MAYBE

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

Strict Trs:
  { length(nil()) -> 0()
  , length(cons(x, l)) -> s(length(l))
  , lt(x, 0()) -> false()
  , lt(0(), s(y)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , head(nil()) -> undefined()
  , head(cons(x, l)) -> x
  , tail(nil()) -> nil()
  , tail(cons(x, l)) -> l
  , reverse(l) -> rev(0(), l, nil(), l)
  , rev(x, l, accu, orig) ->
    if(lt(x, length(orig)), x, l, accu, orig)
  , if(false(), x, l, accu, orig) -> accu
  , if(true(), x, l, accu, orig) ->
    rev(s(x), tail(l), cons(head(l), accu), orig) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { length^#(nil()) -> c_1()
  , length^#(cons(x, l)) -> c_2(length^#(l))
  , lt^#(x, 0()) -> c_3()
  , lt^#(0(), s(y)) -> c_4()
  , lt^#(s(x), s(y)) -> c_5(lt^#(x, y))
  , head^#(nil()) -> c_6()
  , head^#(cons(x, l)) -> c_7()
  , tail^#(nil()) -> c_8()
  , tail^#(cons(x, l)) -> c_9()
  , reverse^#(l) -> c_10(rev^#(0(), l, nil(), l))
  , rev^#(x, l, accu, orig) ->
    c_11(if^#(lt(x, length(orig)), x, l, accu, orig),
         lt^#(x, length(orig)),
         length^#(orig))
  , if^#(false(), x, l, accu, orig) -> c_12()
  , if^#(true(), x, l, accu, orig) ->
    c_13(rev^#(s(x), tail(l), cons(head(l), accu), orig),
         tail^#(l),
         head^#(l)) }

and mark the set of starting terms.

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

Strict DPs:
  { length^#(nil()) -> c_1()
  , length^#(cons(x, l)) -> c_2(length^#(l))
  , lt^#(x, 0()) -> c_3()
  , lt^#(0(), s(y)) -> c_4()
  , lt^#(s(x), s(y)) -> c_5(lt^#(x, y))
  , head^#(nil()) -> c_6()
  , head^#(cons(x, l)) -> c_7()
  , tail^#(nil()) -> c_8()
  , tail^#(cons(x, l)) -> c_9()
  , reverse^#(l) -> c_10(rev^#(0(), l, nil(), l))
  , rev^#(x, l, accu, orig) ->
    c_11(if^#(lt(x, length(orig)), x, l, accu, orig),
         lt^#(x, length(orig)),
         length^#(orig))
  , if^#(false(), x, l, accu, orig) -> c_12()
  , if^#(true(), x, l, accu, orig) ->
    c_13(rev^#(s(x), tail(l), cons(head(l), accu), orig),
         tail^#(l),
         head^#(l)) }
Weak Trs:
  { length(nil()) -> 0()
  , length(cons(x, l)) -> s(length(l))
  , lt(x, 0()) -> false()
  , lt(0(), s(y)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , head(nil()) -> undefined()
  , head(cons(x, l)) -> x
  , tail(nil()) -> nil()
  , tail(cons(x, l)) -> l
  , reverse(l) -> rev(0(), l, nil(), l)
  , rev(x, l, accu, orig) ->
    if(lt(x, length(orig)), x, l, accu, orig)
  , if(false(), x, l, accu, orig) -> accu
  , if(true(), x, l, accu, orig) ->
    rev(s(x), tail(l), cons(head(l), accu), orig) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {1,3,4,6,7,8,9,12} by
applications of Pre({1,3,4,6,7,8,9,12}) = {2,5,11,13}. Here rules
are labeled as follows:

  DPs:
    { 1: length^#(nil()) -> c_1()
    , 2: length^#(cons(x, l)) -> c_2(length^#(l))
    , 3: lt^#(x, 0()) -> c_3()
    , 4: lt^#(0(), s(y)) -> c_4()
    , 5: lt^#(s(x), s(y)) -> c_5(lt^#(x, y))
    , 6: head^#(nil()) -> c_6()
    , 7: head^#(cons(x, l)) -> c_7()
    , 8: tail^#(nil()) -> c_8()
    , 9: tail^#(cons(x, l)) -> c_9()
    , 10: reverse^#(l) -> c_10(rev^#(0(), l, nil(), l))
    , 11: rev^#(x, l, accu, orig) ->
          c_11(if^#(lt(x, length(orig)), x, l, accu, orig),
               lt^#(x, length(orig)),
               length^#(orig))
    , 12: if^#(false(), x, l, accu, orig) -> c_12()
    , 13: if^#(true(), x, l, accu, orig) ->
          c_13(rev^#(s(x), tail(l), cons(head(l), accu), orig),
               tail^#(l),
               head^#(l)) }

