MAYBE

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

Strict Trs:
  { prod(xs) -> prodIter(xs, s(0()))
  , prodIter(xs, x) -> ifProd(isempty(xs), xs, x)
  , ifProd(true(), xs, x) -> x
  , ifProd(false(), xs, x) -> prodIter(tail(xs), times(x, head(xs)))
  , isempty(nil()) -> true()
  , isempty(cons(x, xs)) -> false()
  , tail(nil()) -> nil()
  , tail(cons(x, xs)) -> xs
  , times(x, y) -> timesIter(x, y, 0(), 0())
  , head(nil()) -> error()
  , head(cons(x, xs)) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , plus(0(), y) -> y
  , timesIter(x, y, z, u) -> ifTimes(ge(u, x), x, y, z, u)
  , ifTimes(true(), x, y, z, u) -> z
  , ifTimes(false(), x, y, z, u) -> timesIter(x, y, plus(y, z), s(u))
  , ge(x, 0()) -> true()
  , ge(s(x), s(y)) -> ge(x, y)
  , ge(0(), s(y)) -> false()
  , a() -> b()
  , a() -> c() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { prod^#(xs) -> c_1(prodIter^#(xs, s(0())))
  , prodIter^#(xs, x) ->
    c_2(ifProd^#(isempty(xs), xs, x), isempty^#(xs))
  , ifProd^#(true(), xs, x) -> c_3()
  , ifProd^#(false(), xs, x) ->
    c_4(prodIter^#(tail(xs), times(x, head(xs))),
        tail^#(xs),
        times^#(x, head(xs)),
        head^#(xs))
  , isempty^#(nil()) -> c_5()
  , isempty^#(cons(x, xs)) -> c_6()
  , tail^#(nil()) -> c_7()
  , tail^#(cons(x, xs)) -> c_8()
  , times^#(x, y) -> c_9(timesIter^#(x, y, 0(), 0()))
  , head^#(nil()) -> c_10()
  , head^#(cons(x, xs)) -> c_11()
  , timesIter^#(x, y, z, u) ->
    c_14(ifTimes^#(ge(u, x), x, y, z, u), ge^#(u, x))
  , plus^#(s(x), y) -> c_12(plus^#(x, y))
  , plus^#(0(), y) -> c_13()
  , ifTimes^#(true(), x, y, z, u) -> c_15()
  , ifTimes^#(false(), x, y, z, u) ->
    c_16(timesIter^#(x, y, plus(y, z), s(u)), plus^#(y, z))
  , ge^#(x, 0()) -> c_17()
  , ge^#(s(x), s(y)) -> c_18(ge^#(x, y))
  , ge^#(0(), s(y)) -> c_19()
  , a^#() -> c_20()
  , a^#() -> c_21() }

and mark the set of starting terms.

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

Strict DPs:
  { prod^#(xs) -> c_1(prodIter^#(xs, s(0())))
  , prodIter^#(xs, x) ->
    c_2(ifProd^#(isempty(xs), xs, x), isempty^#(xs))
  , ifProd^#(true(), xs, x) -> c_3()
  , ifProd^#(false(), xs, x) ->
    c_4(prodIter^#(tail(xs), times(x, head(xs))),
        tail^#(xs),
        times^#(x, head(xs)),
        head^#(xs))
  , isempty^#(nil()) -> c_5()
  , isempty^#(cons(x, xs)) -> c_6()
  , tail^#(nil()) -> c_7()
  , tail^#(cons(x, xs)) -> c_8()
  , times^#(x, y) -> c_9(timesIter^#(x, y, 0(), 0()))
  , head^#(nil()) -> c_10()
  , head^#(cons(x, xs)) -> c_11()
  , timesIter^#(x, y, z, u) ->
    c_14(ifTimes^#(ge(u, x), x, y, z, u), ge^#(u, x))
  , plus^#(s(x), y) -> c_12(plus^#(x, y))
  , plus^#(0(), y) -> c_13()
  , ifTimes^#(true(), x, y, z, u) -> c_15()
  , ifTimes^#(false(), x, y, z, u) ->
    c_16(timesIter^#(x, y, plus(y, z), s(u)), plus^#(y, z))
  , ge^#(x, 0()) -> c_17()
  , ge^#(s(x), s(y)) -> c_18(ge^#(x, y))
  , ge^#(0(), s(y)) -> c_19()
  , a^#() -> c_20()
  , a^#() -> c_21() }
Weak Trs:
  { prod(xs) -> prodIter(xs, s(0()))
  , prodIter(xs, x) -> ifProd(isempty(xs), xs, x)
  , ifProd(true(), xs, x) -> x
  , ifProd(false(), xs, x) -> prodIter(tail(xs), times(x, head(xs)))
  , isempty(nil()) -> true()
  , isempty(cons(x, xs)) -> false()
  , tail(nil()) -> nil()
  , tail(cons(x, xs)) -> xs
  , times(x, y) -> timesIter(x, y, 0(), 0())
  , head(nil()) -> error()
  , head(cons(x, xs)) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , plus(0(), y) -> y
  , timesIter(x, y, z, u) -> ifTimes(ge(u, x), x, y, z, u)
  , ifTimes(true(), x, y, z, u) -> z
  , ifTimes(false(), x, y, z, u) -> timesIter(x, y, plus(y, z), s(u))
  , ge(x, 0()) -> true()
  , ge(s(x), s(y)) -> ge(x, y)
  , ge(0(), s(y)) -> false()
  , a() -> b()
  , a() -> c() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of
{3,5,6,7,8,10,11,14,15,17,19,20,21} by applications of
Pre({3,5,6,7,8,10,11,14,15,17,19,20,21}) = {2,4,12,13,16,18}. Here
rules are labeled as follows:

