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)
  , fac(x) -> help(x, 0())
  , help(x, c) -> if(lt(c, x), x, c)
  , if(true(), x, c) -> times(s(c), help(x, s(c)))
  , if(false(), x, c) -> s(0()) }
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))
  , fac^#(x) -> c_4(help^#(x, 0()))
  , help^#(x, c) -> c_5(if^#(lt(c, x), x, c), lt^#(c, x))
  , if^#(true(), x, c) -> c_6(help^#(x, s(c)))
  , if^#(false(), x, c) -> c_7() }

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

We estimate the number of application of {1,2,7} by applications of
Pre({1,2,7}) = {3,5}. 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: fac^#(x) -> c_4(help^#(x, 0()))
    , 5: help^#(x, c) -> c_5(if^#(lt(c, x), x, c), lt^#(c, x))
    , 6: if^#(true(), x, c) -> c_6(help^#(x, s(c)))
    , 7: if^#(false(), x, c) -> c_7() }

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))
  , fac^#(x) -> c_4(help^#(x, 0()))
  , help^#(x, c) -> c_5(if^#(lt(c, x), x, c), lt^#(c, x))
  , if^#(true(), x, c) -> c_6(help^#(x, s(c))) }
Weak DPs:
  { lt^#(x, 0()) -> c_1()
  , lt^#(0(), s(x)) -> c_2()
  , if^#(false(), x, c) -> c_7() }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , fac(x) -> help(x, 0())
  , help(x, c) -> if(lt(c, x), x, c)
  , if(true(), x, c) -> times(s(c), help(x, s(c)))
  , if(false(), x, c) -> s(0()) }
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()
, if^#(false(), x, c) -> c_7() }

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))
  , fac^#(x) -> c_4(help^#(x, 0()))
  , help^#(x, c) -> c_5(if^#(lt(c, x), x, c), lt^#(c, x))
  , if^#(true(), x, c) -> c_6(help^#(x, s(c))) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , fac(x) -> help(x, 0())
  , help(x, c) -> if(lt(c, x), x, c)
  , if(true(), x, c) -> times(s(c), help(x, s(c)))
  , if(false(), x, c) -> s(0()) }
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: fac^#(x) -> c_4(help^#(x, 0()))
     -->_1 help^#(x, c) -> c_5(if^#(lt(c, x), x, c), lt^#(c, x)) :3
  
  3: help^#(x, c) -> c_5(if^#(lt(c, x), x, c), lt^#(c, x))
     -->_1 if^#(true(), x, c) -> c_6(help^#(x, s(c))) :4
     -->_2 lt^#(s(x), s(y)) -> c_3(lt^#(x, y)) :1
  
  4: if^#(true(), x, c) -> c_6(help^#(x, s(c)))
     -->_1 help^#(x, c) -> c_5(if^#(lt(c, x), x, c), lt^#(c, x)) :3
  

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_4(help^#(x, 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))
  , help^#(x, c) -> c_5(if^#(lt(c, x), x, c), lt^#(c, x))
  , if^#(true(), x, c) -> c_6(help^#(x, s(c))) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , fac(x) -> help(x, 0())
  , help(x, c) -> if(lt(c, x), x, c)
  , if(true(), x, c) -> times(s(c), help(x, s(c)))
  , if(false(), x, c) -> s(0()) }
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) }

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))
  , help^#(x, c) -> c_5(if^#(lt(c, x), x, c), lt^#(c, x))
  , if^#(true(), x, c) -> c_6(help^#(x, s(c))) }
Weak Trs:
  { lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

None of the processors succeeded.

Details of failed attempt(s):
-----------------------------
1) 'matrices' failed due to the following reason:
   
   None of the processors succeeded.
   
   Details of failed attempt(s):
   -----------------------------
   1) 'matrix interpretation of dimension 4' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   2) 'matrix interpretation of dimension 3' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   3) 'matrix interpretation of dimension 3' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   4) 'matrix interpretation of dimension 2' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   5) 'matrix interpretation of dimension 2' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   6) 'matrix interpretation of dimension 1' failed due to the
      following reason:
      
      The input cannot be shown compatible
   

2) 'empty' failed due to the following reason:
   
   Empty strict component of the problem is NOT empty.


Arrrr..