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

Strict DPs:
  { length^#(cons(x, l)) -> c_2(length^#(l))
  , lt^#(s(x), s(y)) -> c_5(lt^#(x, y))
  , reverse^#(l) -> c_10(rev^#(0(), l, nil(), l))
  , rev^#(x, l, accu, orig) ->
    c_11(if^#(lt(x, length(orig)), x, l, accu, orig),
         lt^#(x, length(orig)),
         length^#(orig))
  , if^#(true(), x, l, accu, orig) ->
    c_13(rev^#(s(x), tail(l), cons(head(l), accu), orig),
         tail^#(l),
         head^#(l)) }
Weak DPs:
  { length^#(nil()) -> c_1()
  , lt^#(x, 0()) -> c_3()
  , lt^#(0(), s(y)) -> c_4()
  , head^#(nil()) -> c_6()
  , head^#(cons(x, l)) -> c_7()
  , tail^#(nil()) -> c_8()
  , tail^#(cons(x, l)) -> c_9()
  , if^#(false(), x, l, accu, orig) -> c_12() }
Weak Trs:
  { length(nil()) -> 0()
  , length(cons(x, l)) -> s(length(l))
  , lt(x, 0()) -> false()
  , lt(0(), s(y)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , head(nil()) -> undefined()
  , head(cons(x, l)) -> x
  , tail(nil()) -> nil()
  , tail(cons(x, l)) -> l
  , reverse(l) -> rev(0(), l, nil(), l)
  , rev(x, l, accu, orig) ->
    if(lt(x, length(orig)), x, l, accu, orig)
  , if(false(), x, l, accu, orig) -> accu
  , if(true(), x, l, accu, orig) ->
    rev(s(x), tail(l), cons(head(l), accu), orig) }
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.

{ length^#(nil()) -> c_1()
, lt^#(x, 0()) -> c_3()
, lt^#(0(), s(y)) -> c_4()
, head^#(nil()) -> c_6()
, head^#(cons(x, l)) -> c_7()
, tail^#(nil()) -> c_8()
, tail^#(cons(x, l)) -> c_9()
, if^#(false(), x, l, accu, orig) -> c_12() }

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

Strict DPs:
  { length^#(cons(x, l)) -> c_2(length^#(l))
  , lt^#(s(x), s(y)) -> c_5(lt^#(x, y))
  , reverse^#(l) -> c_10(rev^#(0(), l, nil(), l))
  , rev^#(x, l, accu, orig) ->
    c_11(if^#(lt(x, length(orig)), x, l, accu, orig),
         lt^#(x, length(orig)),
         length^#(orig))
  , if^#(true(), x, l, accu, orig) ->
    c_13(rev^#(s(x), tail(l), cons(head(l), accu), orig),
         tail^#(l),
         head^#(l)) }
Weak Trs:
  { length(nil()) -> 0()
  , length(cons(x, l)) -> s(length(l))
  , lt(x, 0()) -> false()
  , lt(0(), s(y)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , head(nil()) -> undefined()
  , head(cons(x, l)) -> x
  , tail(nil()) -> nil()
  , tail(cons(x, l)) -> l
  , reverse(l) -> rev(0(), l, nil(), l)
  , rev(x, l, accu, orig) ->
    if(lt(x, length(orig)), x, l, accu, orig)
  , if(false(), x, l, accu, orig) -> accu
  , if(true(), x, l, accu, orig) ->
    rev(s(x), tail(l), cons(head(l), accu), orig) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { if^#(true(), x, l, accu, orig) ->
    c_13(rev^#(s(x), tail(l), cons(head(l), accu), orig),
         tail^#(l),
         head^#(l)) }