  DPs:
    { 1: prod^#(xs) -> c_1(prodIter^#(xs, s(0())))
    , 2: prodIter^#(xs, x) ->
         c_2(ifProd^#(isempty(xs), xs, x), isempty^#(xs))
    , 3: ifProd^#(true(), xs, x) -> c_3()
    , 4: ifProd^#(false(), xs, x) ->
         c_4(prodIter^#(tail(xs), times(x, head(xs))),
             tail^#(xs),
             times^#(x, head(xs)),
             head^#(xs))
    , 5: isempty^#(nil()) -> c_5()
    , 6: isempty^#(cons(x, xs)) -> c_6()
    , 7: tail^#(nil()) -> c_7()
    , 8: tail^#(cons(x, xs)) -> c_8()
    , 9: times^#(x, y) -> c_9(timesIter^#(x, y, 0(), 0()))
    , 10: head^#(nil()) -> c_10()
    , 11: head^#(cons(x, xs)) -> c_11()
    , 12: timesIter^#(x, y, z, u) ->
          c_14(ifTimes^#(ge(u, x), x, y, z, u), ge^#(u, x))
    , 13: plus^#(s(x), y) -> c_12(plus^#(x, y))
    , 14: plus^#(0(), y) -> c_13()
    , 15: ifTimes^#(true(), x, y, z, u) -> c_15()
    , 16: ifTimes^#(false(), x, y, z, u) ->
          c_16(timesIter^#(x, y, plus(y, z), s(u)), plus^#(y, z))
    , 17: ge^#(x, 0()) -> c_17()
    , 18: ge^#(s(x), s(y)) -> c_18(ge^#(x, y))
    , 19: ge^#(0(), s(y)) -> c_19()
    , 20: a^#() -> c_20()
    , 21: a^#() -> c_21() }

