c_7(plus^#(x, y))
  , fac^#(x) -> c_8(loop^#(x, s(0()), s(0())))
  , loop^#(x, c, y) -> c_9(if^#(lt(x, c), x, c, y), lt^#(x, c))
  , if^#(true(), x, c, y) -> c_10()
  , if^#(false(), x, c, y) ->
    c_11(loop^#(x, s(c), times(y, s(c))), times^#(y, s(c))) }

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))
  , times^#(0(), y) -> c_4()
  , times^#(s(x), y) -> c_5(plus^#(y, times(x, y)), times^#(x, y))
  , plus^#(0(), y) -> c_6()
  , plus^#(s(x), y) -> c_7(plus^#(x, y))
  , fac^#(x) -> c_8(loop^#(x, s(0()), s(0())))
  , loop^#(x, c, y) -> c_9(if^#(lt(x, c), x, c, y), lt^#(x, c))
  , if^#(true(), x, c, y) -> c_10()
  , if^#(false(), x, c, y) ->
    c_11(loop^#(x, s(c), times(y, s(c))), times^#(y, s(c))) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , times(0(), y) -> 0()
  , times(s(x), y) -> plus(y, times(x, y))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y))
  , fac(x) -> loop(x, s(0()), s(0()))
  , loop(x, c, y) -> if(lt(x, c), x, c, y)
  , if(true(), x, c, y) -> y
  , if(false(), x, c, y) -> loop(x, s(c), times(y, s(c))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {1,2,4,6,10} by
applications of Pre({1,2,4,6,10}) = {3,5,7,9,11}. 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: times^#(0(), y) -> c_4()
    , 5: times^#(s(x), y) -> c_5(plus^#(y, times(x, y)), times^#(x, y))
    , 6: plus^#(0(), y) -> c_6()
    , 7: plus^#(s(x), y) -> c_7(plus^#(x, y))
    , 8: fac^#(x) -> c_8(loop^#(x, s(0()), s(0())))
    , 9: loop^#(x, c, y) -> c_9(if^#(lt(x, c), x, c, y), lt^#(x, c))
    , 10: if^#(true(), x, c, y) -> c_10()
    , 11: if^#(false(), x, c, y) ->
          c_11(loop^#(x, s(c), times(y, s(c))), times^#(y, s(c))) }

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))
  , times^#(s(x), y) -> c_5(plus^#(y, times(x, y)), times^#(x, y))
  , plus^#(s(x), y) -> c_7(plus^#(x, y))
  , fac^#(x) -> c_8(loop^#(x, s(0()), s(0())))
  , loop^#(x, c, y) -> c_9(if^#(lt(x, c), x, c, y), lt^#(x, c))
  , if^#(false(), x, c, y) ->
    c_11(loop^#(x, s(c), times(y, s(c))), times^#(y, s(c))) }
Weak DPs:
  { lt^#(x, 0()) -> c_1()
  , lt^#(0(), s(x)) -> c_2()
  , times^#(0(), y) -> c_4()
  , plus^#(0(), y) -> c_6()
  , if^#(true(), x, c, y) -> c_10() }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , times(0(), y) -> 0()
  , times(s(x), y) -> plus(y, times(x, y))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y))
  , fac(x) -> loop(x, s(0()), s(0()))
  , loop(x, c, y) -> if(lt(x, c), x, c, y)
  , if(true(), x, c, y) -> y
  , if(false(), x, c, y) -> loop(x, s(c), times(y, s(c))) }
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()
, times^#(0(), y) -> c_4()
, plus^#(0(), y) -> c_6()
, if^#(true(), x, c, y) -> 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))
  , times^#(s(x), y) -> c_5(plus^#(y, times(x, y)), times^#(x, y))
  , plus^#(s(x), y) -> c_7(plus^#(x, y))
  , fac^#(x) -> c_8(loop^#(x, s(0()), s(0())))
  , loop^#(x, c, y) -> c_9(if^#(lt(x, c), x, c, y), lt^#(x, c))
  , if^#(false(), x, c, y) ->
    c_11(loop^#(x, s(c), times(y, s(c))), times^#(y, s(c))) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , times(0(), y) -> 0()
  , times(s(x), y) -> plus(y, times(x, y))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y))
  , fac(x) -> loop(x, s(0()), s(0()))
  , loop(x, c, y) -> if(lt(x, c), x, c, y)
  , if(true(), x, c, y) -> y
  , if(false(), x, c, y) -> loop(x, s(c), times(y, s(c))) }
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: times^#(s(x), y) -> c_5(plus^#(y, times(x, y)), times^#(x, y))
     -->_1 plus^#(s(x), y) -> c_7(plus^#(x, y)) :3
     -->_2 times^#(s(x), y) ->
           c_5(plus^#(y, times(x, y)), times^#(x, y)) :2
  