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

Strict DPs:
  { length^#(cons(x, l)) -> c_1(length^#(l))
  , lt^#(s(x), s(y)) -> c_2(lt^#(x, y))
  , reverse^#(l) -> c_3(rev^#(0(), l, nil(), l))
  , rev^#(x, l, accu, orig) ->
    c_4(if^#(lt(x, length(orig)), x, l, accu, orig),
        lt^#(x, length(orig)),
        length^#(orig))
  , if^#(true(), x, l, accu, orig) ->
    c_5(rev^#(s(x), tail(l), cons(head(l), accu), orig)) }
Weak Trs:
  { length(nil()) -> 0()
  , length(cons(x, l)) -> s(length(l))
  , lt(x, 0()) -> false()
  , lt(0(), s(y)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , head(nil()) -> undefined()
  , head(cons(x, l)) -> x
  , tail(nil()) -> nil()
  , tail(cons(x, l)) -> l
  , reverse(l) -> rev(0(), l, nil(), l)
  , rev(x, l, accu, orig) ->
    if(lt(x, length(orig)), x, l, accu, orig)
  , if(false(), x, l, accu, orig) -> accu
  , if(true(), x, l, accu, orig) ->
    rev(s(x), tail(l), cons(head(l), accu), orig) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { length(nil()) -> 0()
    , length(cons(x, l)) -> s(length(l))
    , lt(x, 0()) -> false()
    , lt(0(), s(y)) -> true()
    , lt(s(x), s(y)) -> lt(x, y)
    , head(nil()) -> undefined()
    , head(cons(x, l)) -> x
    , tail(nil()) -> nil()
    , tail(cons(x, l)) -> l }

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

Strict DPs:
  { length^#(cons(x, l)) -> c_1(length^#(l))
  , lt^#(s(x), s(y)) -> c_2(lt^#(x, y))
  , reverse^#(l) -> c_3(rev^#(0(), l, nil(), l))
  , rev^#(x, l, accu, orig) ->
    c_4(if^#(lt(x, length(orig)), x, l, accu, orig),
        lt^#(x, length(orig)),
        length^#(orig))
  , if^#(true(), x, l, accu, orig) ->
    c_5(rev^#(s(x), tail(l), cons(head(l), accu), orig)) }
Weak Trs:
  { length(nil()) -> 0()
  , length(cons(x, l)) -> s(length(l))
  , lt(x, 0()) -> false()
  , lt(0(), s(y)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , head(nil()) -> undefined()
  , head(cons(x, l)) -> x
  , tail(nil()) -> nil()
  , tail(cons(x, l)) -> l }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: length^#(cons(x, l)) -> c_1(length^#(l))
     -->_1 length^#(cons(x, l)) -> c_1(length^#(l)) :1
  
  2: lt^#(s(x), s(y)) -> c_2(lt^#(x, y))
     -->_1 lt^#(s(x), s(y)) -> c_2(lt^#(x, y)) :2
  
  3: reverse^#(l) -> c_3(rev^#(0(), l, nil(), l))
     -->_1 rev^#(x, l, accu, orig) ->
           c_4(if^#(lt(x, length(orig)), x, l, accu, orig),
               lt^#(x, length(orig)),
               length^#(orig)) :4
  
  4: rev^#(x, l, accu, orig) ->
     c_4(if^#(lt(x, length(orig)), x, l, accu, orig),
         lt^#(x, length(orig)),
         length^#(orig))
     -->_1 if^#(true(), x, l, accu, orig) ->
           c_5(rev^#(s(x), tail(l), cons(head(l), accu), orig)) :5
     -->_2 lt^#(s(x), s(y)) -> c_2(lt^#(x, y)) :2
     -->_3 length^#(cons(x, l)) -> c_1(length^#(l)) :1
  
  5: if^#(true(), x, l, accu, orig) ->
     c_5(rev^#(s(x), tail(l), cons(head(l), accu), orig))
     -->_1 rev^#(x, l, accu, orig) ->
           c_4(if^#(lt(x, length(orig)), x, l, accu, orig),
               lt^#(x, length(orig)),
               length^#(orig)) :4
  

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).

  { reverse^#(l) -> c_3(rev^#(0(), l, nil(), l)) }


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

Strict DPs:
  { length^#(cons(x, l)) -> c_1(length^#(l))
  , lt^#(s(x), s(y)) -> c_2(lt^#(x, y))
  , rev^#(x, l, accu, orig) ->
    c_4(if^#(lt(x, length(orig)), x, l, accu, orig),
        lt^#(x, length(orig)),
        length^#(orig))
  , if^#(true(), x, l, accu, orig) ->
    c_5(rev^#(s(x), tail(l), cons(head(l), accu), orig)) }
Weak Trs:
  { length(nil()) -> 0()
  , length(cons(x, l)) -> s(length(l))
  , lt(x, 0()) -> false()
  , lt(0(), s(y)) -> true()
  , lt(s(x), s(y)) -> lt(x, y)
  , head(nil()) -> undefined()
  , head(cons(x, l)) -> x
  , tail(nil()) -> nil()
  , tail(cons(x, l)) -> l }
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..