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

Strict DPs:
  { prod^#(xs) -> c_1(prodIter^#(xs, s(0())))
  , prodIter^#(xs, x) ->
    c_2(ifProd^#(isempty(xs), xs, x), isempty^#(xs))
  , ifProd^#(false(), xs, x) ->
    c_4(prodIter^#(tail(xs), times(x, head(xs))),
        tail^#(xs),
        times^#(x, head(xs)),
        head^#(xs))
  , times^#(x, y) -> c_9(timesIter^#(x, y, 0(), 0()))
  , timesIter^#(x, y, z, u) ->
    c_14(ifTimes^#(ge(u, x), x, y, z, u), ge^#(u, x))
  , plus^#(s(x), y) -> c_12(plus^#(x, y))
  , ifTimes^#(false(), x, y, z, u) ->
    c_16(timesIter^#(x, y, plus(y, z), s(u)), plus^#(y, z))
  , ge^#(s(x), s(y)) -> c_18(ge^#(x, y)) }
Weak DPs:
  { ifProd^#(true(), xs, x) -> c_3()
  , isempty^#(nil()) -> c_5()
  , isempty^#(cons(x, xs)) -> c_6()
  , tail^#(nil()) -> c_7()
  , tail^#(cons(x, xs)) -> c_8()
  , head^#(nil()) -> c_10()
  , head^#(cons(x, xs)) -> c_11()
  , plus^#(0(), y) -> c_13()
  , ifTimes^#(true(), x, y, z, u) -> c_15()
  , ge^#(x, 0()) -> c_17()
  , ge^#(0(), s(y)) -> c_19()
  , a^#() -> c_20()
  , a^#() -> c_21() }
Weak Trs:
  { prod(xs) -> prodIter(xs, s(0()))
  , prodIter(xs, x) -> ifProd(isempty(xs), xs, x)
  , ifProd(true(), xs, x) -> x
  , ifProd(false(), xs, x) -> prodIter(tail(xs), times(x, head(xs)))
  , isempty(nil()) -> true()
  , isempty(cons(x, xs)) -> false()
  , tail(nil()) -> nil()
  , tail(cons(x, xs)) -> xs
  , times(x, y) -> timesIter(x, y, 0(), 0())
  , head(nil()) -> error()
  , head(cons(x, xs)) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , plus(0(), y) -> y
  , timesIter(x, y, z, u) -> ifTimes(ge(u, x), x, y, z, u)
  , ifTimes(true(), x, y, z, u) -> z
  , ifTimes(false(), x, y, z, u) -> timesIter(x, y, plus(y, z), s(u))
  , ge(x, 0()) -> true()
  , ge(s(x), s(y)) -> ge(x, y)
  , ge(0(), s(y)) -> false()
  , a() -> b()
  , a() -> 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.

{ ifProd^#(true(), xs, x) -> c_3()
, isempty^#(nil()) -> c_5()
, isempty^#(cons(x, xs)) -> c_6()
, tail^#(nil()) -> c_7()
, tail^#(cons(x, xs)) -> c_8()
, head^#(nil()) -> c_10()
, head^#(cons(x, xs)) -> c_11()
, plus^#(0(), y) -> c_13()
, ifTimes^#(true(), x, y, z, u) -> c_15()
, ge^#(x, 0()) -> c_17()
, ge^#(0(), s(y)) -> c_19()
, a^#() -> c_20()
, a^#() -> c_21() }

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

Strict DPs:
  { prod^#(xs) -> c_1(prodIter^#(xs, s(0())))
  , prodIter^#(xs, x) ->
    c_2(ifProd^#(isempty(xs), xs, x), isempty^#(xs))
  , ifProd^#(false(), xs, x) ->
    c_4(prodIter^#(tail(xs), times(x, head(xs))),
        tail^#(xs),
        times^#(x, head(xs)),
        head^#(xs))
  , times^#(x, y) -> c_9(timesIter^#(x, y, 0(), 0()))
  , timesIter^#(x, y, z, u) ->
    c_14(ifTimes^#(ge(u, x), x, y, z, u), ge^#(u, x))
  , plus^#(s(x), y) -> c_12(plus^#(x, y))
  , ifTimes^#(false(), x, y, z, u) ->
    c_16(timesIter^#(x, y, plus(y, z), s(u)), plus^#(y, z))
  , ge^#(s(x), s(y)) -> c_18(ge^#(x, y)) }
Weak Trs:
  { prod(xs) -> prodIter(xs, s(0()))
  , prodIter(xs, x) -> ifProd(isempty(xs), xs, x)
  , ifProd(true(), xs, x) -> x
  , ifProd(false(), xs, x) -> prodIter(tail(xs), times(x, head(xs)))
  , isempty(nil()) -> true()
  , isempty(cons(x, xs)) -> false()
  , tail(nil()) -> nil()
  , tail(cons(x, xs)) -> xs
  , times(x, y) -> timesIter(x, y, 0(), 0())
  , head(nil()) -> error()
  , head(cons(x, xs)) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , plus(0(), y) -> y
  , timesIter(x, y, z, u) -> ifTimes(ge(u, x), x, y, z, u)
  , ifTimes(true(), x, y, z, u) -> z
  , ifTimes(false(), x, y, z, u) -> timesIter(x, y, plus(y, z), s(u))
  , ge(x, 0()) -> true()
  , ge(s(x), s(y)) -> ge(x, y)
  , ge(0(), s(y)) -> false()
  , a() -> b()
  , a() -> c() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  { prodIter^#(xs, x) ->
    c_2(ifProd^#(isempty(xs), xs, x), isempty^#(xs))
  , ifProd^#(false(), xs, x) ->
    c_4(prodIter^#(tail(xs), times(x, head(xs))),
        tail^#(xs),
        times^#(x, head(xs)),
        head^#(xs)) }