  3: plus^#(s(x), y) -> c_7(plus^#(x, y))
     -->_1 plus^#(s(x), y) -> c_7(plus^#(x, y)) :3
  
  4: fac^#(x) -> c_8(loop^#(x, s(0()), s(0())))
     -->_1 loop^#(x, c, y) ->
           c_9(if^#(lt(x, c), x, c, y), lt^#(x, c)) :5
  
  5: loop^#(x, c, y) -> c_9(if^#(lt(x, c), x, c, y), lt^#(x, c))
     -->_1 if^#(false(), x, c, y) ->
           c_11(loop^#(x, s(c), times(y, s(c))), times^#(y, s(c))) :6
     -->_2 lt^#(s(x), s(y)) -> c_3(lt^#(x, y)) :1
  
  6: if^#(false(), x, c, y) ->
     c_11(loop^#(x, s(c), times(y, s(c))), times^#(y, s(c)))
     -->_1 loop^#(x, c, y) ->
           c_9(if^#(lt(x, c), x, c, y), lt^#(x, c)) :5
     -->_2 times^#(s(x), y) ->
           c_5(plus^#(y, times(x, y)), times^#(x, y)) :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).

  { fac^#(x) -> c_8(loop^#(x, s(0()), s(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))
  , times^#(s(x), y) -> c_5(plus^#(y, times(x, y)), times^#(x, y))
  , plus^#(s(x), y) -> c_7(plus^#(x, y))
  , loop^#(x, c, y) -> c_9(if^#(lt(x, c), x, c, y), lt^#(x, c))
  , if^#(false(), x, c, y) ->
    c_11(loop^#(x, s(c), times(y, s(c))), times^#(y, s(c))) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , times(0(), y) -> 0()
  , times(s(x), y) -> plus(y, times(x, y))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y))
  , fac(x) -> loop(x, s(0()), s(0()))
  , loop(x, c, y) -> if(lt(x, c), x, c, y)
  , if(true(), x, c, y) -> y
  , if(false(), x, c, y) -> loop(x, s(c), times(y, s(c))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { lt(x, 0()) -> false()
    , lt(0(), s(x)) -> true()
    , lt(s(x), s(y)) -> lt(x, y)
    , times(0(), y) -> 0()
    , times(s(x), y) -> plus(y, times(x, y))
    , plus(0(), y) -> y
    , plus(s(x), y) -> s(plus(x, y)) }

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))
  , times^#(s(x), y) -> c_5(plus^#(y, times(x, y)), times^#(x, y))
  , plus^#(s(x), y) -> c_7(plus^#(x, y))
  , loop^#(x, c, y) -> c_9(if^#(lt(x, c), x, c, y), lt^#(x, c))
  , if^#(false(), x, c, y) ->
    c_11(loop^#(x, s(c), times(y, s(c))), times^#(y, s(c))) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , times(0(), y) -> 0()
  , times(s(x), y) -> plus(y, times(x, y))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..