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

Strict DPs:
  { prod^#(xs) -> c_1(prodIter^#(xs, s(0())))
  , prodIter^#(xs, x) -> c_2(ifProd^#(isempty(xs), xs, x))
  , ifProd^#(false(), xs, x) ->
    c_3(prodIter^#(tail(xs), times(x, head(xs))), times^#(x, head(xs)))
  , times^#(x, y) -> c_4(timesIter^#(x, y, 0(), 0()))
  , timesIter^#(x, y, z, u) ->
    c_5(ifTimes^#(ge(u, x), x, y, z, u), ge^#(u, x))
  , plus^#(s(x), y) -> c_6(plus^#(x, y))
  , ifTimes^#(false(), x, y, z, u) ->
    c_7(timesIter^#(x, y, plus(y, z), s(u)), plus^#(y, z))
  , ge^#(s(x), s(y)) -> c_8(ge^#(x, y)) }
Weak Trs:
  { prod(xs) -> prodIter(xs, s(0()))
  , prodIter(xs, x) -> ifProd(isempty(xs), xs, x)
  , ifProd(true(), xs, x) -> x
  , ifProd(false(), xs, x) -> prodIter(tail(xs), times(x, head(xs)))
  , isempty(nil()) -> true()
  , isempty(cons(x, xs)) -> false()
  , tail(nil()) -> nil()
  , tail(cons(x, xs)) -> xs
  , times(x, y) -> timesIter(x, y, 0(), 0())
  , head(nil()) -> error()
  , head(cons(x, xs)) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , plus(0(), y) -> y
  , timesIter(x, y, z, u) -> ifTimes(ge(u, x), x, y, z, u)
  , ifTimes(true(), x, y, z, u) -> z
  , ifTimes(false(), x, y, z, u) -> timesIter(x, y, plus(y, z), s(u))
  , ge(x, 0()) -> true()
  , ge(s(x), s(y)) -> ge(x, y)
  , ge(0(), s(y)) -> false()
  , a() -> b()
  , a() -> c() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { isempty(nil()) -> true()
    , isempty(cons(x, xs)) -> false()
    , tail(nil()) -> nil()
    , tail(cons(x, xs)) -> xs
    , times(x, y) -> timesIter(x, y, 0(), 0())
    , head(nil()) -> error()
    , head(cons(x, xs)) -> x
    , plus(s(x), y) -> s(plus(x, y))
    , plus(0(), y) -> y
    , timesIter(x, y, z, u) -> ifTimes(ge(u, x), x, y, z, u)
    , ifTimes(true(), x, y, z, u) -> z
    , ifTimes(false(), x, y, z, u) -> timesIter(x, y, plus(y, z), s(u))
    , ge(x, 0()) -> true()
    , ge(s(x), s(y)) -> ge(x, y)
    , ge(0(), s(y)) -> false() }

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

Strict DPs:
  { prod^#(xs) -> c_1(prodIter^#(xs, s(0())))
  , prodIter^#(xs, x) -> c_2(ifProd^#(isempty(xs), xs, x))
  , ifProd^#(false(), xs, x) ->
    c_3(prodIter^#(tail(xs), times(x, head(xs))), times^#(x, head(xs)))
  , times^#(x, y) -> c_4(timesIter^#(x, y, 0(), 0()))
  , timesIter^#(x, y, z, u) ->
    c_5(ifTimes^#(ge(u, x), x, y, z, u), ge^#(u, x))
  , plus^#(s(x), y) -> c_6(plus^#(x, y))
  , ifTimes^#(false(), x, y, z, u) ->
    c_7(timesIter^#(x, y, plus(y, z), s(u)), plus^#(y, z))
  , ge^#(s(x), s(y)) -> c_8(ge^#(x, y)) }
Weak Trs:
  { isempty(nil()) -> true()
  , isempty(cons(x, xs)) -> false()
  , tail(nil()) -> nil()
  , tail(cons(x, xs)) -> xs
  , times(x, y) -> timesIter(x, y, 0(), 0())
  , head(nil()) -> error()
  , head(cons(x, xs)) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , plus(0(), y) -> y
  , timesIter(x, y, z, u) -> ifTimes(ge(u, x), x, y, z, u)
  , ifTimes(true(), x, y, z, u) -> z
  , ifTimes(false(), x, y, z, u) -> timesIter(x, y, plus(y, z), s(u))
  , ge(x, 0()) -> true()
  , ge(s(x), s(y)) -> ge(x, y)
  , ge(0(), s(y)) -> false() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: prod^#(xs) -> c_1(prodIter^#(xs, s(0())))
     -->_1 prodIter^#(xs, x) -> c_2(ifProd^#(isempty(xs), xs, x)) :2
  
  2: prodIter^#(xs, x) -> c_2(ifProd^#(isempty(xs), xs, x))
     -->_1 ifProd^#(false(), xs, x) ->
           c_3(prodIter^#(tail(xs), times(x, head(xs))),
               times^#(x, head(xs))) :3
  
  3: ifProd^#(false(), xs, x) ->
     c_3(prodIter^#(tail(xs), times(x, head(xs))), times^#(x, head(xs)))
     -->_2 times^#(x, y) -> c_4(timesIter^#(x, y, 0(), 0())) :4
     -->_1 prodIter^#(xs, x) -> c_2(ifProd^#(isempty(xs), xs, x)) :2
  
  4: times^#(x, y) -> c_4(timesIter^#(x, y, 0(), 0()))
     -->_1 timesIter^#(x, y, z, u) ->
           c_5(ifTimes^#(ge(u, x), x, y, z, u), ge^#(u, x)) :5
  
  5: timesIter^#(x, y, z, u) ->
     c_5(ifTimes^#(ge(u, x), x, y, z, u), ge^#(u, x))
     -->_2 ge^#(s(x), s(y)) -> c_8(ge^#(x, y)) :8
     -->_1 ifTimes^#(false(), x, y, z, u) ->
           c_7(timesIter^#(x, y, plus(y, z), s(u)), plus^#(y, z)) :7
  
  6: plus^#(s(x), y) -> c_6(plus^#(x, y))
     -->_1 plus^#(s(x), y) -> c_6(plus^#(x, y)) :6
  
  7: ifTimes^#(false(), x, y, z, u) ->
     c_7(timesIter^#(x, y, plus(y, z), s(u)), plus^#(y, z))
     -->_2 plus^#(s(x), y) -> c_6(plus^#(x, y)) :6
     -->_1 timesIter^#(x, y, z, u) ->
           c_5(ifTimes^#(ge(u, x), x, y, z, u), ge^#(u, x)) :5
  
  8: ge^#(s(x), s(y)) -> c_8(ge^#(x, y))
     -->_1 ge^#(s(x), s(y)) -> c_8(ge^#(x, y)) :8
  

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).

  { prod^#(xs) -> c_1(prodIter^#(xs, s(0()))) }


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

Strict DPs:
  { prodIter^#(xs, x) -> c_2(ifProd^#(isempty(xs), xs, x))
  , ifProd^#(false(), xs, x) ->
    c_3(prodIter^#(tail(xs), times(x, head(xs))), times^#(x, head(xs)))
  , times^#(x, y) -> c_4(timesIter^#(x, y, 0(), 0()))
  , timesIter^#(x, y, z, u) ->
    c_5(ifTimes^#(ge(u, x), x, y, z, u), ge^#(u, x))
  , plus^#(s(x), y) -> c_6(plus^#(x, y))
  , ifTimes^#(false(), x, y, z, u) ->
    c_7(timesIter^#(x, y, plus(y, z), s(u)), plus^#(y, z))
  , ge^#(s(x), s(y)) -> c_8(ge^#(x, y)) }
Weak Trs:
  { isempty(nil()) -> true()
  , isempty(cons(x, xs)) -> false()
  , tail(nil()) -> nil()
  , tail(cons(x, xs)) -> xs
  , times(x, y) -> timesIter(x, y, 0(), 0())
  , head(nil()) -> error()
  , head(cons(x, xs)) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , plus(0(), y) -> y
  , timesIter(x, y, z, u) -> ifTimes(ge(u, x), x, y, z, u)
  , ifTimes(true(), x, y, z, u) -> z
  , ifTimes(false(), x, y, z, u) -> timesIter(x, y, plus(y, z), s(u))
  , ge(x, 0()) -> true()
  , ge(s(x), s(y)) -> ge(x, y)
  , ge(0(), s(y)) -> false() }
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..